Statistics 2 Week 2 Notes: Independence and Functions of One Variable (Part 1 of 2)

01

Independence

From Independent Events to Independent Random Variables

When does knowing one random variable tell you nothing about another? The definition and the product rule.

This chapter asks one question: when does knowing the value of one random variable tell you nothing about another? The answer is independence. We carry the idea over from events, state the definition, and see that it can be tested cell by cell on a joint PMF.

1.1Recap: independent events

Two events $A$ and $B$ are independent when $P(A \cap B) = P(A)\,P(B)$ . Knowing that $B$ happened does not change the probability of $A$ .

A random variable defines events too: $\{X = 3\}$ , $\{Y \le 1\}$ , $\{X \text{ is even}\}$ . So the idea of independence extends naturally from events to random variables.

1.2The definition

Independent random variables

Let $X$ and $Y$ be random variables on the same probability space, with ranges $T_X$ and $T_Y$ . $X$ and $Y$ are independent if every event defined using $X$ alone is independent of every event defined using $Y$ alone. Equivalently, with joint PMF $f_{XY}$ :

f_{XY}(t_1, t_2) = f_X(t_1)\, f_Y(t_2) \quad \text{for all } t_1 \in T_X,\ t_2 \in T_Y

Remember. The word all matters. The product rule must hold for every pair (t₁, t₂), not just a few of them.

1.3Compare with the factorisation that always holds

For any two random variables the joint PMF factors as a marginal times a conditional. Independence is the special case where the conditional equals the marginal:

	Joint PMF	Conditional PMF
General (always)	$f_{XY}(t_1,t_2) = f_X(t_1)\, f_{Y\mid X=t_1}(t_2)$	$f_{Y\mid X=t_1}(t_2) = \dfrac{f_{XY}(t_1,t_2)}{f_X(t_1)}$
Independent	$f_{XY}(t_1,t_2) = f_X(t_1)\, f_Y(t_2)$	$f_{Y\mid X=t_1}(t_2) = f_Y(t_2)$

In words

For independent X and Y the joint PMF is the product of the marginals, and every conditional PMF equals the marginal PMF. Learning X = t₁ does not change the distribution of Y at all.

1.4Going deeper

Symmetric (extra). $X$ independent of $Y$ means $Y$ independent of $X$ . The condition $f_{XY} = f_X f_Y$ treats both the same way, so it also gives $f_{X\mid Y=t_2} = f_X$ .
Support is a full grid (extra). If $X$ and $Y$ are independent, then $f_{XY}(t_1,t_2) > 0$ exactly when $f_X(t_1) > 0$ and $f_Y(t_2) > 0$ . So the non-zero cells always form a complete rectangle. A zero cell sitting where both its row total and column total are positive can never happen. This is the fast dependence test of Chapter 4.
Why "same probability space" (extra). The joint PMF only makes sense if both random variables are defined on the same experiment, such as the same random pick of a student.

1.5Exam tips and common mistakes

Exam tips

Write the target first: $f_{XY}(t_1,t_2) \overset{?}{=} f_X(t_1)\,f_Y(t_2)$ , then compute the two marginals from the table.
If a question gives independence, use it to fill in cells: each unknown joint value is a row marginal times a column marginal.

Common mistakes

Calling variables independent after checking only one or two cells.
Judging independence from the marginals alone: the same marginals fit both independent and dependent tables (Chapter 2, Example 3).
Mixing up independent with identically distributed: independence is about the joint PMF, not whether X and Y have the same marginal.

Practice questions (graded-assignment style)

Q1Using independence

Numerical Answer Type (NAT).

$X$ and $Y$ are independent with $f_X(2) = 0.3$ and $f_Y(5) = 0.4$ . Find $f_{XY}(2, 5)$ .

Solution

Idea

Joint is the product of the marginals.

$f_{XY}(2,5) = 0.3 \times 0.4 = 0.12$ .
Answer: 0.12

Q2Conditional equals marginal

Numerical Answer Type (NAT).

$X$ and $Y$ are independent, $f_X(1) = 0.4$ and $f_{Y\mid X=1}(0) = 0.7$ . Find $f_{XY}(1, 0)$ .

Solution

Idea

Always $f_{XY} = f_X \cdot f_{Y\mid X}$ .

The factorisation always holds: $f_{XY}(1,0) = 0.4 \times 0.7 = 0.28$ .
By independence $f_{Y\mid X=1}(0) = f_Y(0) = 0.7$ , so the same product is also $f_X(1) f_Y(0)$ .
Answer: 0.28

Q3Which equation?

Multiple Choice Question (MCQ): choose the one correct option.

Which equation is the definition of independence of X and Y?

1

f_{XY}(t_1,t_2) = f_X(t_1) + f_Y(t_2)

2

f_{XY}(t_1,t_2) = f_X(t_1)\, f_Y(t_2)

for all

t_1, t_2

3

f_X = f_Y

4

f_{XY}(t_1,t_2) = f_X(t_1)\, f_Y(t_2)

for one pair

t_1, t_2

Solution

Idea

The product rule must hold for every pair.

Option 2 is the definition. Option 4 only asks for one pair, which is not enough.
Options 1 and 3 are not the definition (a sum is not a product, and equal marginals do not mean independent).
Answer: option 2.

Q4Consequences of independence

Multiple Select Question (MSQ): select all that apply.

Suppose X and Y are independent. Choose the correct statements.

1

Every conditional PMF

f_{Y\mid X=t_1}

equals

f_Y

2

The joint PMF is the product of the marginals

3

f_X

must equal

f_Y

4

Learning

X

changes the distribution of

Y

Solution

Idea

Recall the two forms of the definition.

Statements 1 and 2 are the two equivalent forms.
Statement 3 is false: the marginals can be completely different. Statement 4 is the opposite of independence.
Answer: options 1 and 2.

Next · IndependenceChecking Independence From a Joint PMF Table

02

Independence

Checking Independence From a Joint PMF Table

Compare the joint PMF with the product of its marginals, with one independent and two dependent tables.

In practice the joint PMF comes as a table, and we test independence by comparing it with the product of its own marginals.

2.1How to read the table

In every table below, columns are values $t_1$ of $X$ , and rows are values $t_2$ of $Y$ . The last column holds $f_Y$ (the row sums) and the last row holds $f_X$ (the column sums).

Add each row to get f_Y and each column to get f_X. Check that both add up to 1.
For a cell, multiply its column marginal and its row marginal, and compare with the entry: $f_X(t_1)\, f_Y(t_2)$ against $f_{XY}(t_1,t_2)$ .
One failing cell means dependent. Only when every cell matches are the variables independent.

Figure 1. How to decide independence from a joint table

2.2Example 2: dependent

Figure 2. Example 2 in 32nds: the joint PMF and the product of the marginals

The other three cells also fail ( $\tfrac{8}{32}$ against $\tfrac{9}{32}$ , $\tfrac{4}{32}$ against $\tfrac{5}{32}$ , $\tfrac{4}{32}$ against $\tfrac{3}{32}$ ). Both tables add up to $\tfrac{32}{32}$ , so a mismatch in one cell is made up by the others.

2.3The conditional view

Independence also says the conditional PMF of $Y$ given $X = t_1$ equals the marginal for every $t_1$ . Example 1 passes; Example 2 fails: $f_{Y\mid X=0} = (\tfrac45, \tfrac15)$ and $f_{Y\mid X=1} = (\tfrac23, \tfrac13)$ , neither of which is $f_Y = (\tfrac34, \tfrac14)$ .

Figure 3. Knowing X changes the distribution of Y in Example 2 but not in Example 1

2.4Going deeper

Fewer checks are enough in principle (extra). For an $m \times n$ table, once the cells in the first $m-1$ rows and $n-1$ columns match the product, the last row and column match automatically, because the marginals force the totals. In the exam still check every cell unless it is cheaper to find one failing cell.
Use common denominators (extra). Comparing 99/576 with 96/576 is easier than comparing decimals. Convert both sides to one denominator before deciding that two products differ.

2.5Exam tips and common mistakes

Exam tips

Pick the cell with the simplest numbers first (often the top-left); if it fails you are done.
When the table has a parameter, find the value that fixes one cell, then check that every other cell agrees.

Common mistakes

Using a row total as a column marginal. Multiply the total of the cell's column (f_X) by the total of its row (f_Y).
Stopping at a cell that matches and declaring independence.
Solving for the parameter from one cell but not checking the others.

Practice questions (graded-assignment style)

Q1Decide independence

Multiple Choice Question (MCQ): choose the one correct option.

The joint PMF has $f_{XY}(0,0) = 0.1,\ f_{XY}(1,0) = 0.3,\ f_{XY}(0,1) = 0.4,\ f_{XY}(1,1) = 0.2$ (first argument $X$ , second $Y$ ). Are $X$ and $Y$ independent?

1

Yes

2

No

3

Cannot be decided from this table

4

Only if the marginals are equal

Solution

Idea

Compute both marginals and test one cell.

$f_X(0) = 0.1 + 0.4 = 0.5$ and $f_Y(0) = 0.1 + 0.3 = 0.4$ .
$f_X(0)\,f_Y(0) = 0.2 \ne 0.1 = f_{XY}(0,0)$ , so the variables are dependent.
Answer: option 2.

Q2Make it independent

Numerical Answer Type (NAT).

The joint PMF is $f_{XY}(0,0) = a,\ f_{XY}(1,0) = 0.2 - a,\ f_{XY}(0,1) = 0.5 - a,\ f_{XY}(1,1) = 0.3 + a$ . For what value of $a$ are $X$ and $Y$ independent?

Solution

Idea

Find the marginals, then fix one cell.

$f_X(0) = a + (0.5 - a) = 0.5$ and $f_Y(0) = a + (0.2 - a) = 0.2$ , whatever $a$ is.
Independence needs $a = f_X(0) f_Y(0) = 0.5 \times 0.2 = 0.1$ .
Check the others at $a = 0.1$ : $0.1 = 0.5 \times 0.2$ , $0.4 = 0.5 \times 0.8$ , $0.4 = 0.5 \times 0.8$ . All match.
Answer: 0.1

Q3Example 2 facts

Multiple Select Question (MSQ): select all that apply.

For the joint PMF of Example 2 ( $f_{XY}(0,0) = \tfrac12,\ f_{XY}(1,0) = \tfrac14,\ f_{XY}(0,1) = f_{XY}(1,1) = \tfrac18$ ), choose the correct statements.

1

f_X(0) = \tfrac58

2

f_Y(0) = \tfrac34

3

X

and

Y

are independent

4

f_X(0)\,f_Y(0) = \tfrac{15}{32}

Solution

Idea

Add the columns and rows.

$f_X(0) = \tfrac12 + \tfrac18 = \tfrac58$ , $f_Y(0) = \tfrac12 + \tfrac14 = \tfrac34$ , and their product is $\tfrac{15}{32}$ .
The product differs from $f_{XY}(0,0) = \tfrac{16}{32}$ , so statement 3 is false.
Answer: options 1, 2 and 4.

Q4Which x?

Numerical Answer Type (NAT).

In Example 3 take $x = \tfrac18$ . What is $f_{XY}(1, 0)$ , the cell with $\tfrac12 - x$ ? (Give it as a decimal.)

Solution

Idea

Compute $\tfrac12 - \tfrac18$ .

$\tfrac12 - \tfrac18 = \tfrac38 = 0.375$ , not $\tfrac14$ , so at this $x$ the variables are dependent.
Answer: 0.375

Next · IndependenceIndependence in a Real Setting: A Random 2-Digit Number

03

Independence

Independence in a Real Setting: A Random 2-Digit Number

A counting example where a natural pair of variables turns out to be dependent.

Here independence is tested on real numbers instead of a ready-made table. We build the joint distribution by counting, and find that a very natural pair of digits-based random variables is not independent.

3.1The experiment

Example 4Units digit and remainder mod 4

A number from 00 to 99 is picked uniformly at random. Let X = the units digit and Y = the remainder when the number is divided by 4. Are X and Y independent?

Solution

Marginals. Each units digit appears in exactly 10 of the 100 numbers, and each remainder appears in exactly 25 (since 4 divides 100).
So $X \sim \text{Uniform}\{0,1,\dots,9\}$ and $Y \sim \text{Uniform}\{0,1,2,3\}$ .
Look for an impossible pair. If the units digit is 1, the number is odd, so it cannot be a multiple of 4.
Hence $f_{XY}(1,0) = 0$ , but $f_X(1)\,f_Y(0) = \tfrac1{10} \cdot \tfrac14 = \tfrac1{40} \ne 0$ .
X and Y are not independent. The units digit fixes whether the number is odd or even, and that restricts the remainder mod 4.

3.2The whole joint table, by counting

Counting how many of the 100 numbers have each (units digit, remainder) pair gives the full picture. Each cell of the figure shows how many numbers fall there.

Figure 4. Numbers 00 to 99 by units digit and remainder mod 4

For an even units digit the number is even, so the remainder is 0 or 2 (5 numbers each). For example units digit 2 and remainder 2: 2, 22, 42, 62, 82.
For an odd units digit the remainder is 1 or 3 (5 numbers each).
In an independent table every cell would hold $100 \times \tfrac1{10} \times \tfrac14 = 2.5$ numbers, which is not even a whole number. Here half the cells hold 0 and half hold 5.

Check. The numbers add up: 20 non-zero cells × 5 = 100. Each row total is 25 and each column total is 10, matching the marginals.

3.3Related pairs, to build intuition

Example 5Tens digit and units digit (extra)

For the same random number let X be the tens digit and Y the units digit. Are they independent?

Solution

Both are $\text{Uniform}\{0,\dots,9\}$ , so each marginal value is $\tfrac1{10}$ .
Each pair (tens, units) is exactly one number out of 100, so every cell is $\tfrac1{100}$ .
$\tfrac1{100} = \tfrac1{10} \cdot \tfrac1{10}$ for all 100 cells: independent.

Example 6Units digit and remainder mod 5 (extra)

Now let Y be the remainder when the number is divided by 5. Are the units digit X and Y independent?

Solution

The units digit decides the remainder mod 5: the remainder is the units digit mod 5.
So $f_{XY}(1,0) = 0$ , but $f_X(1) = \tfrac1{10}$ and $f_Y(0) = \tfrac{20}{100} = \tfrac15$ , giving a product $\tfrac1{50} \ne 0$ .
Dependent, and more strongly than in Example 4: here X completely determines Y.

Compare the three: tens and units digits carry no information about each other (independent), the units digit says a little about the remainder mod 4 (dependent), and it says everything about the remainder mod 5 (dependent).

3.4Independence in real modelling

The same thinking applies in modelling. Let X = runs scored in a cricket over and Y = wickets taken in that over. These are not independent: an over with many wickets tends to have fewer runs. Deciding whether an independence assumption is reasonable is a routine part of building a model from real data.

Independent assumptions make life simple. If two variables are independent the joint PMF is just the two marginals multiplied. If they are not, you must model the whole joint table.
Real data is a sanity check (extra). Ask whether one variable physically limits the other, as the units digit limits the remainder. If yes, expect dependence.

3.5Exam tips and common mistakes

Exam tips

Ask first: does one variable restrict the values of the other? If so, look for an impossible pair and use it.
For uniform sample spaces, count cases instead of using probability formulas.

Common mistakes

Checking only the marginals: both uniform marginals look "perfectly independent" but the pair is not.
Forgetting that 4 divides 100, which is why each remainder appears exactly 25 times.

Practice questions (graded-assignment style)

Q1Units digit 1, remainder 0

Numerical Answer Type (NAT).

How many of the numbers 00 to 99 have units digit 1 and leave remainder 0 when divided by 4?

Solution

Idea

Units digit 1 means the number is odd.

An odd number cannot be a multiple of 4, so there are 0 such numbers.
That makes $f_{XY}(1,0) = 0$ , which is the impossible pair used in Example 4.
Answer: 0

Q2Units digit 2, remainder 2

Numerical Answer Type (NAT).

How many of the numbers 00 to 99 have units digit 2 and remainder 2 when divided by 4?

Solution

Idea

List the numbers ending in 2.

The numbers ending in 2 are 2, 12, 22, ..., 92. Their remainders mod 4 alternate 2, 0, 2, 0, ...
The ones with remainder 2 are 2, 22, 42, 62, 82: five numbers, so f_XY(2,2) = 5/100.
Answer: 5

Q3Tens and units digit

Multiple Choice Question (MCQ): choose the one correct option.

For a uniformly random number from 00 to 99, the tens digit and the units digit are:

1

Independent

2

Dependent, because both come from the same number

3

Dependent, because the sum of the digits is fixed

4

Cannot be decided

Solution

Idea

Count the numbers in each (tens, units) cell.

Every (tens, units) pair is exactly one number, so each cell is $\tfrac1{100} = \tfrac1{10} \cdot \tfrac1{10}$ .
Coming from the same experiment does not make variables dependent.
Answer: option 1.

Q4Digits statements

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements about X = units digit and Y = remainder mod 4.

1

f_X(t) = \tfrac1{10}

for each digit

t

2

f_Y(r) = \tfrac14

for each remainder

r

3

f_{XY}(1,0) = \tfrac1{40}

4

f_{XY}(1,0) = 0

Solution

Idea

Use counting and the odd/even argument.

Statements 1 and 2 are the marginals found in Example 4.
Statement 4 is the impossible pair. Statement 3 is the product $f_X(1) f_Y(0)$ , not the joint value.
Answer: options 1, 2 and 4.

Next · IndependenceIndependent vs Dependent: The Working Rules

04

Independence

Independent vs Dependent: The Working Rules

The rules for proving each, the fastest checks, and the "Try these" questions worked out.

This chapter collects the working rules and uses them on the "Try these" questions of the lecture, with every step shown.

4.1The working rules

To show independenceVerify

f_{XY}(t_1,t_2) = f_X(t_1)\,f_Y(t_2)

for every

t_1 \in T_X

,

t_2 \in T_Y

.

To show dependenceOne pair is enough: find

t_1, t_2

with

f_{XY}(t_1,t_2) \ne f_X(t_1)\,f_Y(t_2)

.

4.2The fastest dependence check: a zero cell

Look for a cell with $f_{XY} = 0$ whose row total and column total are both non-zero. Then $f_{XY} = 0$ but $f_X f_Y > 0$ , so the product rule fails and the variables are dependent. This is exactly what happened in Example 4 with the pair (1, 0).

Careful. A zero cell alone is not enough: if its whole row (or column) is zero, the product is also 0 and nothing fails. And the converse is false: a table with no zero cells can still be dependent (Example 2).

4.3A visual shortcut: the rows

X and Y are independent exactly when every row of the joint table is a multiple of the same row, namely $f_X$ . Row $t_2$ must equal $f_Y(t_2)$ times the vector $f_X$ . If two rows have different shapes, the variables are dependent.

Figure 5. Conditional PMF of X in each row: the same shape when independent

Why it works (extra). If $f_{XY}(t_1,t_2) = c(t_2)\,f_X(t_1)$ in every row, then adding over $t_1$ gives $f_Y(t_2) = c(t_2)\cdot 1$ , so $c(t_2) = f_Y(t_2)$ and the product rule holds. Reading it the other way, row $t_2$ divided by its total is the conditional PMF $f_{X\mid Y=t_2}$ , and independence says that is always $f_X$ .

4.4Try these: worked solutions

Example 7Try these Q1: a 3 by 3 table

Check whether X and Y are independent, given the joint PMF below.

Solution

$Y\backslash X$	0	1	2
0	1/6	1/12	1/12
1	1/4	1/8	1/8
2	1/12	1/24	1/24

Row sums. With denominator 24: $f_Y(0) = \tfrac4{24} + \tfrac2{24} + \tfrac2{24} = \tfrac8{24} = \tfrac13$ , $f_Y(1) = \tfrac6{24} + \tfrac3{24} + \tfrac3{24} = \tfrac12$ , $f_Y(2) = \tfrac2{24} + \tfrac1{24} + \tfrac1{24} = \tfrac16$ .
Column sums. $f_X(0) = \tfrac4{24} + \tfrac6{24} + \tfrac2{24} = \tfrac12$ , $f_X(1) = \tfrac2{24} + \tfrac3{24} + \tfrac1{24} = \tfrac14$ , $f_X(2) = \tfrac14$ .
Sanity check. $\tfrac13 + \tfrac12 + \tfrac16 = 1$ and $\tfrac12 + \tfrac14 + \tfrac14 = 1$ .
Products f_X(t₁) f_Y(t₂): row 0 is 1/6, 1/12, 1/12; row 1 is 1/4, 1/8, 1/8; row 2 is 1/12, 1/24, 1/24.
Compare cell by cell. All nine products match the given joint PMF exactly.
Row shortcut. Every row is a multiple of $f_X = (\tfrac12,\tfrac14,\tfrac14)$ : row 0 is $\tfrac13$ of it, row 1 is $\tfrac12$ of it, row 2 is $\tfrac16$ of it.
Answer: X and Y are independent.

Example 8Try these Q2(a): a zero cell

Decide whether X and Y are independent for this joint PMF.

Solution

$Y\backslash X$	0	1	2
0	0	1/8	1/8
1	1/8	1/8	1/8
2	1/8	1/8	1/8

Spot the zero cell. f_XY(0,0) = 0. Now check whether its row and column have non-zero totals.
Marginals. $f_X(0) = 0 + \tfrac18 + \tfrac18 = \tfrac14$ and $f_Y(0) = 0 + \tfrac18 + \tfrac18 = \tfrac14$ .
Compare. $f_{XY}(0,0) = 0$ but $f_X(0)\,f_Y(0) = \tfrac14 \cdot \tfrac14 = \tfrac1{16}$ . This is the "zero cell with non-zero marginals" case.
Answer (a): dependent.

Example 9Try these Q2(b): a near miss

Decide whether X and Y are independent for this joint PMF.

Solution

$Y\backslash X$	0	1	2
0	1/6	1/12	1/8
1	1/4	1/8	1/8
2	1/24	1/24	1/24

Marginals (denominator 24): $f_X(0) = \tfrac4{24} + \tfrac6{24} + \tfrac1{24} = \tfrac{11}{24}$ , $f_X(1) = \tfrac6{24}$ , $f_X(2) = \tfrac7{24}$ ; $f_Y(0) = \tfrac9{24}$ , $f_Y(1) = \tfrac{12}{24}$ , $f_Y(2) = \tfrac3{24}$ . Both sets add to $\tfrac{24}{24} = 1$ .
Test the cell $(0,0)$ . $f_X(0)\,f_Y(0) = \tfrac{11}{24} \cdot \tfrac9{24} = \tfrac{99}{576}$ .
Conclude. $f_{XY}(0,0) = \tfrac16 = \tfrac{96}{576} \ne \tfrac{99}{576}$ , so the product rule fails at $(0,0)$ .
Row shortcut agrees. Row 0 is $(\tfrac16,\tfrac1{12},\tfrac18) \propto (4,2,3)$ , while row 1 is $(\tfrac14,\tfrac18,\tfrac18) \propto (2,1,1)$ . Different shapes, so dependent.
Answer (b): dependent.

This table is very close to the independent one in Q1: only two cells differ (top-right and bottom-left). Near misses like this are why every cell must be checked before claiming independence.

Example 10Sum of two coins (extra)

Two fair coins give $X_1, X_2 \in \{0,1\}$ (1 for heads), independent of each other. Let $X = X_1$ and $Z = X_1 + X_2$ . Are $X$ and $Z$ independent?

Solution

$Z\backslash X$	0	1	$f_Z$
0	1/4	0	1/4
1	1/4	1/4	1/2
2	0	1/4	1/4
$f_X$	1/2	1/2	1

The four equally likely outcomes give the table above.
Zero cell. $f_{XZ}(1,0) = 0$ (if the first coin is heads, the sum cannot be 0), but $f_X(1) = \tfrac12$ and $f_Z(0) = \tfrac14$ , so $f_X(1) f_Z(0) = \tfrac18 \ne 0$ .
Dependent. The sum depends on the first coin even though the two coins are independent. Independence of the parts does not carry over to a variable built from them.

4.5Marginals never decide it

Many different joint PMFs share the same marginals. Example 1 and Example 3 (with x ≠ 1/4) have identical marginals but only one is independent.
The marginals tell you what the independent table would be (their product). You still have to compare it with the actual joint table.

4.6Exam tips and common mistakes

Exam tips

Order of attack: (1) look for a zero cell with non-zero marginals; (2) test one or two simple cells; (3) only if all pass, check the rest or use the row shortcut.
Keep a common denominator for the whole table (24 in Q1 and Q2).

Common mistakes

Declaring independence because two marginals look similar or a few cells match.
Treating a zero cell as proof of dependence even when its row or column total is also 0.
Dividing a row by the wrong total when using the row shortcut: divide by that row's own total.

Practice questions (graded-assignment style)

Q1Fastest check

Multiple Choice Question (MCQ): choose the one correct option.

A joint table has a cell equal to 0 whose row total and column total are both positive. Then X and Y are:

1

Independent

2

Dependent

3

Independent only if the marginals are uniform

4

Cannot say

Solution

Idea

Compare the cell with the product of its marginals.

The cell is 0 but the product of two positive marginals is positive, so the product rule fails.
Answer: option 2.

Q2Row shortcut

Multiple Choice Question (MCQ): choose the one correct option.

A joint table (rows Y = 0, 1, 2; columns X = 0, 1, 2) has rows (0.06, 0.09, 0.15), (0.10, 0.15, 0.25) and (0.04, 0.06, 0.10). Are X and Y independent?

1

Yes

2

No, because the rows have different totals

3

No, because there are three rows

4

Cannot say

Solution

Idea

Rows must be multiples of the same vector.

Row 0 is $0.03 \times (2,3,5)$ , row 1 is $0.05 \times (2,3,5)$ , row 2 is $0.02 \times (2,3,5)$ : the same shape.
The table adds up: $0.3 + 0.5 + 0.2 = 1$ , and $f_X = (0.2, 0.3, 0.5)$ is that shape. So the product rule holds in every cell: independent. Different row totals are fine; only the shape must match.
Answer: option 1.

Q3Fill in using independence

Numerical Answer Type (NAT).

X and Y are independent, $f_X(0) = 0.5$ , $f_Y(1) = 0.4$ . Find $f_{XY}(0, 1)$ .

Solution

Idea

Multiply the two marginals.

$0.5 \times 0.4 = 0.2$ .
Answer: 0.2

Q4Independent or not

Multiple Select Question (MSQ): select all that apply.

Which of the following joint tables (rows Y, columns X) show independent X and Y?

1

Row 0: 1/6, 1/12, 1/12; Row 1: 1/4, 1/8, 1/8; Row 2: 1/12, 1/24, 1/24

2

Row 0: 0, 1/8, 1/8; Row 1: 1/8, 1/8, 1/8; Row 2: 1/8, 1/8, 1/8

3

Row 0: 1/6, 1/12, 1/8; Row 1: 1/4, 1/8, 1/8; Row 2: 1/24, 1/24, 1/24

4

Row 0: 1/4, 1/4; Row 1: 1/4, 1/4

Solution

Idea

Test each: zero cell, product of marginals, row shortcut.

Table 1 (Q1) and table 4 (Example 1) are independent.
Table 2 has a zero cell with non-zero marginals; table 3 fails at the cell (0,0): 99/576 against 96/576.
Answer: options 1 and 4.

Next · RevisionIndependence: Revision Sheet

05

Revision

Independence: Revision Sheet

One page to revise everything in this section.

The big picture

Independent events to variables

Events: $P(A\cap B) = P(A)P(B)$ .
Variables: every event from X alone is independent of every event from Y alone.

The test

$f_{XY}(t_1,t_2) = f_X(t_1)\,f_Y(t_2)$ for all pairs.
Equivalent: every conditional $f_{Y\mid X=t_1}$ equals $f_Y$ .

Reading a table

Columns $t_1$ of $X$ , rows $t_2$ of $Y$ ; last column $f_Y$ , last row $f_X$ .
One failing cell means dependent; all cells passing means independent.

Parameter tables

Marginals can be the same for independent and dependent tables.
Example 3 is independent exactly when $x = \tfrac14$ .

Digits example

Units digit and remainder mod 4: $f_{XY}(1,0) = 0$ but $f_X(1) f_Y(0) = \tfrac1{40}$ , so dependent.
Tens and units digits: independent (extra).

Fast checks

Zero cell with non-zero row and column totals: dependent.
Independent exactly when every row is a multiple of the same vector f_X.

Revision sheet

Idea	Key point
Definition	$f_{XY}(t_1,t_2) = f_X(t_1)\,f_Y(t_2)$ for all $t_1, t_2$
Conditional view	Independent $\iff f_{Y\mid X=t_1} = f_Y$ for every $t_1$
Proving independence	Check every cell of the joint table
Proving dependence	One failing cell is enough
Fast dependence test	A zero cell with non-zero row and column totals
Row shortcut	Independent $\iff$ all rows are multiples of the same vector $f_X$
Marginals alone	Never decide independence: many joint PMFs share the same marginals

Week 2 roadmap

Chapters	Topic	Where
1 to 5	Independence of two random variables	this note
6 to 10	Independence of multiple random variables	this note
11 to 16	Functions of one random variable	this note
17 to 22	Functions of two random variables and visualization	Week 2 · Part 2 of 2
23 to 28	Functions of multiple random variables	Week 2 · Part 2 of 2
29 to 35	Sums of random variables and convolution	Week 2 · Part 2 of 2
36 onwards	Minimum and maximum of random variables (coming)	Week 2 · Part 2 of 2

This note has Chapters 1 to 16 (independence, and functions of one variable). Chapters 17 onwards are in the next note, Week 2 · Part 2 of 2; more chapters will be added as they are sent.

Common mistakes

Checking only a few cells

Independence needs the product rule at every cell. Dependence needs just one failing cell.

Deciding from the marginals

Independent and dependent tables can share the same marginals. Always compare with the joint table.

Swapping row and column marginals

For a cell take its column total $f_X$ times its row total $f_Y$ .

Zero cell with a zero row or column

If the whole row or column is 0 the product is 0 as well, so no failure.

Independent parts, independent function

X₁, X₂ independent does not make X₁ and X₁ + X₂ independent.

Independent vs identically distributed

Same marginals say nothing about independence.

Next · Many variablesIndependence for n Random Variables

06

Many variables

Independence for n Random Variables

The definition for n variables and the three-tosses example.

Chapters 1 to 5 dealt with two random variables. Often there are three, ten or a hundred (all the tosses of a coin, all the digits of a number). This chapter carries independence over to n random variables.

6.1Definition for n random variables

Independent random variables (n of them)

Let $X_1, X_2, \dots, X_n$ be random variables on one probability space, with the range of $X_i$ written $T_{X_i}$ . They are independent if events defined using different $X_i$ are mutually independent. Equivalently:

f_{X_1X_2\cdots X_n}(t_1, \dots, t_n) = f_{X_1}(t_1)\, f_{X_2}(t_2) \cdots f_{X_n}(t_n) \quad \text{for all } t_i \in T_{X_i}

The joint PMF must split into the product of all n marginals, at every point $(t_1, \dots, t_n)$ .

Remember. For n = 2 this is exactly the definition from Chapter 1. The new thing is that the product now has n factors, one per variable.

6.2Example: three tosses of a fair coin

Figure 6. The 8 equally likely outcomes of three fair tosses

6.3Going deeper

Number of points to check (extra). The product rule must hold at all $|T_{X_1}| \cdots |T_{X_n}|$ points: $2^3 = 8$ for three tosses, $10 \times 10 \times 5 = 500$ in Try these Q1. This is why a structural argument (separate digits, separate tosses) is much quicker than checking cell by cell.
Same idea, n tosses (extra). For $n$ fair tosses every outcome has probability $\tfrac1{2^n}$ , and so does the product of $n$ factors of $\tfrac12$ . For $n = 5$ the outcome (1,0,1,1,0) has probability $\tfrac1{32}$ .

6.4Exam tips and common mistakes

Exam tips

If the variables come from separate, repeated experiments (tosses, rolls, trials), independence is built in. Say so and multiply.
To disprove independence of $n$ variables find one point $(t_1,\dots,t_n)$ where the joint PMF differs from the product.

Common mistakes

Checking only some of the points of the product rule.
Using only two-at-a-time checks for three or more variables. The next chapters show why that fails.

Practice questions (graded-assignment style)

Q1Five tosses

Numerical Answer Type (NAT).

Five fair coin tosses are independent. What is the probability of the exact outcome (heads, tails, heads, heads, tails)?

Solution

Idea

Multiply five factors of 1/2.

$(\tfrac12)^5 = \tfrac1{32} = 0.03125$ .
Answer: 0.03125

Q2What must be checked?

Multiple Choice Question (MCQ): choose the one correct option.

To show $X_1, X_2, X_3$ are independent from the joint PMF we must check:

1

Only that each pair is independent

2

f = f_{X_1} f_{X_2} f_{X_3}

at every point

(t_1,t_2,t_3)

3

f = f_{X_1} f_{X_2} f_{X_3}

at one point

4

Only that the three marginals are equal

Solution

Idea

The product rule must hold everywhere.

Option 2 is the definition. Pairwise checks alone are not enough (the next chapters show a counterexample).
Answer: option 2.

Q3Independent tosses

Multiple Select Question (MSQ): select all that apply.

Three fair tosses give $X_1, X_2, X_3$ as above. Choose the correct statements.

1

f_{X_1X_2X_3}(1,0,1) = \tfrac18

2

f_{X_1}(1) = \tfrac12

3

f_{X_1X_2X_3}(1,1,1) = \tfrac12

4

The 8 joint values add up to 1

Solution

Idea

Every outcome has probability 1/8.

Each of the 8 outcomes has probability $\tfrac18$ , which adds to 1.
Statement 3 is false: $\tfrac12$ is a single marginal, not the joint value.
Answer: options 1, 2 and 4.

Next · Many variablesA Random 3-Digit Number

07

Many variables

A Random 3-Digit Number

Two independent pairs and one dependent pair from the digits of a number.

A random 3-digit number is a good test bed: its digits are independent of each other, but quantities built from the same digit are not.

7.1The experiment

Example 12First digit, parity and last digit

A number from 000 to 999 is chosen uniformly at random. Let $X$ = first digit from the left, $Y$ = the number modulo 2, $Z$ = first digit from the right. Are $X, Y, Z$ independent?

Solution

Marginals. $X \sim \text{Uniform}\{0,\dots,9\}$ , $Z \sim \text{Uniform}\{0,\dots,9\}$ , $Y \sim \text{Uniform}\{0,1\}$ .
Pair $(X,Z)$ . Fixing the first and last digit leaves 10 choices for the middle digit, so $f_{XZ}(t_1,t_3) = \tfrac{10}{1000} = \tfrac1{100} = \tfrac1{10}\cdot\tfrac1{10} = f_X(t_1) f_Z(t_3)$ for every pair: $X$ and $Z$ are independent.
Pair $(X,Y)$ . Fixing the first digit and the parity leaves 10 middle digits and 5 last digits of the right parity: $10 \times 5 = 50$ numbers. So $f_{XY}(t_1,t_2) = \tfrac{50}{1000} = \tfrac1{20} = \tfrac1{10}\cdot\tfrac12 = f_X(t_1) f_Y(t_2)$ : $X$ and $Y$ are independent.
Pair $(Y,Z)$ . The parity of the number is decided by its last digit. A number ending in 2 is even, so it can never have remainder 1 mod 2: $f_{YZ}(1,2) = 0$ , but $f_Y(1) f_Z(2) = \tfrac12 \cdot \tfrac1{10} = \tfrac1{20} \ne 0$ . $Y$ and $Z$ are dependent.
Conclusion. Since even one pair fails, $X, Y, Z$ are not independent.

Figure 7. Which digit feeds which variable

7.2Why one dependent pair is enough

If X, Y, Z were independent, then every pair among them would be independent too (Chapter 9 proves this by summing out the third variable). So a single dependent pair, here (Y, Z), rules out independence of the whole collection immediately. There was no need to look at the triple joint PMF at all.

7.3Going deeper

Rule of thumb (extra). Quantities built from different digits are independent (the digits are chosen independently). Quantities that read the same digit are a warning sign for dependence, as Y and Z both read the last digit.
But sharing is not a verdict (extra). Take $X$ = first digit and $W = (\text{first digit} + \text{middle digit}) \bmod 2$ . They share the first digit, yet for every pair $f_{XW} = \tfrac{50}{1000} = \tfrac1{10}\cdot\tfrac12$ : independent. The middle digit, which is uniform and independent, scrambles the parity completely. Always compute.
Why 1000 numbers (extra). 000 to 999 includes numbers like 007 and 042, so every number has exactly three digits and each digit position is uniform.

7.4Exam tips and common mistakes

Exam tips

Count numbers: fix the variable values, then count how many numbers fit, and divide by 1000.
To disprove, find an impossible combination like an odd number ending in 2.

Common mistakes

Concluding that all three are independent after checking only (X, Z) and (X, Y).
Forgetting that "number modulo 2" depends only on the last digit.

Practice questions (graded-assignment style)

Q1f_XZ(3, 7)

Numerical Answer Type (NAT).

For the random number of Example 12, find $f_{XZ}(3, 7)$ , the probability that the first digit is 3 and the last digit is 7.

Solution

Idea

Count numbers of the form 3_7.

The numbers 307, 317, ..., 397 are 10 numbers, so the probability is 10/1000 = 0.01.
Answer: 0.01

Q2f_XY(4, 0)

Numerical Answer Type (NAT).

Find $f_{XY}(4, 0)$ : first digit 4 and the number is even.

Solution

Idea

Fix the first digit, then choose the last digit to be even.

Middle digit: 10 choices. Last digit: 5 even choices. So 10 × 5 = 50 numbers and the probability is 50/1000 = 0.05.
Answer: 0.05

Q3The dependent pair

Multiple Choice Question (MCQ): choose the one correct option.

In Example 12 which pair is dependent?

1

(X, Z)

2

(X, Y)

3

(Y, Z)

4

None

Solution

Idea

Which two variables read the same digit?

$Y$ (the parity) and $Z$ (the last digit) both depend on the last digit: $f_{YZ}(1,2) = 0$ but $f_Y(1) f_Z(2) = \tfrac1{20}$ .
Answer: option 3.

Q4Conclusions

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements about X, Y, Z of Example 12.

1

X and Z are independent

2

X and Y are independent

3

X, Y, Z are independent

4

Y and Z are dependent

Solution

Idea

One dependent pair rules out the whole collection.

Statements 1, 2 and 4 are the three pair results. Statement 3 is false because the pair (Y, Z) is dependent.
Answer: options 1, 2 and 4.

Next · Many variablesEven Parity: Pairwise Independent, Not Independent

08

Many variables

Even Parity: Pairwise Independent, Not Independent

The key counterexample: every pair unrelated, the triple tightly tied.

Can three variables be such that any two look completely unrelated, and yet the three together are strongly tied? Yes: this is the most important counterexample of the week.

8.1The even parity PMF

Example 13Even parity

Let $(X_1, X_2, X_3)$ have the joint PMF below. It is called even parity because every possible outcome contains an even number of 1s. Each $X_i \sim \text{Uniform}\{0,1\}$ (each value appears in two rows). Are $X_1, X_2, X_3$ independent? Are they pairwise independent?

Solution

$t_1$	$t_2$	$t_3$	$f_{X_1X_2X_3}$
0	0	0	1/4
0	1	1	1/4
1	0	1	1/4
1	1	0	1/4

Every pair is independent. Summing out the third variable, each pair has the same joint table (rows $X_j$ , columns $X_i$ ): every cell is $\tfrac14$ , and every marginal is $\tfrac12$ .
$X_j\backslash X_i$ 0 1 $f_{X_j}$

0 1/4 1/4 1/2

1 1/4 1/4 1/2

$f_{X_i}$ 1/2 1/2 1

For $(X_1,X_2)$ , $(X_1,X_3)$ and $(X_2,X_3)$ every cell is $\tfrac14 = \tfrac12 \cdot \tfrac12$ .
But the three together are not independent. Take $(X_1,X_2,X_3) = (0,0,1)$ , which has an odd number of 1s: $P(X_1=0, X_2=0, X_3=1) = 0$ , while $P(X_1=0)P(X_2=0)P(X_3=1) = \tfrac18$ .

$X_j\backslash X_i$	0	1	$f_{X_j}$
0	1/4	1/4	1/2
1	1/4	1/4	1/2
$f_{X_i}$	1/2	1/2	1

Figure 8. The joint PMF of the even parity triple

Pairwise independence does not imply independence

In even parity, any two variables look completely unrelated, yet the first two determine the third: $X_3 = (X_1 + X_2) \bmod 2$ .

8.2The direction that does hold

Figure 9. Independence implies pairwise independence, but not the reverse

Independent $\Rightarrow$ pairwise independent. Every sub-collection of independent variables is independent.
Pairwise independent $\not\Rightarrow$ independent. Even parity is a counterexample, and Try these Q2 gives another.

8.3Going deeper

Conditional view (extra). Given $X_1 = 0$ and $X_2 = 0$ , the third variable is forced: $P(X_3 = 1 \mid X_1 = 0, X_2 = 0) = 0$ , whereas $P(X_3 = 1) = \tfrac12$ . Independence would have said that learning $X_1, X_2$ tells nothing about $X_3$ , but here it tells everything.
Why the pair tables hide it (extra). A pair table only sees two coordinates at a time; the constraint "even number of 1s" lives in all three together.
Where it appears (extra). Parity checks are used to detect errors in data: the last bit is chosen so that the number of 1s is even, which is exactly this structure.

8.4Exam tips and common mistakes

Exam tips

To disprove independence of three variables look for a point with joint probability 0 but a positive product (an impossible outcome).
To show pairwise independence sum out the third variable and test each pair table.

Common mistakes

Believing that "every pair independent" is the same as "independent".
Forgetting that outcomes with probability 0 are points of the grid too. They must also satisfy the product rule, and an impossible outcome with a positive product (like (0,0,1) in even parity) is the clearest failure.

Practice questions (graded-assignment style)

Q1f(1, 1, 0)

Numerical Answer Type (NAT).

In the even parity PMF, what is the probability of the outcome (X1, X2, X3) = (1, 1, 0)?

Solution

Idea

It has two 1s, an even number.

(1,1,0) is one of the four allowed outcomes, so its probability is 1/4 = 0.25.
Answer: 0.25

Q2f(1, 1, 1)

Numerical Answer Type (NAT).

In the even parity PMF, what is f(1, 1, 1)?

Solution

Idea

Count the 1s.

(1,1,1) has three 1s, an odd number, so it is forbidden: probability 0. An independent triple would give 1/8.
Answer: 0

Q3Pairwise versus independent

Multiple Choice Question (MCQ): choose the one correct option.

For the even parity triple, which statement is true?

1

Independent and pairwise independent

2

Pairwise independent but not independent

3

Independent but not pairwise independent

4

Neither

Solution

Idea

Check pairs, then the triple.

Every pair table has all cells 1/4, so every pair is independent. The triple fails at (0,0,1).
Answer: option 2.

Q4Implications

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements.

1

If X1, X2, X3 are independent then every pair among them is independent

2

If every pair among X1, X2, X3 is independent then they are independent

3

One dependent pair rules out independence of X1, X2, X3

4

Even parity is a counterexample to statement 2

Solution

Idea

Which way does the implication run?

Statements 1 and 3 are the same fact, used in two directions. Statement 2 is false, and even parity shows it (statement 4).
Answer: options 1, 3 and 4.

Next · Many variablesUseful Consequences and Try These

09

Many variables

Useful Consequences and Try These

Sub-collections, functions of independent variables, i.i.d., and the two lecture questions solved.

Two useful consequences of independence, the notion of i.i.d. variables, and the two "Try these" questions of the lecture, worked out step by step.

9.1Sub-collections of independent variables

If $X_1, \dots, X_n$ are independent, then every pair (indeed every sub-collection) is independent: just sum the product formula over the other variables. For three variables, summing out $t_3$ :

f_{X_1X_2}(t_1,t_2) = \sum_{t_3} f_{X_1}(t_1) f_{X_2}(t_2) f_{X_3}(t_3) = f_{X_1}(t_1) f_{X_2}(t_2) \sum_{t_3} f_{X_3}(t_3) = f_{X_1}(t_1) f_{X_2}(t_2)

The last sum is 1. So one dependent pair immediately rules out independence of the whole collection.

9.2Functions of independent variables

Functions of independent variables are independent

If $X$ and $Y$ are independent, then $g(X)$ and $h(Y)$ are independent for any functions $g, h$ . Reason: for values $a, b$ ,

P(g(X)=a,\ h(Y)=b) = \sum_{x: g(x)=a}\ \sum_{y: h(y)=b} f_X(x) f_Y(y) = \Big(\sum_{x: g(x)=a} f_X(x)\Big)\Big(\sum_{y: h(y)=b} f_Y(y)\Big) = P(g(X)=a)\,P(h(Y)=b)

For example, if $X$ and $Y$ are independent then $X^2$ and $e^Y$ are independent, and so are $X \bmod 2$ and $\max(Y, 0)$ .

Example 14A numeric check of the function rule (extra)

Let $X \sim \text{Uniform}\{0,1,2,3\}$ and $Y \sim \text{Uniform}\{-1,0,1\}$ be independent. Check that $g(X) = X \bmod 2$ and $h(Y) = \max(Y,0)$ are independent at two points.

Solution

$g(X) = 0$ for $X \in \{0, 2\}$ , so $P(g=0) = \tfrac12$ ; $h(Y) = 0$ for $Y \in \{-1, 0\}$ , so $P(h=0) = \tfrac23$ ; $h(Y) = 1$ only for $Y = 1$ , so $P(h=1) = \tfrac13$ .
$P(g=0, h=0) = P(X \in \{0,2\})\,P(Y \in \{-1,0\}) = \tfrac12 \cdot \tfrac23 = \tfrac13 = P(g=0)P(h=0)$ .
$P(g=1, h=1) = \tfrac12 \cdot \tfrac13 = \tfrac16 = P(g=1)P(h=1)$ . The same works for the other points.

Careful. The rule is about a function of X alone and a function of Y alone. Two functions of the same variable are usually not independent: for X uniform on {−1, 0, 1}, X and X² have P(X = 0, X² = 1) = 0 but P(X = 0) P(X² = 1) = 1/3 · 2/3 = 2/9.

9.3i.i.d. variables

i.i.d.A collection

X_1, \dots, X_n

is independent and identically distributed (i.i.d.) if the variables are independent and all share one distribution. Repeated tosses, rolls and trials are the standard examples.

Example 15Three i.i.d. biased tosses (extra)

Three independent tosses each land heads with probability $\tfrac13$ (so $X_i \sim$ Bernoulli $(\tfrac13)$ , i.i.d.). Find the probability of the outcome (1, 1, 0).

Solution

By independence multiply the marginals: $f(1)\,f(1)\,f(0) = \tfrac13 \cdot \tfrac13 \cdot \tfrac23$ .
$= \tfrac{2}{27}$ .
Because the tosses are identically distributed, the same formula p^k (1 − p)^(3 − k) works for any outcome with k heads. The 8 values add up to 1.

9.4Try these: worked solutions

Example 16Try these Q1: hundreds digit, tens digit and N mod 5

A number from 000 to 999 is chosen uniformly at random. Let $X$ = the hundreds digit, $Y$ = the tens digit, and $Z$ = the remainder when the number is divided by 5. Are $X, Y, Z$ independent?

Solution

Describe the number by its digits. Write $N = 100A + 10B + C$ . Choosing $N$ uniformly from 1000 numbers is the same as choosing the digit triple $(A,B,C)$ uniformly from $10^3$ triples, each with probability $\tfrac1{1000}$ . Here $X = A$ , $Y = B$ .
Simplify $Z$ . $100A + 10B$ is always a multiple of 5, so $Z = N \bmod 5 = C \bmod 5$ depends only on the last digit: $C \in \{0,5\} \to Z = 0$ , $C \in \{1,6\} \to Z = 1$ , and so on.
Marginals. $X, Y \sim \text{Uniform}\{0,\dots,9\}$ . Each $z \in \{0,1,2,3,4\}$ comes from exactly two last digits ( $z$ and $z+5$ ), so $f_Z(z) = \tfrac{2}{10} = \tfrac15$ , $Z \sim \text{Uniform}\{0,1,2,3,4\}$ .
Joint PMF. Fix $X = a$ , $Y = b$ , $Z = z$ . Then $A = a$ , $B = b$ , and $C$ must be $z$ or $z+5$ : exactly 2 numbers. So $f_{XYZ}(a,b,z) = \tfrac{2}{1000} = \tfrac1{500}$ .
Compare. $f_X(a)f_Y(b)f_Z(z) = \tfrac1{10} \cdot \tfrac1{10} \cdot \tfrac15 = \tfrac1{500}$ , equal for every $(a,b,z)$ : all $10 \times 10 \times 5 = 500$ points.
Conclude and contrast. $X, Y, Z$ are independent. They are built from three different digits. In Example 12, "mod 2" and "last digit" both used the last digit, which created the dependence.
Answer: yes, X, Y and Z are independent.

Figure 10. Z = N mod 5 is decided by the last digit alone: two last digits for each value of Z

Example 17Try these Q2: a product of two signs

Let $X$ and $Y$ be independent, each equal to $+1$ or $-1$ with probability $\tfrac12$ . Define $Z = XY$ . Show that $X, Y, Z$ are pairwise independent but not independent.

Solution

$x$	$y$	$z = xy$	$f_{XYZ}$
1	1	1	1/4
1	−1	−1	1/4
−1	1	−1	1/4
−1	−1	1	1/4

Joint PMF. $(X,Y)$ takes 4 equally likely values and each determines $Z$ : the table above.
Marginals. $Z = 1$ in two rows and $Z = -1$ in two rows, so $Z$ is also $\pm1$ with probability $\tfrac12$ each. Same for $X$ and $Y$ .
Pair $(X,Y)$ is independent by assumption.
Pair $(X,Z)$ . Given $X = x$ , the event $Z = z$ means $Y = z/x = xz$ . So $P(X=x, Z=z) = P(X=x, Y=xz) = \tfrac14 = \tfrac12 \cdot \tfrac12 = P(X=x)P(Z=z)$ for all four choices of $x, z \in \{1,-1\}$ . Independent.
Pair $(Y,Z)$ . The identical argument with the roles of $X$ and $Y$ swapped. Independent.
All three together. Take $X = 1, Y = 1, Z = -1$ . It is impossible, because $1 \cdot 1 = 1$ : $P(X=1, Y=1, Z=-1) = 0 \ne \tfrac12 \cdot \tfrac12 \cdot \tfrac12 = \tfrac18$ .
Conclude. Every pair is independent, but the triple is not: any two of the values fix the third. This is the same structure as even parity, written with $\pm1$ instead of $0/1$ .
Answer: pairwise independent, but X, Y, Z are not independent.

Figure 11. The joint PMF of (X, Y, Z = XY)

9.5Exam tips and common mistakes

Exam tips

Structure first: if every variable comes from its own digit, toss or trial, independence follows without a table.
To build a "pairwise but not independent" example, let the third variable be a function of the other two that is balanced against each of them separately (parity, product of signs).

Common mistakes

Applying the function rule to two functions of the same variable (X and X²).
Forgetting that Z = N mod 5 only sees the last digit, so it is independent of the first two digits.
Using i.i.d. when the tosses are independent but have different biases: identically distributed means the same distribution.

Practice questions (graded-assignment style)

Q1Joint value in Q1

Numerical Answer Type (NAT).

In Try these Q1, what is f_XYZ(a, b, z) for any digits a, b and any z in {0,1,2,3,4}? Give a decimal.

Solution

Idea

Two numbers out of 1000.

Exactly two numbers fit: 2/1000 = 0.002 = 1/500.
Answer: 0.002

Q2Mod 2 instead of mod 5

Numerical Answer Type (NAT).

Let X = hundreds digit, Y = tens digit and W = N mod 2 for a number from 000 to 999. What is f_XYW(a, b, w)? Give a decimal.

Solution

Idea

The last digit must have the right parity.

Five last digits have the right parity, so 5/1000 = 0.005 = 1/10 · 1/10 · 1/2: independent, like Q1.
Answer: 0.005

Q3Product of signs

Multiple Choice Question (MCQ): choose the one correct option.

For Z = XY with X, Y independent fair signs, which statement is true?

1

X, Y, Z are independent

2

X, Y, Z are pairwise independent but not independent

3

X and Z are dependent

4

Z is a constant

Solution

Idea

Each pair is balanced; the triple is constrained.

Every pair has all cells $\tfrac14$ , but $P(1,1,-1) = 0 \ne \tfrac18$ .
Answer: option 2.

Q4Functions of independent X and Y

Multiple Select Question (MSQ): select all that apply.

X and Y are independent. Which pairs are guaranteed to be independent?

1

X² and e^Y

2

X mod 2 and max(Y, 0)

3

X and X²

4

X + Y and X − Y

Solution

Idea

The rule needs a function of X alone and a function of Y alone.

Options 1 and 2 are a function of X and a function of Y. Option 3 uses the same variable twice, and option 4 mixes X and Y in both.
Answer: options 1 and 2.

Q5i.i.d. probability

Numerical Answer Type (NAT).

Three i.i.d. tosses each have $P(\text{heads}) = 0.2$ . Find the probability of the outcome $(1,1,0)$ .

Solution

Idea

Multiply 0.2, 0.2 and 0.8.

$0.2 \times 0.2 \times 0.8 = 0.032$ .
Answer: 0.032

Next · RevisionMany variables: Revision Sheet

10

Revision

Many variables: Revision Sheet

One page to revise everything in this section.

The big picture

n variables

$f_{X_1\cdots X_n} = f_{X_1}\cdots f_{X_n}$ at every point.
Equivalent: events from different variables are mutually independent.

Digits example

Example: $X, Z$ independent; $X, Y$ independent; $Y, Z$ dependent (both read the last digit).
One dependent pair rules out the whole triple.

Even parity

Every pair independent, triple not: (0,0,1) has probability 0 but product 1/8.
$X_3 = (X_1 + X_2) \bmod 2$ .

Implications

Independent ⇒ every sub-collection independent ⇒ pairwise independent.
Pairwise independent does not imply independent.

Functions

$X \perp Y \Rightarrow g(X) \perp h(Y)$ .
Not for two functions of the same variable.

Try these

Q1: $X, Y, Z = N \bmod 5$ independent (joint $\tfrac1{500}$ at all 500 points).
Q2: Z = XY is pairwise independent but not independent.

Revision sheet

Idea	Key point
Definition	$f_{X_1\cdots X_n} = f_{X_1}\cdots f_{X_n}$ at every point
Independent ⇒ pairwise	Every sub-collection of independent variables is independent
Pairwise ⇏ independent	Even parity and Z = XY are counterexamples
Disproving	One dependent pair, or one failing point, is enough
Functions	$X \perp Y \Rightarrow g(X) \perp h(Y)$
i.i.d.	Independent and identically distributed

Revision overview

DefinitionProduct of n marginals
At every point
Mutually independent events

ExamplesThree fair tosses: 1/8
3-digit number
Q1: 1/500 everywhere

DisprovingOne dependent pair
One failing point
Impossible outcome

Many variablesWeek 2

ImplicationsIndependent ⇒ pairwise
Pairwise ⇏ independent

CounterexamplesEven parity
Z = XY with signs
Third is fixed by two

ToolsFunctions of independent
i.i.d. variables
Sum out to get pairs

Revision: independence of many variables and why pairwise is not enough

Common mistakes

Pairwise equals independent

Pairwise independence is weaker. Even parity and Z = XY pass every pair test and still fail the triple.

Checking only a few points

For n variables the product rule is needed at every point of the grid.

Shared digits

Variables reading the same digit (parity and last digit) are dependent. Compute, do not assume.

Functions of the same variable

The function rule is for a function of X and a function of Y, not X and X².

Identical is not independent

i.i.d. needs both: independence and one common distribution.

Forgetting the impossible outcome

A point with joint probability 0 but a positive product is the quickest proof of dependence.

Next · Functions of one variableSeeing a PMF: Stem Plots

11

Functions of one variable

Seeing a PMF: Stem Plots

If X is random so is g(X). Stem plots show which values are possible and which are likely.

If $X$ is a random variable, so is $Y = g(X)$ : once $X$ is random, anything computed from it is random too. This chapter answers the question: what is the PMF of $Y$ ? First we need a good way of looking at a PMF.

11.1Stem plots

Stem plotA stem plot draws a vertical line at each value

x

the random variable can take, with height

P(X = x)

. It shows at a glance which values are possible and which are likely.

11.2Two reference distributions

Two distributions are used through the whole part, so it helps to know both well.

(a) Uniform

X \sim \text{Uniform}\{0, 1, \dots, 10\}

:

P(X = x) = \tfrac1{11}

for

x \in \{0,\dots,10\}

. Every stem has the same height.

(b) Binomial

X \sim \text{Binomial}(10, 0.5)

:

P(X = k) = \binom{10}{k}(0.5)^{10} = \dfrac{\binom{10}{k}}{1024}

,

k = 0, 1, \dots, 10

. The heights differ.

$k$	0	1	2	3	4	5	6	7	8	9	10
$\binom{10}{k}$	1	10	45	120	210	252	210	120	45	10	1
Uniform	1/11	1/11	1/11	1/11	1/11	1/11	1/11	1/11	1/11	1/11	1/11
Binomial	1/1024	10/1024	45/1024	120/1024	210/1024	252/1024	210/1024	120/1024	45/1024	10/1024	1/1024

Figure 12. Stem plots: flat for the uniform, bell-shaped for the binomial

Binomial: X = 5 is the most likely value (252/1024); then 4 and 6 (210/1024 each); 3 and 7 (120/1024); and 0 and 10 are the least likely (1/1024 each).
Check: the binomial numerators add up to 1024 (they are the row of Pascal's triangle), and the eleven uniform stems add up to 1.

11.3Reading a stem plot

Example 18Reading probabilities off the stems (extra)

For both reference distributions find $P(X \ge 8)$ and say which value of $X$ is most likely.

Solution

Uniform: the stems at 8, 9, 10 each have height $\tfrac1{11}$ , so $P(X \ge 8) = \tfrac3{11}$ . All values tie for most likely.
Binomial: add the stems at 8, 9, 10: $\tfrac{45 + 10 + 1}{1024} = \tfrac{56}{1024} = \tfrac7{128}$ .
The most likely binomial value is $X = 5$ , the tallest stem, with $\tfrac{252}{1024}$ .

11.4Going deeper

Stems are not the whole story (extra). A stem plot only shows heights. To get a probability of several values, add stems. To get the shape of the distribution of Y, we need to know which stems of X land where: that is the table method of the next chapter.
Symmetry (extra). The binomial stems are mirror images about X = 5: P(X = k) = P(X = 10 − k). This symmetry is what produces the pile-up pairs in the many-to-one example.

11.5Exam tips and common mistakes

Exam tips

Keep the binomial numerators 1, 10, 45, 120, 210, 252 in mind; they are reused in every example of this chapter.
Always check that your probabilities add up to 1 (or the numerators to 1024).

Common mistakes

Reading a stem height as a value of X. The horizontal position is the value; the height is the probability.
Writing the binomial probabilities without the 1024 denominator.

Practice questions (graded-assignment style)

Q1A binomial stem

Numerical Answer Type (NAT).

For $X \sim \text{Binomial}(10, 0.5)$ , what is $1024 \times P(X = 3)$ ?

Solution

Idea

$\binom{10}{3}$ .

$P(X = 3) = \binom{10}{3}/1024 = 120/1024$ , so the answer is 120.
Answer: 120

Q2Tail probability

Numerical Answer Type (NAT).

For $X \sim \text{Uniform}\{0,\dots,10\}$ find $11 \times P(X \ge 8)$ .

Solution

Idea

Count the stems at 8, 9, 10.

Three stems of height 1/11, so the probability is 3/11 and the answer is 3.
Answer: 3

Q3Which plot?

Multiple Choice Question (MCQ): choose the one correct option.

Which distribution has a stem plot with all stems of the same height?

1

Binomial(10, 0.5)

2

Uniform{0, ..., 10}

3

Both

4

Neither

Solution

Idea

Recall the two reference plots.

The uniform gives every value the same probability, 1/11. The binomial is bell-shaped.
Answer: option 2.

Q4Binomial stems

Multiple Select Question (MSQ): select all that apply.

For $X \sim \text{Binomial}(10, 0.5)$ choose the correct statements.

1

X = 5 is the most likely value

2

P(X = 4) = P(X = 6)

3

P(X = 0) = P(X = 5)

4

The stems add up to 1

Solution

Idea

Symmetry about 5.

P(X = 0) is 1/1024 but P(X = 5) is 252/1024, so statement 3 is false. The others are true.
Answer: options 1, 2 and 4.

Next · Functions of one variableThe Table Method

12

Functions of one variable

The Table Method

How to find the PMF of any function of a random variable.

Given the PMF of $X$ , the table method finds the PMF of $Y = g(X)$ for any function $g$ , one-to-one or many-to-one.

12.1The method

The table method

To find the distribution of $Y = g(X)$ :

Step I. Make a table with columns $x$ , $P(X = x)$ and $y = g(x)$ .
Step II. Read off the range of $Y$ (the distinct values in the $y$ column), and for each $y$ add up the probabilities of all rows with that $y$ :

P(Y = y) = \sum_{x:\, g(x) = y} P(X = x)

Why use it

It is simple, works for any function, and makes mistakes easy to spot: every row of X must appear in exactly one group, so the final probabilities must still add up to 1.

12.2A small worked example

Example 19Squaring a three-valued variable (extra)

$X$ takes the values $-1, 0, 1$ with probabilities $0.2, 0.5, 0.3$ . Find the PMF of $Y = X^2$ .

Solution

$x$	$P(X=x)$	$y = x^2$
−1	0.2	1
0	0.5	0
1	0.3	1

Step I: the table above.
Step II: the range of $Y$ is $\{0, 1\}$ .
$P(Y=1) = P(X=-1) + P(X=1) = 0.2 + 0.3 = 0.5$ (two rows give $y = 1$ ).
$P(Y=0) = P(X=0) = 0.5$ .
Check: 0.5 + 0.5 = 1.

Figure 13. Rows with the same y are added together

12.3Going deeper

Values outside the range (extra). If $y$ is not in the range of $Y$ , then $P(Y = y) = 0$ : no row of the table produces it.
The range can be smaller (extra). Y never has more distinct values than X. It has the same number exactly when g is one-to-one on the values of X.
Two kinds of function. One-to-one functions only move the probabilities (Chapter 13). Many-to-one functions make probabilities pile up (Chapter 14).

12.4Exam tips and common mistakes

Exam tips

Always write the full table, even when it is long. Then colour or tick the rows by their y value before adding.
Finish by adding all the new probabilities: the total must be 1.

Common mistakes

Listing the same y twice in the answer instead of merging its rows.
Forgetting a row, so the total is less than 1.

Practice questions (graded-assignment style)

Q1P(Y = 1)

Numerical Answer Type (NAT).

$X$ takes $-1, 0, 1$ with probabilities $0.2, 0.5, 0.3$ . Find $P(X^2 = 1)$ .

Solution

Idea

Which $x$ give $x^2 = 1$ ?

x = −1 and x = 1: 0.2 + 0.3 = 0.5.
Answer: 0.5

Q2Distinct values

Numerical Answer Type (NAT).

$X$ is uniform on $\{1, 2, 3, 4\}$ . How many distinct values does $Y = X^2$ take?

Solution

Idea

Square each value.

The squares are 1, 4, 9, 16: all different, so 4 values, each with probability 1/4.
Answer: 4

Q3The formula

Multiple Choice Question (MCQ): choose the one correct option.

In the table method, P(Y = y) equals:

1

P(X = y)

2

\sum_{x:\,g(x)=y} P(X = x)

3

g(P(X = x))

4

\sum_x P(X = x)

Solution

Idea

Add the rows that land on y.

Option 2 is the general formula; it reduces to P(X = x) when exactly one x gives y.
Answer: option 2.

Q4Table method facts

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements.

1

It works for any function g

2

The new probabilities must add up to 1

3

The range of Y can be larger than the range of X

4

Rows with the same y are added

Solution

Idea

Every row of X lands in exactly one group.

The range of Y has at most as many values as the range of X, so statement 3 is false.
Answer: options 1, 2 and 4.

Next · Functions of one variableOne-to-One Functions

13

Functions of one variable

One-to-One Functions

The probabilities stay; only their locations move.

If different $x$ always give different $y$ , each $y$ comes from exactly one $x$ , so $P(Y = g(x)) = P(X = x)$ . The probabilities are unchanged; they just move to new locations.

13.1Example: Y = X − 5

Figure 14. Y = X − 5: the same shapes, slid 5 units to the left

13.2Example: Y = 2X

Correction (extra). The PDF titles this example "Y = 2X", but its table (1, 2, 4, 8, ..., 1024) and range show Y = 2^X. Plain Y = 2X would give the values 0, 2, 4, ..., 20. The method is the same, and the note uses 2^X as in the table.

Figure 15. Y = 2^X: the same probabilities at very uneven locations

The takeaway

A one-to-one function never changes the list of probabilities; it only changes where they sit. Shifting ( $X - 5$ ) keeps the spacing; a non-linear map like $2^X$ squashes small values together and spreads large ones apart.

13.3Going deeper

Reflection (extra). For $Y = -X$ the stems are mirrored: $P(Y = -k) = P(X = k)$ . Still one-to-one, so probabilities are unchanged. The binomial is symmetric, so its picture looks the same.
One-to-one on the values that occur (extra). $g(x) = x^2$ is not one-to-one on all integers, but it is one-to-one on $\{1,2,3,4\}$ : the values 1, 4, 9, 16 are all different, so $P(Y = x^2) = P(X = x)$ .
Summing a shifted PMF (extra). The sum of all probabilities is still 1: no stem is added or removed.

13.4Exam tips and common mistakes

Exam tips

If a question asks for P(Y = y) with a one-to-one g, solve g(x) = y for x and read P(X = x).
Check one-to-one by listing the y values: if all are different, you are done.

Common mistakes

Changing the probabilities along with the locations. They stay exactly the same.
Assuming a non-linear map keeps equal spacing.

Practice questions (graded-assignment style)

Q1Shifted binomial

Numerical Answer Type (NAT).

$X \sim \text{Binomial}(10, 0.5)$ and $Y = X - 5$ . What is $1024 \times P(Y = -2)$ ?

Solution

Idea

Which x gives y = −2?

$y = -2$ comes from $x = 3$ , and $1024 \times P(X = 3) = 120$ .
Answer: 120

Q2Powers of two

Numerical Answer Type (NAT).

$X \sim \text{Binomial}(10, 0.5)$ and $Y = 2^X$ . What is $1024 \times P(Y = 32)$ ?

Solution

Idea

$2^5 = 32$ .

$Y = 32$ means $X = 5$ , whose probability is $252/1024$ .
Answer: 252

Q3What changes?

Multiple Choice Question (MCQ): choose the one correct option.

A one-to-one function g is applied to X. What changes in the PMF?

1

The probabilities

2

The locations of the stems

3

Both

4

Neither

Solution

Idea

One x for each y.

The same probabilities sit at the new locations g(x).
Answer: option 2.

Q4One-to-one maps

Multiple Select Question (MSQ): select all that apply.

$X$ is Uniform $\{0,\dots,10\}$ . For which $g$ is $Y = g(X)$ still uniform on 11 values?

1

g(x) = x - 5

2

g(x) = 2^x

3

g(x) = (x-5)^2

4

g(x) = -x

Solution

Idea

Which maps are one-to-one on 0, ..., 10?

$(x-5)^2$ is many-to-one (4 and 6 both give 1), so the result is not uniform on 11 values.
Answer: options 1, 2 and 4.

Next · Functions of one variableMany-to-One Functions

14

Functions of one variable

Many-to-One Functions

Probabilities pile up and the shape of the PMF can change.

When several $x$ values map to the same $y$ , their probabilities pile up on that $y$ . The shape of the PMF can change completely.

14.1Example: Y = (X − 5)²

Figure 17. Y = (X − 5)²: the flat uniform is no longer flat; the most likely binomial value moves from 5 (for X) to y = 1 (for Y)

Check on the PDF caption. The PDF says the most likely binomial value "moves from 0 to 1". The most likely value of X is 5 (which maps to y = 0, with 252/1024), but the most likely value of Y is y = 1 with 420/1024, because two x values land there. Both are in the table; only the wording differs.

14.2Another many-to-one example

Example 23Remainders (extra)

Find the PMF of $Y = X \bmod 3$ for the uniform, and of $Y = X \bmod 2$ for the binomial.

Solution

Uniform, mod 3. $y = 0$ from $x \in \{0,3,6,9\}$ : 4 values; $y = 1$ from $\{1,4,7,10\}$ : 4 values; $y = 2$ from $\{2,5,8\}$ : 3 values. So $P(Y=0) = P(Y=1) = \tfrac4{11}$ and $P(Y=2) = \tfrac3{11}$ (total $\tfrac{11}{11}$ ).
Binomial, mod 2. $Y = 0$ when $X$ is even: $\tfrac{1 + 45 + 210 + 210 + 45 + 1}{1024} = \tfrac{512}{1024} = \tfrac12$ , so $P(Y=1) = \tfrac12$ too. (Even and odd numbers of heads are equally likely.)

14.3Going deeper

Unevenness appears (extra). The uniform input had equal stems, but the output stems are 1/11 and 2/11. The more x values collapse onto one y, the taller its stem.
Information is lost (extra). From Y = 1 alone you cannot tell whether X was 4 or 6. A one-to-one function loses nothing; a many-to-one function does.
Symmetric input, folded output (extra). Because the binomial is symmetric about 5, squaring the distance from 5 folds the plot in half and doubles every stem except the one at 0.

14.4Exam tips and common mistakes

Exam tips

Circle matching y values in the table before adding, so no row is missed or used twice.
Mention which x values feed each y, as in "P(Y = 1) = P(X = 4) + P(X = 6)".

Common mistakes

Doubling y = 0 as well. It comes from the single value x = 5.
Leaving the uniform answer as 1/11 each because the input was uniform.

Practice questions (graded-assignment style)

Q1P(Y = 4), binomial

Numerical Answer Type (NAT).

$X \sim \text{Binomial}(10, 0.5)$ , $Y = (X-5)^2$ . What is $1024 \times P(Y = 4)$ ?

Solution

Idea

Which x give y = 4?

$x = 3$ and $x = 7$ : $120 + 120 = 240$ .
Answer: 240

Q2P(Y = 25), uniform

Numerical Answer Type (NAT).

$X$ is uniform on $\{0,\dots,10\}$ and $Y = (X-5)^2$ . What is $11 \times P(Y = 25)$ ?

Solution

Idea

x = 0 and x = 10.

Two values of $x$ give $y = 25$ , each with probability $\tfrac1{11}$ : $\tfrac2{11}$ , so the answer is 2.
Answer: 2

Q3Most likely value of Y

Multiple Choice Question (MCQ): choose the one correct option.

For the binomial $X$ and $Y = (X-5)^2$ , the most likely value of $Y$ is:

1

0

2

1

3

4

25

Solution

Idea

Compare 252, 420, 240, 90, 20, 2.

P(Y = 1) = 420/1024 is the largest. P(Y = 0) = 252/1024 comes from the single value x = 5.
Answer: option 2.

Q4Many-to-one facts

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements about $Y = (X-5)^2$ with uniform $X$ .

1

P(Y = 0) = \tfrac1{11}

2

P(Y = 9) = \tfrac2{11}

3

Y

is uniform on its range

4

The probabilities of

Y

add up to 1

Solution

Idea

Count the x values for each y.

P(Y = 0) = 1/11 and the others are 2/11, so Y is not uniform on its range (statement 3 is false).
Answer: options 1, 2 and 4.

Next · Try theseTry These: Clipping Functions

15

Try these

Try These: Clipping Functions

The two lecture questions solved step by step.

The two "Try these" questions of the lecture, solved step by step. Both are clipping functions: everything beyond a cut-off collapses onto one value.

15.1Try these Q1

Example 24Try these Q1: max(X, 0)

Let $X \sim \text{Uniform}\{-5, -4, \dots, 5\}$ and $f(x) = x$ for $x > 0$ , $f(x) = 0$ for $x \le 0$ . Find the distribution of $Y = f(X)$ .

Solution

PMF of $X$ . There are 11 values from $-5$ to $5$ , so $P(X = x) = \tfrac1{11}$ for each.
Table method. Rows x = −5, ..., 5 each with 1/11; y = 0 for the six values −5 to 0, and y = x for 1 to 5.
Range of $Y$ . The distinct $y$ values are $\{0,1,2,3,4,5\}$ .
Add up probabilities for each $y$ . The value $y = 0$ comes from six $x$ values ( $-5$ to $0$ ); each positive $y$ comes from exactly one $x$ : $P(Y=0) = 6 \cdot \tfrac1{11} = \tfrac6{11}$ , and $P(Y=k) = \tfrac1{11}$ for $k = 1,\dots,5$ .
Check. $\tfrac6{11} + 5 \cdot \tfrac1{11} = 1$ . ✓

$y$	0	1	2	3	4	5
$P(Y=y)$	6/11	1/11	1/11	1/11	1/11	1/11

$f$ is many-to-one on the non-positive values (they all collapse to 0) and one-to-one on the positive values. Answer: $P(Y=0) = \tfrac6{11}$ and $P(Y=k) = \tfrac1{11}$ for $k = 1,\dots,5$ .

15.2Try these Q2

Example 25Try these Q2: max(X, 5)

Let $X \sim \text{Uniform}\{-500, -499, \dots, 500\}$ and $f(x) = \max(x, 5)$ . Find the distribution of $Y = f(X)$ .

Solution

PMF of $X$ . The values run from $-500$ to $500$ : that is $500 + 500 + 1 = 1001$ values, so $P(X = x) = \tfrac1{1001}$ .
Understand $f$ . $\max(x, 5)$ replaces anything below 5 by 5 and leaves values of 5 or more unchanged.
Range of $Y$ . $Y \in \{5, 6, 7, \dots, 500\}$ .
Count how many $x$ give $y = 5$ . Every $x$ from $-500$ to $5$ inclusive: $5 - (-500) + 1 = 506$ values, so $P(Y = 5) = \tfrac{506}{1001}$ .
The other values. Each $y \in \{6,\dots,500\}$ comes only from $x = y$ , so $P(Y = y) = \tfrac1{1001}$ .
Check. There are $500 - 6 + 1 = 495$ such values: $\tfrac{506}{1001} + \tfrac{495}{1001} = \tfrac{1001}{1001} = 1$ . ✓
Answer: P(Y = 5) = 506/1001, and P(Y = y) = 1/1001 for y = 6, 7, ..., 500.

The stem plot of Y is one very tall stem at 5 (just over half the probability) followed by a flat run of tiny stems.

Figure 18. Both answers: one tall stem, then a flat run

15.3Going deeper

Clipping is everywhere (extra). max(X, c) models a floor (for example a minimum charge) and min(X, c) models a cap. In both, a whole tail of values collapses onto one value, which gets a big stem.
Counting the collapsed values (extra). When the values are consecutive integers from a to b, there are b − a + 1 of them. That is why the count is 5 − (−500) + 1 = 506 here and not 505.
Cap example (extra). For X uniform on {1, ..., 10} and Y = min(X, 4): Y = 1, 2, 3 each have 1/10, and Y = 4 has 7/10 (x = 4, ..., 10).

15.4Exam tips and common mistakes

Exam tips

For clipping, find the cut-off value first, then count how many x values are at or beyond it.
Test your answer by adding all probabilities: 506 + 495 = 1001.

Common mistakes

Off-by-one errors when counting the collapsed values (506, not 505).
Giving the clipped value the probability of one x only.

Practice questions (graded-assignment style)

Q1Cap at 4

Numerical Answer Type (NAT).

X is uniform on {1, 2, ..., 10} and Y = min(X, 4). Find P(Y = 4).

Solution

Idea

Which x give y = 4?

x = 4, 5, ..., 10 are 7 values, so P(Y = 4) = 7/10 = 0.7.
Answer: 0.7

Q2Absolute value

Numerical Answer Type (NAT).

$X$ is uniform on $\{-5,\dots,5\}$ and $Y = |X|$ . What is $11 \times P(Y = 3)$ ?

Solution

Idea

x = 3 and x = −3.

Two values of x give y = 3, so P(Y = 3) = 2/11 and the answer is 2.
Answer: 2

Q3Try these Q1 value

Numerical Answer Type (NAT).

In Try these Q1, what is $11 \times P(Y = 0)$ ?

Solution

Idea

Count x from −5 to 0.

Six values of x collapse to 0, so P(Y = 0) = 6/11.
Answer: 6

Q4Try these Q2 value

Numerical Answer Type (NAT).

In Try these Q2, what is 1001 × P(Y = 5)?

Solution

Idea

Count x from −500 to 5.

5 − (−500) + 1 = 506 values collapse to 5.
Answer: 506

Q5Clipping facts

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements about Y = max(X, 5) for X uniform on {−500, ..., 500}.

1

Y takes 496 distinct values

2

P(Y = 5) is a little over 1/2

3

P(Y = 6) = 1/1001

4

Y is uniform

Solution

Idea

Range 5, ..., 500 has 496 values.

Y is not uniform: one value has probability 506/1001 and the rest 1/1001.
Answer: options 1, 2 and 3.

Next · RevisionFunctions of one variable: Revision Sheet

16

Revision

Functions of one variable: Revision Sheet

One page to revise everything in this section.

The big picture

Stem plots

A stem of height $P(X=x)$ at each value $x$ .
Uniform: flat stems. Binomial(10, 0.5): bell-shaped, tallest at 5.

Table method

Columns $x$ , $P(X=x)$ , $y = g(x)$ ; group rows by $y$ .
$P(Y=y) = \sum_{x:\,g(x)=y} P(X=x)$ .

One-to-one

Same probabilities, new locations: $P(Y=g(x)) = P(X=x)$ .
$X - 5$ slides the stems; $2^X$ spreads them unevenly.

Many-to-one

Probabilities pile up and the shape can change.
$(X-5)^2$ : uniform becomes 1/11, 2/11, ...; binomial mode moves to 1.

Clipping

$\max(X, c)$ : every value beyond the cut collapses onto one value.
Q1: 6/11 at 0. Q2: 506/1001 at 5.

Revision sheet

Idea	Key point
Stem plot	A stem of height $P(X=x)$ at each value $x$
Table method	Columns $x,\ P(X=x),\ g(x)$ ; group rows by $g(x)$
General formula	$P(Y=y) = \sum_{x:\,g(x)=y} P(X=x)$
One-to-one	Same probabilities, new locations: $P(Y=g(x)) = P(X=x)$
Many-to-one	Probabilities pile up; the shape of the PMF can change
Clipping, e.g. $\max(X,c)$	All values beyond the cut collapse onto one value

Revision overview

See a PMFStem plots
Uniform is flat
Binomial is a bell

Table methodx, P(X=x), g(x)
Group by y
Add probabilities

CheckNew PMF adds to 1
Every row used once

Functions of one variableWeek 2

One-to-oneX − 5 slides
2^X stretches
Probabilities unchanged

Many-to-one(X − 5)² folds
Pairs add up
Shape changes

Clippingmax(X, 0): 6/11 at 0
max(X, 5): 506/1001

Revision: functions of one random variable in one view

Common mistakes

Merging rows wrongly

Add only the rows with the same y. Check that every row of X is used exactly once.

Changing probabilities for one-to-one g

They stay the same; only the locations move.

Doubling the stem at y = 0

In (X − 5)², y = 0 comes from the single value x = 5.

Off-by-one when counting

From a to b inclusive there are b − a + 1 values (506, not 505).

Assuming uniform stays uniform

A many-to-one function can make the output uneven, as in 1/11, 2/11, ...

Forgetting to check the total

The final PMF must add up to 1; if not, a row was missed or counted twice.

Next: Chapters 17 to 35 (functions of two and of several variables, contours, sums and convolution) are in the second note, Week 2 · Part 2 of 2.

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Statistics forData Science 2

Chapters

From Independent Events to Independent Random Variables

1.1Recap: independent events

1.2The definition

1.3Compare with the factorisation that always holds

1.4Going deeper

1.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Checking Independence From a Joint PMF Table

2.1How to read the table

2.2Example 2: dependent

2.3The conditional view

2.4Going deeper

2.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Independence in a Real Setting: A Random 2-Digit Number

3.1The experiment

3.2The whole joint table, by counting

3.3Related pairs, to build intuition

3.4Independence in real modelling

3.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Independent vs Dependent: The Working Rules

4.1The working rules

4.2The fastest dependence check: a zero cell

4.3A visual shortcut: the rows

4.4Try these: worked solutions

4.5Marginals never decide it

4.6Exam tips and common mistakes

Practice questions (graded-assignment style)

Independence: Revision Sheet

The big picture

Revision sheet

Week 2 roadmap

Common mistakes

Independence for n Random Variables

6.1Definition for n random variables

6.2Example: three tosses of a fair coin

6.3Going deeper

6.4Exam tips and common mistakes

Practice questions (graded-assignment style)

A Random 3-Digit Number

7.1The experiment

7.2Why one dependent pair is enough

7.3Going deeper

7.4Exam tips and common mistakes

Practice questions (graded-assignment style)

Even Parity: Pairwise Independent, Not Independent

8.1The even parity PMF

8.2The direction that does hold

8.3Going deeper

8.4Exam tips and common mistakes

Practice questions (graded-assignment style)

Useful Consequences and Try These

9.1Sub-collections of independent variables

9.2Functions of independent variables

9.3i.i.d. variables

9.4Try these: worked solutions

9.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Many variables: Revision Sheet

The big picture

Revision sheet

Common mistakes

Seeing a PMF: Stem Plots

11.1Stem plots

11.2Two reference distributions

11.3Reading a stem plot

11.4Going deeper

11.5Exam tips and common mistakes

Practice questions (graded-assignment style)

The Table Method

12.1The method

12.2A small worked example

12.3Going deeper

12.4Exam tips and common mistakes

Practice questions (graded-assignment style)

One-to-One Functions

13.1Example: Y = X − 5

Statistics for
Data Science 2