Week 1.4: Multiple Random Variables (n > 2)

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# Week 1.4: Multiple Random Variables (n > 2) > **Prerequisites:** Joint PMF ([Week 1.1: Joint Probability Mass Function (Joint PMF)](/notes/01-foundation-bsma1004-stats-2-week01-01-joint-pmf)), Conditional ([Week 1.3: Conditional Distributions](/notes/01-foundation-bsma1004-stats-2-week01-03-conditional-distributio...

Week 1.4: Multiple Random Variables (n > 2)

Prerequisites: Joint PMF (Week 1.1: Joint Probability Mass Function (Joint PMF)), Conditional (Week 1.3: Conditional Distributions) Core question: How do we handle joint distributions of three or more random variables?

1. Extension to n Dimensions

Everything from two RVs extends naturally. For

X_1, \dots, X_n

with joint PMF

f_{X_1\dots X_n}

: Joint PMF:

f_{X_1\dots X_n}(t_1,\dots,t_n) = \mathbb{P}(X_1=t_1,\dots,X_n=t_n)

. Marginalisation: Sum over everything you don't want.

f_{X_1}(t) = \sum_{t_2}\cdots\sum_{t_n} f_{X_1\dots X_n}(t,t_2,\dots,t_n).

Joint marginal:

f_{X_1X_2}(t_1,t_2) = \sum_{t_3}\cdots\sum_{t_n} f_{X_1\dots X_n}(t_1,t_2,t_3,\dots,t_n)

. Chain rule factorisation:

f_{X_1\dots X_n} = f_{X_1 \mid X_2,\dots,X_n} \cdot f_{X_2 \mid X_3,\dots,X_n} \cdots f_{X_n}.

2. Example: Triple Coin Toss

$X_1$	$X_2$	$X_3$	$f_{X_1X_2X_3}$
0	0	0	1/8
0	0	1	1/8
0	1	0	1/8
0	1	1	1/8
1	0	0	1/8
1	0	1	1/8
1	1	0	1/8
1	1	1	1/8

Marginal of $X_1$ : Sum over

X_2, X_3

f_{X_1}(0) = 4 \times 1/8 = 1/2

f_{X_1}(1) = 1/2

. Joint marginal of $(X_1,X_3)$ : Sum over

X_2

$f_{X_1X_3}(0,0) = f(0,0,0) + f(0,1,0) = 1/8+1/8 = 1/4$
$f_{X_1X_3}(0,1) = f(0,0,1) + f(0,1,1) = 1/4$
$f_{X_1X_3}(1,0) = f(1,0,0) + f(1,1,0) = 1/4$
$f_{X_1X_3}(1,1) = f(1,0,1) + f(1,1,1) = 1/4$

3. Even Parity Example (Pairwise but not Mutual Independence)

X_1, X_2, X_3

each Uniform

\{0,1\}

, sum is even.

$X_1$	$X_2$	$X_3$	Prob
0	0	0	1/4
0	1	1	1/4
1	0	1	1/4
1	1	0	1/4

Any pair is independent, but the triple is dependent (not all 8 combinations possible).

4. Practice Questions

Q1 (Easy)

For the triple coin toss, find

f_{X_2 \mid X_1=1, X_3=0}(t)

Full Solution

$f_{X_1X_3}(1,0) = 1/4$ . $f_{X_1X_2X_3}(1,0,0) = 1/8$ , $f_{X_1X_2X_3}(1,1,0) = 1/8$ .

$f_{X_2 \mid X_1=1, X_3=0}(0) = \frac{1/8}{1/4} = 1/2$ , $f_{X_2 \mid X_1=1, X_3=0}(1) = \frac{1/8}{1/4} = 1/2$ .

Next topic: Week 2.1: Functions of Random Variables — Functions of random variables. Join Discord PreviousWeek 1.3: Conditional DistributionsNextWeek 2.1: Functions of Random Variables

What this note is really saying

Week 1.4: Multiple Random Variables (n > 2)

1. Extension to n Dimensions

2. Example: Triple Coin Toss

3. Even Parity Example (Pairwise but not Mutual Independence)

4. Practice Questions

Q1 (Easy)

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