Conditional Distributions and Independence
Last modified — 13 Nov 2026
By the end of this lecture, students are anticipated to be able to:
Calculate and utilize conditional distributions
Determine whether random variables are independent based on their conditional and joint distributions
Let \(Y\) be a random variable, where \(X\) is a discrete random variable. Suppose that \(\mathbb{P}(X = x) >0\) for some \(x\). Then the conditional distribution of \(Y\) given \(X = x\) assigns to each set \(A \subset {\mathbb{R}}\) the probability \[\mathbb{P}(Y \in A \ \vert\ X = x) = \frac{\mathbb{P}(Y \in A, X = x)}{\mathbb{P}(X = x)}.\]
Here, we do not specify whether \(Y\) is continuous or discrete.
If \(X\) and \(Y\) are both discrete random variables, then the conditional PMF of \(Y\) given \(X = x\) is defined by \[p_{Y|X}(y \ \vert\ x) = \frac{\mathbb{P}(Y = y, X = x)}{\mathbb{P}(X = x)} = \frac{p_{X,Y}(x,y)}{p_X(x)}.\]
Note: we can rearrange this equation to get \[p_{X,Y}(x,y) = p_{Y|X}(y \ \vert\ x)p_X(x)\]
The joint PMF of \(W\) and \(V\) is
| \(W\ \backslash\ V\) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 1/36 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 2 | 0 | 2/36 | 1/36 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 3 | 0 | 0 | 2/36 | 2/36 | 1/36 | 0 | 0 | 0 | 0 | 0 | 0 |
| 4 | 0 | 0 | 0 | 2/36 | 2/36 | 2/36 | 1/36 | 0 | 0 | 0 | 0 |
| 5 | 0 | 0 | 0 | 0 | 2/36 | 2/36 | 2/36 | 2/36 | 1/36 | 0 | 0 |
| 6 | 0 | 0 | 0 | 0 | 0 | 2/36 | 2/36 | 2/36 | 2/36 | 2/36 | 1/36 |
To condition on \(W\), we look at a particular row, while to condition on \(V\), we look at a particular column. Then renormalize by the sum of the row/column.
| \(W\ \backslash\ V\) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 1/36 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 2 | 0 | 2/36 | 1/36 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 3 | 0 | 0 | 2/36 | 2/36 | 1/36 | 0 | 0 | 0 | 0 | 0 | 0 |
| 4 | 0 | 0 | 0 | 2/36 | 2/36 | 2/36 | 1/36 | 0 | 0 | 0 | 0 |
| 5 | 0 | 0 | 0 | 0 | 2/36 | 2/36 | 2/36 | 2/36 | 1/36 | 0 | 0 |
| 6 | 0 | 0 | 0 | 0 | 0 | 2/36 | 2/36 | 2/36 | 2/36 | 2/36 | 1/36 |
Given the above table, what is \(\mathbb{P}(W = 5 \ \vert\ V = 8)\)
If \(X\) and \(Y\) are jointly absolutely continuous random variables, then the conditional density of \(Y\) given \(X = x\), is the function \[f_{Y|X}(y \ \vert\ x) = \frac{f_{X,Y}(x,y)}{f_X(x)},\] valid for any \(y\in{\mathbb{R}}\), and all \(x\) such that \(f_X(x) > 0\).
Note: we can rearrange this equation to get \[f_{X,Y}(x,y) = f_{Y|X}(y \ \vert\ x)f_X(x)\]
Let \(X\) and \(Y\) be jointly absolutely continuous random variables with joint PDF \(f_{X,Y}\). The conditional distribution of \(Y\) given \(X = x\) assigns to each set \(A \subset {\mathbb{R}}\) the probability \[\mathbb{P}(a \le Y \le b \ \vert\ X = x) = \int_a^b f_{Y|X}(y\ \vert\ x)\, \mathsf{d}y,\] valid for all \(x\) such that \(f_X(x) > 0\).
Let \(X\) and \(Y\) be jointly continuous random variables with joint PDF \[f_{X,Y}(x,y) = \frac{1}{x} e^{-x} I_{\{0 \le y \le x\}}(x,y).\]
Let \(X\) and \(Y\) be independent with \(X \sim {\mathrm{Exp}}(1)\) and \(Y\sim {\mathrm{Exp}}(2)\). Let \(S = X + Y\). Find the conditional density \(f_{X|S}(X = x | S = s)\)
Recall that if \(X\) and \(Y\) are independent, we have: \(\mathbb{P}(Y \in A \ \vert\ X = x) = P(Y \in A)\) This implies:
\[p_{Y|X}(y \ \vert\ x) = p_Y(y)\] and \[f_{Y|X}(y \ \vert\ x) = f_Y(y)\] when \(X\) and \(Y\) are independent.
You may have:
HANDWRITTEN Cheat Sheet
Non-graphing, non-programmable calculator
Writing materials
Phones and laptops put away, please! You can have 20 minutes.
No class Monday November 9. No checkpoint on Friday, either!
Midterm 2 is on Friday November 20
Stat 302 - Winter 2025/26