Joint Distributions
Last modified — 13 Nov 2026
By the end of this lecture, students are anticipated to be able to:
The joint CDF of the two random variables \(X\) and \(Y\) is defined by \[F_{X, Y} ( a, b ) = \mathbb{P}( X \le a, \, \ Y \le b).\]
Recall: \[\left\{ X \le a \, , \ Y \le b \right\} = \left\{ X \le a \right\} \cap \left\{ Y \le b \right\}.\]
Let \(X\) and \(Y\) be two random variables with joint CDF \(F_{X, Y}(x, y)\).
If \(X\) and \(Y\) are both discrete, then the joint PMF of \((X, Y)\) is defined by \[p_{X, Y} ( a, b ) = \mathbb{P}( X = a, \, Y = b).\]
The random variables \(X\) and \(Y\) are jointly (absolutely) continuous if there exists a density function \(f_{X, Y}(x, y)\) such that for any set \(A \subset \mathbb{R}^2\), we have \[\mathbb{P}\left( (X, Y) \in A \right) = \iint_A f_{X, Y}(x, y) \mathsf{d}x \mathsf{d}y.\]
Find the joint PMF of \(X\) and \(Y\).
This will involve finding all of the joint probabilities for each combination of the dice:
| \(f_{X,Y}(x, y)\) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 1/36 | 2/36 | 2/36 | 2/36 | 2/36 | 2/36 |
| 2 | 0 | 1/36 | 2/36 | 2/36 | 2/36 | 2/36 |
| 3 | 0 | 0 | 1/36 | 2/36 | 2/36 | 2/36 |
| 4 | 0 | 0 | 0 | 1/36 | 2/36 | 2/36 |
| 5 | 0 | 0 | 0 | 0 | 1/36 | 2/36 |
| 6 | 0 | 0 | 0 | 0 | 0 | 1/36 |
Suppose \(\mathbb{P}(X=x, Y=y) = c · (x + 2y + 1)I_{\{0,1,2\}}(x)I_{\{0,1\}}(y)\)
What value of c will make this a valid joint PMF?
What is \(\mathbb{P}(X = 0, Y = 1)\)?
Let \(X\) and \(Y\) be jointly continuous, with joint density function
\[ f(x,y) = \begin{cases} 4x^2y + 2y^5 & 0\le x \le 1, 0\le y \le 1\\ 0 &\text{otherwise} \end{cases} \]
Verify that this is a density function.
What is \(\mathbb{P}(0 \le X \le 0.5, 0 \le y \le 0.75)\)?
The joint PDF is a surface:
\[ f(x,y) = \begin{cases} 4x^2y + 2y^5 & 0\le x \le 1, 0\le y \le 1\\ 0 &\text{otherwise} \end{cases} \]
Bivariate Normal Distribution: Given \(\mu _1, \mu_2, \sigma_1, \sigma_2, \rho \in {\mathbb{R}}\) where \(\sigma_1, \sigma_2 > 0\) and \(-1\le \rho \le 1\), the bivariate normal distribution \(\mathcal{N}(\mu,\mu_2, \sigma_1, \sigma_2)\), is given by:
\[\begin{aligned} &f_{X_1, X_2}(x_1, x_2)\\ &= \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1 - \rho^2}} \exp\left\{ -\frac{1}{2(1 - \rho^2)} \left(\left(\frac{x_1-\mu_1}{\sigma_1}\right)^2 - 2\rho \frac{(x_1-\mu_1)(x_2-\mu_2)}{\sigma_1\sigma_2} + \left(\frac{x_2-\mu_2}{\sigma_2}\right)^2 \right) \right\}. \end{aligned}\]
When \(\mu_1 = \mu_2 = \rho = 0\) and \(\sigma_1 = \sigma_2 = 1\), this is the standard bivariate normal distribution.

Let \(f_{X,Y}(x,y) = e^{-x-y}I_{[0, \infty)}(x)I_{[0, \infty)}(y)\)
What is the probability \(\mathbb{P}(X < Y)\)?
You may have:
HANDWRITTEN Cheat Sheet
Non-graphing calculator
Writing materials
Phones and laptops put away, please! You can have 15 minutes.
Stat 302 - Winter 2025/26