Bivariate Discrete Random Variables
Data Science and A.I. Lecture Series
By Bindeshwar Singh Kushwaha, PostNetwork Academy
Definition
Let \( X \) and \( Y \) be two discrete random variables defined on the sample space \( S \) of a random experiment. Then, the function \( (X, Y) \) defined on the same sample space is called a two-dimensional discrete random variable.
- \( (X, Y) \) is a two-dimensional random variable if its possible values are finite or countably infinite.
- Each value of \( X \) and \( Y \) is represented as a point \( (x, y) \) in the XY-plane.
- Example: Consider placing three balls \( b_1, b_2, b_3 \) randomly in three cells.
- The number of balls in a cell and the number of occupied cells form discrete random variables.
Possible Outcomes of Placing Three Balls in Three Cells
\[ X (\text{Number of balls in Cell 1}) \in \{0,1,2,3\} \]
\[ Y (\text{Number of occupied cells}) \in \{1,2,3\} \]
| Arrangement | Cell 1 | Cell 2 | Cell 3 |
|---|---|---|---|
| 1 | \( b_1 \) | \( b_2 \) | \( b_3 \) |
| 2 | \( b_1 \) | \( b_3 \) | \( b_2 \) |
| 3 | \( b_2 \) | \( b_1 \) | \( b_3 \) |
| 4 | \( b_2 \) | \( b_3 \) | \( b_1 \) |
| 5 | \( b_3 \) | \( b_1 \) | \( b_2 \) |
| 6 | \( b_3 \) | \( b_2 \) | \( b_1 \) |
Joint Probability Mass Function
The joint probability mass function (PMF) \( p(x,y) \) is defined as:
\[ p(x,y) = P(X = x, Y = y) \]
| \( X \backslash Y \) | 1 | 2 | 3 | \( P(X) \) |
|---|---|---|---|---|
| 0 | \( \frac{2}{27} \) | \( \frac{6}{27} \) | 0 | \( \frac{8}{27} \) |
| 1 | 0 | \( \frac{6}{27} \) | \( \frac{6}{27} \) | \( \frac{12}{27} \) |
| 2 | 0 | \( \frac{6}{27} \) | 0 | \( \frac{6}{27} \) |
| 3 | \( \frac{1}{27} \) | 0 | 0 | \( \frac{1}{27} \) |
| \( P(Y) \) | \( \frac{3}{27} \) | \( \frac{18}{27} \) | \( \frac{6}{27} \) | 1 |
Marginal Probability Distributions
\[ P(X=x) = \sum_{y} P(X=x, Y=y) \]
\[ P(X=0) = \frac{8}{27}, \quad P(X=1) = \frac{12}{27}, \quad P(X=2) = \frac{6}{27}, \quad P(X=3) = \frac{1}{27} \]
\[ P(Y=y) = \sum_{x} P(X=x, Y=y) \]
\[ P(Y=1) = \frac{3}{27}, \quad P(Y=2) = \frac{18}{27}, \quad P(Y=3) = \frac{6}{27} \]
Conditional Probability Mass Function
The conditional probability mass function is given by:
\[ P(X=x \mid Y=y) = \frac{P(X=x, Y=y)}{P(Y=y)} \]
Example: Conditional PMF of \( X \) given \( Y=2 \)
\[ P(X=0 \mid Y=2) = \frac{6}{18} = \frac{1}{3}, \quad P(X=1 \mid Y=2) = \frac{6}{18} = \frac{1}{3}, \quad P(X=2 \mid Y=2) = \frac{6}{18} = \frac{1}{3} \]
Video
Reach PostNetwork Academy
- Website: www.postnetwork.co
- YouTube Channel: www.youtube.com/@postnetworkacademy
- Facebook Page: www.facebook.com/postnetworkacademy
- LinkedIn Page: www.linkedin.com/company/postnetworkacademy