Joint probability distribution function : Exercises

Introduction

A pair of discrete random variables $X$ and $Y$ has a joint probability mass function in which $$ f_{XY}(x,y) = P(X=x \wedge Y=y) $$ The following exercises get you to manipulate these objects and to extract marginal distributions from joint distributions.

Example problems

Click on the problems to reveal the solution

Problem 1

Consider a pair of discrete random variables $X$ and $Y$. What formulae would you use to calculate the conditional probability $P(X=4|Y=1)$ and the conditional expectation $\mathbb{E}(X|Y=1)$

Probability $X=4$ given $Y=1$ is given by: \[ P(X=4|Y=1) = \frac{P(X=4 \wedge Y=1)}{P(Y=1)} \]
Expectation of $X$ given $Y=1$ is given by: \[ \mathbb{E}(X|Y=1) = \sum_{i=0}^\infty x_i P(X=i|Y=1) \]

Problem 2

Consider the two random variables $X$ and $Y$ with $X \in \{0,1,2\}$ and $Y \in \{1,2,4\}$. The following table gives the join probability mass for the variables (with $X$ along the top):

$f_{XY}(x,y)$ 0 1 2

1 0.1 0.1 0.0

2 0.2 0.1 0.1

4 0.0 0.2 0.2

Calculate the values of (i) $\mathbb{E}(X)$, (ii) $\mathbb{E}(Y)$ and (iii) $\textrm{cov}(X,Y)$. Then calculate a table of values for the conditional probability mass $f_{X|Y}(x,y)$ and use these conditional probabilities to calculate the values of $\mathbb{E}(X|Y)$ for $y=1,2,4$.

$$ \mathbb{E}(X) = 1 \qquad \mathbb{E}(Y) = 2.6 \qquad \textrm{cov}(X,Y) = 0.5 $$
The table of conditional probability masses is as follows:

$f_{X\|Y}(x,y)$	0	1	2
1	0.5	0.5	0.0
2	0.5	0.25	0.25
4	0.0	0.5	0.5

which ensures that the conditional expectations are: $$ \mathbb{E}(X|Y=1) = 0.5 \quad \mathbb{E}(X|Y=2) = 0.75 \quad \mathbb{E}(X|Y=4) = 1.5 $$

Problem 3

The random variables $X$ and $Y$ are independent and have probability masses given by: \[ \begin{aligned} P(X=1) &= 0.2 \quad P(X=2) = 0.2 \quad P(X=3) = 0.6 \\ P(Y=0) &= 0.4 \quad P(Y=1)=0.3 \quad P(Y=2) = 0.3 \end{aligned} \] Calculate the values of $\mathbb{E}(X)$, $\mathbb{E}(Y)$ and $\mathbb{E}(XY)$.

$$ \mathbb{E}(X) = 2.4 \qquad \mathbb{E}(Y) = 0.9 \qquad \mathbb{E}(XY)= 2.16 $$