Question

Consider discrete random variables \( X \) and \( Y \) with probabilities as follows:

\( P(X=0 \text{ and } Y=0) = \frac{1}{4}, \)
\( P(X=1 \text{ and } Y=0) = \frac{1}{8}, \)
\( P(X=0 \text{ and } Y=1) = \frac{1}{2}, \)
\( P(X=1 \text{ and } Y=1) = \frac{1}{8}. \)

Given \( X = 1 \), the expected value of \( Y \) is

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{8} \)
• D. \( \frac{1}{3} \)

Hint:

To compute conditional expectation \( E[Y \mid X=x] \), use the conditional probabilities derived from joint and marginal probabilities: \( E[Y \mid X=x] = \sum_y y \cdot P(Y=y \mid X=x) \).

Answer 1

The Correct Option is B

We are given the joint probabilities: \[ P(X=1, Y=0) = \frac{1}{8}, \quad P(X=1, Y=1) = \frac{1}{8} \] First, calculate the marginal probability \( P(X=1) \): \[ P(X=1) = P(X=1, Y=0) + P(X=1, Y=1) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \] Now compute the conditional probabilities: \[ P(Y=0 \mid X=1) = \frac{P(X=1, Y=0)}{P(X=1)} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2} \] \[ P(Y=1 \mid X=1) = \frac{P(X=1, Y=1)}{P(X=1)} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2} \] Now, compute the expected value of \( Y \) given \( X = 1 \): \[ E[Y \mid X = 1] = 0 \cdot P(Y=0 \mid X=1) + 1 \cdot P(Y=1 \mid X=1) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2} \]