Question

Let $(X, Y)$ have the joint probability mass function \[ f(x, y) = \begin{cases} \binom{x + 1}{y} \binom{16}{x} \left(\frac{1}{6}\right)^y \left(\frac{5}{6}\right)^{x + 1 - y} \left(\frac{1}{2}\right)^{16}, & y = 0, 1, \ldots, x + 1; \, x = 0, 1, \ldots, 16, \\ 0, & \text{otherwise.} \end{cases} \] 

Then which of the following statements is/are TRUE? 
 

• A. $E(Y) = \dfrac{3}{2}$
• B. $\mathrm{Var}(Y) = \dfrac{49}{36}$
• C. $E(XY) = \dfrac{37}{3}$
• D. $\mathrm{Var}(X) = 3$

Hint:

For joint discrete variables, use the law of total expectation: $E(Y) = E[E(Y|X)]$, and similarly for mixed terms like $E(XY)$.

Answer 1

The Correct Option is A, B

Analysis:

This can be rewritten as: $$f(x,y) = \binom{16}{x}\left(\frac{1}{2}\right)^{16} \cdot \binom{x+1}{y}\left(\frac{1}{6}\right)^y\left(\frac{5}{6}\right)^{x+1-y}$$

Step 1: Identify marginal distribution of $X$

$$P(X = x) = \sum_{y=0}^{x+1} f(x,y) = \binom{16}{x}\left(\frac{1}{2}\right)^{16} \sum_{y=0}^{x+1} \binom{x+1}{y}\left(\frac{1}{6}\right)^y\left(\frac{5}{6}\right)^{x+1-y}$$

By binomial theorem: $\sum_{y=0}^{x+1} \binom{x+1}{y}\left(\frac{1}{6}\right)^y\left(\frac{5}{6}\right)^{x+1-y} = 1$

Therefore: $$P(X = x) = \binom{16}{x}\left(\frac{1}{2}\right)^{16}$$

This means $X \sim \text{Binomial}(16, 1/2)$

Step 2: Identify conditional distribution of $Y|X$

Given $X = x$: $$P(Y = y | X = x) = \binom{x+1}{y}\left(\frac{1}{6}\right)^y\left(\frac{5}{6}\right)^{x+1-y}$$

This means $Y|X = x \sim \text{Binomial}(x+1, 1/6)$

Step 3: Calculate $E(Y)$

Using law of total expectation: $$E(Y) = E[E(Y|X)] = E\left[\frac{x+1}{6}\right] = \frac{1}{6}E(X+1) = \frac{1}{6}(E(X) + 1)$$

Since $X \sim \text{Binomial}(16, 1/2)$: $E(X) = 16 \cdot \frac{1}{2} = 8$

$$E(Y) = \frac{1}{6}(8 + 1) = \frac{9}{6} = \frac{3}{2}$$

(A) is TRUE

Step 4: Calculate $\text{Var}(Y)$

Using law of total variance: $$\text{Var}(Y) = E[\text{Var}(Y|X)] + \text{Var}[E(Y|X)]$$

$$\text{Var}(Y|X) = (x+1) \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{5(x+1)}{36}$$

$$E[\text{Var}(Y|X)] = \frac{5}{36}E(X+1) = \frac{5}{36} \cdot 9 = \frac{45}{36} = \frac{5}{4}$$

$$E(Y|X) = \frac{x+1}{6}$$

$$\text{Var}[E(Y|X)] = \text{Var}\left(\frac{X+1}{6}\right) = \frac{1}{36}\text{Var}(X) = \frac{1}{36} \cdot 4 = \frac{1}{9}$$

(Since $\text{Var}(X) = 16 \cdot \frac{1}{2} \cdot \frac{1}{2} = 4$)

$$\text{Var}(Y) = \frac{5}{4} + \frac{1}{9} = \frac{45}{36} + \frac{4}{36} = \frac{49}{36}$$

(B) is TRUE

Answer: (A) and (B) are true