Question

Let $X_{1}, X_{2}$ be independent random variables each from a discrete probability mass function $P_{X}(x) = \begin{cases} 1/3 & \text{if } x = 0 \\ 2/3 & \text{if } x = 1 \end{cases}, i = 1, 2$. Then the moment generating function of $Y = X_{1} \cdot X_{2}$ is

• A. $\frac{4}{9} + \frac{5}{9}e^{t}$
• B. $\frac{5}{9} + \frac{4}{9}e^{t}$
• C. $\frac{5}{9} + \frac{4}{9}e^{2t}$
• D. $(\frac{1}{3} + \frac{2}{3}e^{t})^{2}$

Hint:

For binary variables $\{0, 1\}$, the product $X_1 X_2$ behaves like an AND gate in logic. It is only 1 when both inputs are 1. Note that Option (4) is the MGF of the sum $X_1 + X_2$, which is a common trap in these types of questions.

Answer 1

The Correct Option is B

To find the Moment Generating Function (MGF) of the product $Y = X_1 X_2$, we must first determine the probability distribution of $Y$.

Step 1: \color{redDetermine the possible values of Y
Since $X_1$ and $X_2$ can only take values 0 or 1:
- If $X_{1}=0, X_{2}=0 \implies Y = 0 \cdot 0 = 0$.
- If $X_{1}=0, X_{2}=1 \implies Y = 0 \cdot 1 = 0$.
- If $X_{1}=1, X_{2}=0 \implies Y = 1 \cdot 0 = 0$.
- If $X_{1}=1, X_{2}=1 \implies Y = 1 \cdot 1 = 1$.
Thus, $Y$ is a Bernoulli-like random variable taking values $\{0, 1\}$.

Step 2: \color{redCalculate the probability $P(Y=1)$
Because $X_1$ and $X_2$ are independent, the probability of their intersection is the product of their individual probabilities.
$P(Y=1) = P(X_{1}=1 \cap X_{2}=1) = P(X_{1}=1) \cdot P(X_{2}=1)$.
$P(Y=1) = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}$.

Step 3: \color{redCalculate the probability $P(Y=0)$
Since the total probability must sum to 1:
$P(Y=0) = 1 - P(Y=1)$.
$P(Y=0) = 1 - \frac{4}{9} = \frac{5}{9}$.

Step 4: \color{redApply the MGF formula
The MGF of a discrete random variable is $M_{Y}(t) = E[e^{tY}] = \sum e^{ty} P(Y=y)$.
$M_{Y}(t) = e^{t(0)} P(Y=0) + e^{t(1)} P(Y=1)$.
$M_{Y}(t) = 1 \cdot (\frac{5}{9}) + e^{t} \cdot (\frac{4}{9})$.
$M_{Y}(t) = \frac{5}{9} + \frac{4}{9} e^{t}$.
This matches Option (2).