Question

Let \(X\) and \(Y\) be two independent discrete random variables such that the moment generating functions of \(X\) and \(X+Y\) are given by \[ M_X(t)=\frac{1+2e^{-t}+3e^{2t}}{6},\quad t\in\mathbb{R}, \] and \[ M_{X+Y}(t)=\frac{2+e^{t}+3e^{3t}}{6},\quad t\in\mathbb{R}, \] respectively. Then which of the following statements is correct?

• A. \(P(XY=0)=\dfrac{1}{6}\)
• B. \(P(XY=2)=\dfrac{1}{6}\)
• C. \(E(X)=0\)
• D. \(\mathrm{Var}(Y)=1\)

Hint:

Read the distribution of X directly off its MGF, then divide the MGF of X+Y by the MGF of X to find the MGF of Y. A constant random variable equal to c has MGF e^{ct}.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept.
A moment generating function of the form \(M(t)=\sum_i p_i e^{x_i t}\) directly reveals the distribution of a discrete random variable: each exponent \(x_i\) is a value the variable can take, and its coefficient \(p_i\) is the probability of that value. We first read off the distribution of \(X\), then use independence to recover the distribution of \(Y\) from \(M_{X+Y}=M_X\cdot M_Y\).

Step 2: Read off the distribution of \(X\).
\[ M_X(t)=\frac{1+2e^{-t}+3e^{2t}}{6}=\frac{1}{6}e^{0\cdot t}+\frac{2}{6}e^{(-1)t}+\frac{3}{6}e^{2t} \]
So \(X\) takes the values \(-1,0,2\) with probabilities \[ P(X=-1)=\frac{2}{6}=\frac{1}{3},\quad P(X=0)=\frac{1}{6},\quad P(X=2)=\frac{3}{6}=\frac{1}{2} \] (These add to \(1\), as a check.)

Step 3: Use independence to find \(M_Y(t)\).
Since \(X,Y\) are independent, \(M_{X+Y}(t)=M_X(t)\,M_Y(t)\), so \[ M_Y(t)=\frac{M_{X+Y}(t)}{M_X(t)}=\frac{2+e^{t}+3e^{3t}}{1+2e^{-t}+3e^{2t}} \]

Step 4: Guess and verify a simple form for \(Y\).
Try \(Y=1\) with probability \(1\) (a constant/degenerate random variable), so \(M_Y(t)=e^{t}\). Then \[ M_X(t)\cdot e^{t}=\frac{e^{t}+2+3e^{3t}}{6}=\frac{2+e^{t}+3e^{3t}}{6}=M_{X+Y}(t) \] This matches the given \(M_{X+Y}(t)\) exactly, so \(Y=1\) with probability \(1\).

Step 5: Compute \(XY\) and check option (A).
Since \(Y=1\) always, \(XY=X\). So \[ P(XY=0)=P(X=0)=\frac{1}{6} \] This matches option (A) exactly.

Step 6: Check option (B).
\[ P(XY=2)=P(X=2)=\frac{1}{2}\neq\frac{1}{6} \] So (B) is FALSE.

Step 7: Check option (C).
\[ E(X)=(-1)\left(\frac{1}{3}\right)+0\left(\frac{1}{6}\right)+2\left(\frac{1}{2}\right)=-\frac{1}{3}+0+1=\frac{2}{3}\neq0 \] So (C) is FALSE.

Step 8: Check option (D).
Since \(Y\) always equals \(1\), it has no spread at all, so \(\mathrm{Var}(Y)=0\neq1\). So (D) is FALSE.

Final Answer:
\(P(XY=0)=\dfrac{1}{6}\) is the correct statement. \[ \boxed{P(XY=0)=\tfrac{1}{6}} \]