Question

The Moment Generating Function (MGF) of random variable $X$ is given by $M_{X}(t)=(\frac{e^{-t}+e^{t}}{2})^{3}, t\ge0$ then $P(|X|>1)$ is

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

Hint:

When an MGF is a power of a sum of exponentials like $(\sum p_i e^{t x_i})^n$, the variable is a sum of $n$ i.i.d. variables. Recognizing the Bernoulli form $\frac{e^t + e^{-t}}{2}$ (Rademacher distribution) makes these discrete probability problems much faster to solve.

Answer 1

The Correct Option is B

To find the probability $P(|X|>1)$, we first need to identify the probability distribution of the random variable $X$ from its given Moment Generating Function (MGF).

Step 1: \color{redAnalyze the MGF Structure
The given MGF is $M_{X}(t) = \left(\frac{e^{-t} + e^{t}}{2}\right)^{3}$.
This can be rewritten as:
$M_{X}(t) = \left(\frac{1}{2}e^{-1t} + \frac{1}{2}e^{1t}\right)^{3}$.
This is in the form of the MGF for a Binomial distribution or a sum of independent Bernoulli trials. Specifically, if a random variable $Z$ takes values $-1$ and $1$ with probability $1/2$ each, its MGF is $\frac{1}{2}e^{-t} + \frac{1}{2}e^{t}$.
Thus, $X$ is the sum of 3 independent random variables $Z_1, Z_2, Z_3$, where $P(Z_i = 1) = 1/2$ and $P(Z_i = -1) = 1/2$.

Step 2: \color{redDetermine the Range and Probabilities of X
The possible values for $X = Z_1 + Z_2 + Z_3$ are:
- $(-1) + (-1) + (-1) = -3$
- $(-1) + (-1) + (1) = -1$ (occurs in 3 ways: $\{-1,-1,1\}, \{-1,1,-1\}, \{1,-1,-1\}$)
- $(-1) + (1) + (1) = 1$ (occurs in 3 ways: $\{-1,1,1\}, \{1,-1,1\}, \{1,1,-1\}$)
- $(1) + (1) + (1) = 3$

Step 3: \color{redCalculate the Probability Mass Function
Using the binomial expansion $(\frac{1}{2} + \frac{1}{2})^3$:
$P(X = -3) = \binom{3}{0} (1/2)^3 = 1/8$
$P(X = -1) = \binom{3}{1} (1/2)^3 = 3/8$
$P(X = 1) = \binom{3}{2} (1/2)^3 = 3/8$
$P(X = 3) = \binom{3}{3} (1/2)^3 = 1/8$

Step 4: \color{redEvaluate P(|X| > 1)
The condition $|X| > 1$ means $X > 1$ or $X < -1$.
Looking at our set of possible values $\{-3, -1, 1, 3\}$, the values that satisfy this are $X = 3$ and $X = -3$.
$P(|X| > 1) = P(X = 3) + P(X = -3)$
$P(|X| > 1) = 1/8 + 1/8 = 2/8$
Thus, the probability is $2/8$.