Question

Let $X_{1}$ and $X_{2}$ be i.i.
D. Bernoulli (p), $0

• A. $(2-p)(1-p)^2$
• B. $p(2+p)(1-p)^2$
• C. $p^2(1-p^2)$
• D. $p(2-p)(1-p)^2$

Hint:

The maximum of Bernoulli variables is 1 if at least one variable is 1. Its success probability is the complement of "all are zeros," i.e., $1 - q^n$. The variance follows the standard $p'q'$ formula.

Answer 1

The Correct Option is D

Let $Y = \max(X_1, X_2)$. Since $X_1, X_2$ are Bernoulli random variables, they take values $0$ or $1$. Consequently, $Y$ also takes only values $0$ or $1$, making $Y$ a Bernoulli random variable itself.

Step 1: \color{redFind the Probability Mass Function of Y
$Y = 0$ if and only if both $X_1 = 0$ and $X_2 = 0$.
Since $X_1, X_2$ are i.i.d.:
$P(Y = 0) = P(X_1 = 0) \cdot P(X_2 = 0) = (1-p) \cdot (1-p) = (1-p)^2$.

Step 2: \color{redDetermine the Success Probability for Y
Let $P(Y=1) = p'$.
$p' = 1 - P(Y=0) = 1 - (1-p)^2$.
Expanding this:
$p' = 1 - (1 - 2p + p^2) = 2p - p^2 = p(2-p)$.

Step 3: \color{redCalculate Variance of a Bernoulli Variable
For any Bernoulli random variable with success probability $p'$, the variance is $p'(1-p')$.
Here, $p' = p(2-p)$ and $1-p' = (1-p)^2$.

Step 4: \color{redAssemble the Final Expression
$Var(Y) = p'(1-p')$
$Var(Y) = [p(2-p)] \cdot [(1-p)^2]$
$Var(Y) = p(2-p)(1-p)^2$.
This matches Option (4).