Question

If $X$ and $Y$ are independent non-degenerated random variables then $Var(XY)=Var(X)\cdot Var(Y)$ iff

• A. $E(X)\ne0$ and $E(Y)\ne0$
• B. $E(X)=0$ and $E(Y)\ne0$
• C. $E(X)\ne0$ and $E(Y)=0$
• D. $E(X)=0$ and $E(Y)=0$

Hint:

Memory aid: The variance of a product is the product of the variances ONLY when both variables are centered at zero (mean = 0). If the means are non-zero, the variance of the product is always larger than the product of the variances.

Answer 1

The Correct Option is D

We need to examine the general formula for the variance of the product of two independent random variables and find the condition under which it simplifies to the product of their variances.

Step 1: \color{redApply the General Formula for Var(XY)
For two independent random variables $X$ and $Y$, the variance of their product is:
$Var(XY) = E[(XY)^2] - [E(XY)]^2$
Since $X$ and $Y$ are independent, $E[(XY)^2] = E(X^2)E(Y^2)$ and $E(XY) = E(X)E(Y)$.
$Var(XY) = E(X^2)E(Y^2) - [E(X)E(Y)]^2$

Step 2: \color{redRelate Moments to Variances
Recall that $E(X^2) = Var(X) + [E(X)]^2$ and $E(Y^2) = Var(Y) + [E(Y)]^2$.
Substitute these into the expression:
$Var(XY) = (Var(X) + [E(X)]^2)(Var(Y) + [E(Y)]^2) - [E(X)]^2[E(Y)]^2$

Step 3: \color{redExpand the Expression
$Var(XY) = Var(X)Var(Y) + Var(X)[E(Y)]^2 + Var(Y)[E(X)]^2 + [E(X)]^2[E(Y)]^2 - [E(X)]^2[E(Y)]^2$
The last two terms cancel out:
$Var(XY) = Var(X)Var(Y) + Var(X)[E(Y)]^2 + Var(Y)[E(X)]^2$

Step 4: \color{redSet the Condition
The question asks when $Var(XY) = Var(X)Var(Y)$.
This requires the extra terms to be zero:
$Var(X)[E(Y)]^2 + Var(Y)[E(X)]^2 = 0$
Since $X$ and $Y$ are non-degenerate, their variances $Var(X)$ and $Var(Y)$ are strictly positive ($> 0$).
The only way for the sum of two non-negative terms to be zero is if both terms are zero:
$Var(X)[E(Y)]^2 = 0 \implies E(Y) = 0$
$Var(Y)[E(X)]^2 = 0 \implies E(X) = 0$
Therefore, both $E(X)$ and $E(Y)$ must be $0$.