Question

Let \(X\) and \(Y\) be two independent random variables. \(X\) follows \(Bernoulli(p=0.3)\) distribution and \(Y\) follows \(Normal(\mu=0, \sigma^{2}=100)\) distribution.

Which of the following options is the variance of \((2X-1)Y\)?

• A. 100
• B. 90
• C. 49
• D. 21

Hint:

Notice that \(2X-1\) only takes values \(-1\) or \(1\), so its square is always 1; use this with independence to find \(E[W]\) and \(E[W^{2}]\) for \(W=(2X-1)Y\).

Answer 1

The Correct Option is A

Step 1: Use the general variance formula for a product of two independent variables.
For independent $U$ and $V$: $\text{Var}(UV) = \text{Var}(U)\text{Var}(V) + \text{Var}(U)(E[V])^{2} + \text{Var}(V)(E[U])^{2}$.
Here take $U = 2X-1$ and $V = Y$.

Step 2: Find the moments of $U = 2X-1$.
$X$ is Bernoulli(0.3), so $\text{Var}(X) = p(1-p) = 0.3 \times 0.7 = 0.21$ and $E[X]=0.3$.
For a linear transform, $\text{Var}(2X-1) = 4\,\text{Var}(X) = 4(0.21) = 0.84$, and $E[2X-1] = 2(0.3)-1 = -0.4$.

Step 3: Find the moments of $V = Y$.
$Y$ is $Normal(0,100)$, so $\text{Var}(Y)=100$ and $E[Y]=0$.

Step 4: Substitute into the product-variance formula.
$$\text{Var}(UV) = \text{Var}(U)\text{Var}(V) + \text{Var}(U)(E[V])^{2} + \text{Var}(V)(E[U])^{2}$$ $$= (0.84)(100) + (0.84)(0)^{2} + (100)(-0.4)^{2}$$ $$= 84 + 0 + 100(0.16)$$ $$= 84 + 16 = 100$$

Step 5: Cross-check against the direct computation.
This matches the direct route of noting $(2X-1)^{2}=1$ always, so $E[(2X-1)^{2}Y^{2}] = E[Y^{2}] = 100$ and, since the mean of the product is 0, the variance equals this second moment, 100. Both routes agree.

Conclusion:
The variance of $(2X-1)Y$ is 100, so option (A) is correct. Getting 90, 49 or 21 instead means one of the pieces, such as $\text{Var}(X)$, $E[X]$, or $\text{Var}(Y)$, was plugged into the product-variance formula incorrectly.$$\boxed{100}$$