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: Simplify the factor \((2X-1)\) first.
Let \(Z = 2X - 1\). Since \(X\) is Bernoulli, it only takes the values 0 or 1, so \(Z\) only takes the values \(2(0)-1=-1\) (when \(X=0\), probability 0.7) or \(2(1)-1=1\) (when \(X=1\), probability 0.3).
The key fact here is that \(Z\) only ever equals \(-1\) or \(1\), so \(Z^{2}=1\) with probability 1, no matter what \(X\) turns out to be.

Step 2: Find the mean of \(Z\) and confirm independence from \(Y\).
\(E[Z] = 2E[X] - 1 = 2(0.3) - 1 = -0.4\).
Since \(X\) and \(Y\) are independent, any function of \(X\) (such as \(Z\)) is also independent of \(Y\).

Step 3: Find the mean of \(W = ZY\).
For independent variables, \(E[ZY] = E[Z]E[Y]\).
Since \(Y\) is Normal with mean 0, \(E[Y]=0\), so \(E[W] = (-0.4)(0) = 0\).

Step 4: Find \(E[W^{2}]\).
\[ E[W^{2}] = E[Z^{2}Y^{2}] = E[Z^{2}]\,E[Y^{2}] \] (using independence again, this time between \(Z^{2}\) and \(Y^{2}\)).
We already found \(Z^{2}=1\) always, so \(E[Z^{2}]=1\).
For \(Y\), \(E[Y^{2}] = \text{Var}(Y) + (E[Y])^{2} = 100 + 0 = 100\).
So \(E[W^{2}] = 1 \times 100 = 100\).

Step 5: Compute the variance.
\[ \text{Var}(W) = E[W^{2}] - (E[W])^{2} = 100 - 0^{2} = 100 \]

Final Answer:
The variance of \((2X-1)Y\) is 100, option (A). The other options come from partial or incorrect use of the variance-of-a-product rule; none of them account correctly for the fact that \((2X-1)^{2}\) is identically 1.\[ \boxed{100} \]