# What is the variance of the product of a Bernoulli (0,1) and a normal random variable?

I want to find the variance of the product of a Bernoulli random variable and a normal random variable. I only have an introductory probability background...and here is what I would do normally.

• If I were dealing with the product of a discrete random variable, I would just list all the cases and find their probabilities and get $$E[X], E[X^2],$$ and $$Var(X)$$ from that.
• If I were dealing with a continuous random variable, I would find the CDF, differentiate to get the pdf, and find $$E[X], E[X^2],$$ and $$Var(X)$$ from that.

But I'm not sure where to start with this case, where I have a piecewise continuous pdf. The answer to this question seems very relevant, except how would I work with a piecewise cdf like this?

I appreciate any insight!

• In general if $X$ and $Y$ are independent with means $\mu_X$ and $\mu_Y$ and variances $\sigma^2_X$ and $\sigma^2_Y$, then $XY$ has mean $\mu_X\mu_Y$ and variance $\sigma^2_X\sigma^2_Y + \mu^2_X\sigma^2_Y + \sigma^2_X\mu^2_Y$ Jul 13, 2021 at 20:49

If $$X \sim B(1, p)$$ and $$Y \sim N(\mu, \sigma^2)$$ are independent, you don't have to determine the CDF of $$Z = XY$$ for merely determining its variance. All you need is to use the expectation property $$E(X_1X_2) = E(X_1)E(X_2)$$ if $$X_1$$ and $$X_2$$ are independent.
In detail, you just need to find $$E(Z)$$ and $$E(Z^2)$$, which can be calculated as follows: \begin{align*} & E(Z) = E(XY) = E(X)E(Y) = p \times \mu = p\mu, \\ & E(Z^2) = E(X^2Y^2) = E(X^2)E(Y^2) = p \times(\sigma^2 + \mu^2) = p(\sigma^2 + \mu^2). \end{align*} Therefore, $$\mathrm{Var}(Z) = E(Z^2) - (E(Z))^2 = p(\sigma^2 + \mu^2) - p^2\mu^2 = p\sigma^2 + p(1 - p)\mu^2.$$