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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!

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    $\begingroup$ 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 $\sigma^2_X\sigma^2_Y + \mu^2_X\sigma^2_Y + \sigma^2_X\mu^2_Y$ $\endgroup$
    – Henry
    Jul 13, 2021 at 20:49

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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, $$\mathrm{Var}(Z) = E(Z^2) - (E(Z))^2 = p(\sigma^2 + \mu^2) - p^2\mu^2 = p\sigma^2 + p(1 - p)\mu^2.$$

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  • $\begingroup$ Ah, sorry, I really should have thought more about this before posting. (I actually spent a while thinking, but apparently not about the right thing.) You are completely right. I will accept this answer after 7 minutes when the site allows me to $\endgroup$
    – Curious
    Jul 13, 2021 at 20:38

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