Variance of Bernoulli when success probability varies

Question

Say the success probability $X$ is a random variable with mean $\mu$ and Variance $\sigma^2$ which takes values in $[0,1]$. How can I compute the variance of a random Variable $Y$ which is 1 with probability $X$ and 0 with probability $1-X$. So there are "two layers" of variance.

Answer 1

In general, you solve problems like this using the 'law of iterated variance'.

Let $Y|X \sim \text{Bern}(X)$ and use your stipulated prior mean and variance for the success probability $X$. Using the law of iterated variance, you get:

$$\begin{align} \mathbb{V}(Y) &= \mathbb{V}(\mathbb{E}(Y|X)) + \mathbb{E}(\mathbb{V}(Y|X)) \\[6pt] &= \mathbb{V}(X) + \mathbb{E}(X(1-X)) \\[6pt] &= \mathbb{V}(X) + \mathbb{E}(X) - \mathbb{E}(X^2) \\[6pt] &= \mathbb{E}(X^2) - \mathbb{E}(X)^2 + \mathbb{E}(X) - \mathbb{E}(X^2) \\[6pt] &= \mathbb{E}(X) - \mathbb{E}(X)^2 \\[6pt] &= \mu(1-\mu). \\[6pt] \end{align}$$

As you can see, the variance of $Y$ is determined by $\mu$, and is unaffected by $\sigma$. It turns out that this is true for all the moments of $Y$. In fact, we have $\mathbb{E}(f(Y)) = f(0) + [f(1)-f(0)] \mu$, so the expectation of any function of $Y$ is determined by $\mu$ and is unaffected by $\sigma$.