What is an example of Var(XY) (as opposed to Var(X + Y)) I'm reading Probability: For the Enthusiastic Beginner, and he explains that Var(X + Y) = Var(X) + Var(Y) for independent variables. His example is, if you have a coin toss where H = 1 and T = 0, Var(X) is 1/4, so Var(X + Y) would be two independent coin tosses and so Var(X + Y) = 1/4 + 1/4 = 1/2. That makes sense, I can read Var(X + Y) as "X and Y". But then what does Var(XY) mean in this coin toss example? I don't see how I could multiply one coin toss by another?
 A: Why would you read $\operatorname{Var}(X+Y)$ as variance of $X$ and $Y$? Just take the notation literally: $\operatorname{Var}(X+Y)$ is variance of $X$ plus $Y$. Then, similarly, $\operatorname{Var}(XY)$ is the variance of $X$ times $Y$. To compute it, observe:


*

*$XY$ can only take on 2 values $1$ and $0$

*$XY=1$ when both $X$ and $Y$ are 1; that is, $XY=1$ iff both independent tosses turn heads, which happens with probability $\frac{1}{4}$

*$XY=0$ in the other 3 cases: $(X,Y)\in\{(1,0),(0,1),(0,0)\}$.


Thus, so $XY$ follows the Bernoulli distribution with probability $\frac{1}{4}$ of getting $1$ and probability $1-\frac{1}{4}=\frac{3}{4}$ of getting $0$. It follows that:
$$
E(XY)=\frac{1}{4},\quad\operatorname{Var}(XY)=\frac{1}{4}\frac{3}{4}=\frac{3}{16}.
$$
You can of course compute $E(XY)$ and $\operatorname{Var}(XY)$ directly (without appealing to knowledge of the Bernoulli distribution:
$$
E(XY)=\frac{1}{4}\cdot 1+\frac{3}{4}\cdot 0=\ldots,\quad E[(XY)^2]=\frac{1}{4}\cdot 1^2+\frac{3}{4}\cdot 0^2=\ldots,\\ \operatorname{Var}(XY)=E[(XY)^2]-[E(XY)]^2=\ldots
$$
