# Variance and identical random variables

If $X,Y$ and $Z$ have identical distributions and are independent, and can assume $0$ or $1$, what is $\text{Var}(XYZ)$? So this is either $0$ or $1$. It's $1$ with probability $1/8$ and $0$ with probability $7/8$. This means that $\text{Var}(XYZ) = 7/16$? Or is this not right? How does this use the definition of variance? We could note that $E(X) = E(X^2)$, etc....?

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$Var(X) = E[X^2] - E[X]^2$. Assuming each of the variables take 0 or 1 equally likely, as you noted, $E[XYZ] = E[X^2Y^2Z^2] = 1/8$. So $Var(X) = 1/8 - 1/64 = 7/64$.
You imply that $X, Y$, and $Z$ have equal chance of being 0 or 1. In that case $E((XYZ)^2)=\frac{1}{8}$ and $(E(XYZ))^2=\frac{1}{64}$. So the variance would be $\frac{7}{64}$