# Product of binomial and uniformly distributed variables

I've got a question that I really think should be quite simple to answer, but I just can't see what I'm missing. We have the random variables $X \sim R(0,1)$ and $Z\sim b(1,1/2)$. I want to determine $\operatorname{Cov}(XZ,Z)$. However, as far as I can tell, this is the same as $\operatorname{Cov}(Z,Z)=\operatorname{Var}(Z)=1/4$. The listed answer is $1/8$. What am I misunderstanding here?

• I changed \textit{~} to \sim. Thus you see $X\sim R(0,1)$ rather than $X\textit{~}R(0,1)$. ${}\qquad{}$ – Michael Hardy Aug 1 '14 at 16:35

We are presumably expected to assume that $X$ and $Z$ are independent. That really should have been explicitly specified.

We want to calculate $E(XZ^2)-E(XZ)E(Z)$.

By independence $E(XZ)E(Z)=E(X)E(Z)E(Z)=1/8$.

The first term is $E(X)E(Z^2)$. But $E(Z^2)=1/2$ because $Z^2=Z$, Thus $E(XZ^2)=1/4$.

Our covariance is therefore $1/4-1/8$.

• Yeah, this is what happens when you forget your covariance formulas! – Elswyyr Aug 1 '14 at 15:45
• You probably have used over and over again $\text{Var}(X)=E(X^2)-(E(X))^2$. The covariance formula has exactly the same structure, $\text{Cov}(XY)=E(XY)-E(X)E(Y)$, – André Nicolas Aug 1 '14 at 15:49

The $\operatorname{Cov}(XZ,Z)$ is given by $E[XZ\cdot Z]-E[XZ]E[Z]$ which is the same as $E[X]E[Z^2]-E[X]E[Z]E[Z]$. This gives us $1/2 \cdot 1/2 - 1/2 \cdot 1/4 \cdot\cdot 1/4$ which is equal to $\frac{1}{8}$.