# Law of total covariance for products of random variables

I have two sets of random variables.

$X_i \sim N( \theta, \sigma^2 )$, where $X_i$ are i.i.d.

$Z_j$ which are simply iid binary random variables with success probability $p$.

I want to find $\text{Cov}( X_i Z_j, X_i Z_k )$, $j \neq k$. I tried using the law of total covariance but not sure if my reasoning is valid:

\begin{align} \text{Cov}( X_i Z_j, X_i Z_k ) & = \text{Cov}( \text{E}( X_i Z_j \mid Z ), \text{E}( X_i Z_k \mid Z ) ) + \text{E}( \text{Cov}( X_i Z_j, X_i Z_k \mid Z ) ) \\ & = \theta^2 \text{Cov}( Z_j, Z_k ) + \sigma^2 \text{E}( Z_j Z_k ) \\ & = \sigma^2 p^2 \end{align}

I wasn't sure if it's okay to treat $Z_j$ and $Z_k$ as constants and pull them out in the step $\text{E}( \text{Cov}(X_i Z_j, X_i Z_k \mid Z ) )$, since we're conditioning on them.

• Are the $X$s and $Z$s correlated? – Greenparker Mar 24 '16 at 12:50
• No they can be assumed independent – Jack Mar 24 '16 at 13:35

Your answer is correct, but there is an easier way to do the problem with just the definition of Cov.

\begin{align*} Cov(X_i Z_j,X_iZ_k) & = E(X_i^2Z_jZ_k) - E(X_iZ_j) E(X_iZ_k)\\ & = E(X_i^2)E(Z_j)E(Z_k) - E(X_i)^2E(Z_j)E(Z_k)\\ & = E(Z_j)E(Z_k) \left(E(X_i^2) - E(X_i)^2 \right)\\ & = p^2 \sigma^2. \end{align*}