Sample mean of product of independent variables

Question

If two random variables $X$ and $Z$ are independent, then E($XZ$) = E($X$) E($Z$). But does this equation also hold for the sample mean of $X_{i} Z_{i}$ with i from 1 to n? If not, why?

Answer 1

For two independent random variables, we do have that $E(XZ) = E(X) E(Z)$ but if we have two samples, let's call them sample $x$ and sample $y$, both of size $n$, then we cannot say that $\overline{xy} = \bar{x} * \bar{y}$

Check with a simple example $x=\{1,2,3\}$ and $y=\{4,5,6\}$

We can also see why this happens by looking at the formulas for the means. It is safe to say that, in general: $$\frac1n \sum_{i=1} ^ n x_i y_i \ne \frac1n\sum_{i=1} ^ n x_i * \frac1n \sum_{i=1} ^ n y_i$$

Answer 2

I assume you are asking if

$E\left[\left(\frac{1}{n}\sum_{i=1}^{n}X_{i}\right)\left(\frac{1}{n}\sum_{i=1}^{n}Z_{i}\right)\right]=E\left[\left(\frac{1}{n}\sum_{i=1}^{n}X_{i}\right)\right]E\left[\left(\frac{1}{n}\sum_{i=1}^{n}Z_{i}\right)\right]$

The answer follows from using the fact that $\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{j=1}^{n}b_{j}\right)=\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j}$ on both sides to expand out the expressions. First pull the constants $\frac{1}{n}$ out, then on the left expand out first then take $E$ into the sum, and on the right take $E$ into each sum then expand out. For example the left expression expands out to

$\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}E\left[X_{i}\right]E\left[Z_{j}\right]$