I have two independent random variables $x_1$ and $x_2$, mean $0$ and standard deviation $1$. I want to draw 1 samples from each and take the weighted average. If the weight is a fixed number, I can easily calculate the variance of the average (i.e. var($w_1$$x_1$+$w_2$$x_2$). However, now instead of always fixing the weight, I now use a distribution to determine the multiple for each variable:
Case 1: Uniform distribution, so 0 to N from $x_1$ (the renaming from $x_2$) all with the same prob
Case 2: Binomial with $p=0.5$, B*$x_1$ + (N-B)*$x_2$
Now I want to calculate the variance of the sample average under these two cases. My intuition is that it should be bigger because another layer of randomness is added. But I am having a hard time deriving the solution. Thanks!