Expectation over multiple variable? So I have come across a question asked by my peers. 
Define: $$g:=\sqrt{E[|y_r(t)|^2]}$$
Given that $$y_r(t)=\sqrt{t}\cdot h+k,$$ where $h$ and $k$ are independent random variables with variance $\sigma_h$ and $\sigma_k$ respectively.  
So what is $g$ in this case? The answer key gives: $g=\sqrt{t \sigma_h + \sigma_k}$
.

Well, I think this answer make sense. But my friend ask me, which random variables does it apply the expectation?
I am confuse to his question, I thought when you take the expectation, you take it over all the random variable. But he says this doesn't make sense to him. Now I am confused. 
 A: $\newcommand{\E}{\operatorname{E}}\newcommand{\var}{\operatorname{var}}$
It follows from independence that $\E(hk)=\E(h)\E(k)$.
If $\E(h)=\E(k) = 0$ then $\E(h^2)= \var(h)$ and $\E(k^2)=\var(k)$ and we have
\begin{align}
\E\left((\sqrt{t}\, h + k)^2\right) = \E(th^2 + 2\sqrt t\,hk + k^2) & = t\E(h^2) + 2\sqrt t\E(h)\E(k) + \E(k^2) \\[10pt]
& = t\var(h) + 2\cdot0\cdot0 + \var(k).
\end{align}
Therefore the answer key is right provided $\E(h)=\E(k)=0$.
A: Building on Michael Hardy's answer. I use the notation $\bar h = E(h)$ and $\bar k = E(k)$ for readability. Since $h,k$ are independent, $E(hk) = \bar h \bar k$.
By definition, $E(h^2) = \sigma_h + \bar h^2$ and $E(k^2) = \sigma_k + \bar k^2$. We have,
$$\begin{align*}E((\sqrt{t}h+k)^2)& =E(th^2+2\sqrt{t}hk + k^2)\\ &= tE(h^2)+2\sqrt{t}E(hk)+E(k^2)\\ &= t(\sigma_h + \bar h^2)+2\sqrt{t}\bar h \bar k + \sigma_k + \bar k^2\\ &= (t\sigma_h + \sigma_k) + (\sqrt{t} \bar h + \bar k)^2 \,\text{.}\end{align*}$$
Therefore the answer key is right iff $\sqrt{t}E(h)+E(k)=E(y_r(t)) = 0$.
The expectation is with respect to the product variable $(h,k)$, the distribution of which is determined by those of $h$ and $k$, since they are independent.
