How can I calculate $\mathrm{cov}(aX,aY)$ when $X,Y$ are independent random variables distributed according to the same distribution with a mean of zero and $a$ is a constant? The following is my attempt, but is it true?
We have $$\mathrm{cov}(aX,aY)=E[a^2XY]-E[aX]E[aY]=E[a^2XY]=a^2E[XY]=a^2E[X^2]=a^2\mathrm{var}[X].$$
Can I simply take $a$ outside the expectation operation? I'm also unsure that the step $E[XY]=E[X^2]$ is correct. If $X,Y$ are independent, do we then have $E[XY]=E[X]E[Y]=0$? This just doesn't vibe well with my intuition that both random variables are being scaled by the same number $a$... shouldn't they have some correlation in that case?
$E[XY]$ does not in general equal $E[X^2]$. What we can say is that if $X$ and $Y$ are independent, then $E[XY] = E[X]E[Y]$. So, in this case, we have $$\begin{aligned} \operatorname{cov}(aX,aY) &= E[a^2XY] - E[aX]E[aY] & .\\ &= a^2 E[XY] - a^2E[X]E[Y] & \text{since $E$ is a linear operator} \\ &= a^2 E[XY] & \text{since $E[X] = E[Y] = 0$} \\ &= a^2 E[X]E[Y] & \text{since $X,Y$ are independent} \\ &= 0 & \text{since $E[X] = E[Y] = 0$} \\ \end{aligned}$$ In conclusion, if $X$ and $Y$ are uncorrelated, then so are $aX$ and $aY$.
To help with the intuition, imagine that $X$ and $Y$ are uncorrelated physical measurements of some kind, say lengths in meters. You wouldn't expect them to become correlated if you instead expressed their lengths in feet or inches (i.e. multiplied them both by a constant).