Given $X$~$N(0,\sigma^2_X)$ and $Y$~$N(0,\sigma^2_Y)$ ($X$ and $Y$ are normally distributed random variables with expectations given by $\mu_X=\mu_Y=0$ and variances given by $\sigma^2_X$ and $\sigma^2_Y)$, can anyone help me figure out the expectation and variance of $Z=\sin(X)\cos(Y)$?
NOTE: $X$ and $Y$ are independent. I have found some threads that purport the following:
$E[\cos(X)]=e^{-\sigma^2_X/2}$
$E[\sin(X)]=0$
but i still have trouble using this info to deduce $Z=\sin(X)\cos(Y)$
I just realized the answer is zero. Ayyyyye
EDIT:
I am also curious to know…
$E[\sin X \cos X]$
where $X$~$N(0,\sigma)$. Here, the domain of $X$ is any real number, but any such number actually corresponds to an angle in radians.
Best Answer
Independence of $X$ and $Y$ implies $$\operatorname{E}[Z] = \operatorname{E}[\sin X] \operatorname{E}[\cos Y] = 0,$$ as you already observed.
The variance is more complicated. You'd need to compute $$\operatorname{Var}[Z] = \operatorname{E}[Z^2] - \operatorname{E}[Z]^2 = \operatorname{E}[Z^2] = \operatorname{E}[\sin^2 X \cos^2 Y] = \operatorname{E}[\sin^2 X] \operatorname{E}[\cos^2 Y].$$
To this end, recall that $$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}, \quad \sin^2 \theta = \frac{1 - \cos 2\theta}{2}.$$ So for instance, $$\operatorname{E}[\sin^2 X] = \frac{1 - \operatorname{E}[\cos 2X]}{2} = \frac{1 - e^{-2\sigma_X^2}}{2},$$ since $2X \sim \operatorname{Normal}(0, 4 \sigma_X^2)$. A similar result holds for $\operatorname{E}[\cos^2 Y]$.
All that is left is to demonstrate that $\operatorname{E}[\cos X] = e^{-\sigma_X^2/2}$, which you state without proof. (The fact that $\operatorname{E}[\sin X] = 0$ is an immediate consequence of the fact that $|\sin X| \le 1$ and $\sin (-X) = -\sin X$.) I recommend that you try this as an exercise.