Prove that if $X, Y \sim N(0, 1)$ and $\alpha \sim U[0, 1])$ are independent, then $X\cos{\alpha} + Y\sin{\alpha} \sim N(0, 1)$ Let $X, Y \sim N(0, 1)$ and $\alpha \sim U[0, 1])$, and suppose $X, Y, \alpha$ are independent. Then how to prove that $X\cos{\alpha} + Y\sin{\alpha} \sim N(0, 1)$?
For a constant $\alpha$ the claim is obvious but in the case of $\alpha \sim U[0, 1])$, the problem appears to be trickier.
I think, using the formulas for convolution of probability distributions, the following plan could work:

*

*We find the distribution of $X\cos{\alpha}$ and $Y\sin{\alpha}$ separately.

*We find the distribution of their sum.

But the execution of this plan would probably be overly complicated, and I feel like there must be a better way.
I would appreciate any help with this problem.
 A: Using the characteristic function of $Z=X\cos \alpha+Y\sin \alpha$, we have
\begin{align}
\phi_Z(t) &= E(e^{it(X\cos \alpha+Y\sin \alpha)}) \\
          &= E\left(E(e^{it(X\cos \alpha+Y\sin \alpha)}|\alpha) \right) \\
          &= E\left(E(e^{it(X\cos \alpha)}|\alpha)E(e^{it(Y\sin \alpha)}|\alpha) \right) \\
          &= E\left(e^{-\frac{1}{2}t^2\cos^2 \alpha}e^{-\frac{1}{2}t^2\sin^2 \alpha} \right) \\
    &= E\left(e^{-\frac{1}{2}t^2} \right) \\
&= e^{-\frac{1}{2}t^2}  \\
\end{align}
Because  $\phi_Z(t) = e^{-\frac{1}{2}t^2}$ then $Z$ follows the standard normal distribution $\mathcal{N}(0,1)$.
Remark: whatever distribution $\alpha$ follows, $Z$ still follows the standard normal distribution $\mathcal{N}(0,1)$
A: Compute the characteristic function, since we know the answer for constant $\alpha$
$$
\begin{align}
\phi(t) &= E[\exp(it(X\cos\alpha + Y \sin \alpha))]\\
&= E[E[\exp(it(X\cos\alpha + Y \sin \alpha))| \alpha] ]\\
&= E[\exp(-\frac{1}{2}t^2)]\\
&=\exp(-\frac{1}{2}t^2)
\end{align}
$$
and we are done.
