# Prove that if X,Y∼N(0,1)X, Y \sim N(0, 1) and α∼U[0,1])\alpha \sim U[0, 1]) are independent, then Xcosα+Ysinα∼N(0,1)X\cos{\alpha} + Y\sin{\alpha} \sim N(0, 1)

Let $$X,Y∼N(0,1)X, Y \sim N(0, 1)$$ and $$α∼U[0,1])\alpha \sim U[0, 1])$$, and suppose $$X,Y,αX, Y, \alpha$$ are independent. Then how to prove that $$Xcosα+Ysinα∼N(0,1)X\cos{\alpha} + Y\sin{\alpha} \sim N(0, 1)$$?

For a constant $$α\alpha$$ the claim is obvious but in the case of $$α∼U[0,1])\alpha \sim U[0, 1])$$, the problem appears to be trickier.

I think, using the formulas for convolution of probability distributions, the following plan could work:

1. We find the distribution of $$XcosαX\cos{\alpha}$$ and $$YsinαY\sin{\alpha}$$ separately.
2. 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.

Using the characteristic function of $$Z=Xcosα+YsinαZ=X\cos \alpha+Y\sin \alpha$$, we have
ϕZ(t)=E(eit(Xcosα+Ysinα))=E(E(eit(Xcosα+Ysinα)|α))=E(E(eit(Xcosα)|α)E(eit(Ysinα)|α))=E(e−12t2cos2αe−12t2sin2α)=E(e−12t2)=e−12t2\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 $$ϕZ(t)=e−12t2\phi_Z(t) = e^{-\frac{1}{2}t^2}$$ then $$ZZ$$ follows the standard normal distribution $$N(0,1)\mathcal{N}(0,1)$$.
Remark: whatever distribution $$α\alpha$$ follows, $$ZZ$$ still follows the standard normal distribution $$N(0,1)\mathcal{N}(0,1)$$