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)

Question 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.

Question 2

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 $\mathcal{N}(0,1)$

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)

Answer

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