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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:

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

Swistack
  • 802

2 Answers2

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

NN2
  • 15,892
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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.

nullUser
  • 27,877