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,YN(0,1) and αU[0,1]), and suppose X,Y,α are independent. Then how to prove that Xcosα+YsinαN(0,1)?

For a constant α the claim is obvious but in the case of αU[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 Xcosα and Ysinα 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.

Answer

Using the characteristic function of Z=Xcosα+Ysinα, we have

ϕZ(t)=E(eit(Xcosα+Ysinα))=E(E(eit(Xcosα+Ysinα)|α))=E(E(eit(Xcosα)|α)E(eit(Ysinα)|α))=E(e12t2cos2αe12t2sin2α)=E(e12t2)=e12t2

Because ϕZ(t)=e12t2 then Z follows the standard normal distribution N(0,1).

Remark: whatever distribution α follows, Z still follows the standard normal distribution N(0,1)

Attribution
Source : Link , Question Author : Swistack , Answer Author : NN2

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