$X\cos\beta+Y\sin\beta\sim N(0,1)$, show $X$ and $Y$ are independent $N(0,1)$ random variable Question Let $(X,Y)$ be a random point drawn from a two-dimensional distribution. Suppose that $X\cosβ+Y\sinβ\sim N(0,1)$ for any $β∈ \mathbb{R}$. Show that $X$ and $Y$ are independent $N(0,1)$ random variables.

Attempt to Solution Let $Z=X\cosβ+Y\sinβ\sim N(0,1)$, so using mgf, 
\begin{align}
M_Z(t) & = \exp(1/2(x^2\cos^2\beta+Y^2\sin^2\beta)) \\[8pt]
& = \exp\left(\frac{x^2\cos^2\beta}{2}\right) \exp \left( \frac{y^2 \sin^2 \beta}{2}\right) \\[8pt]
&=M_X(s)M_Y(t), \\[8pt]
\text{and } X & \sim N(0,\cos^2\beta), \quad Y\sim N(0,\sin^2\beta).
\end{align} 
My confusion 


*

*Is my method correct?  

*Can I conclude $X\sim N(0,1), Y\sim N(0,1)$,
since $\beta \in \mathbb{R}$? 

 A: We just need to show that $\text{Cov} (X,Y)=0$. To do this, note that when $\beta = \pi/4$, we have $\frac 1{\sqrt 2} (X+Y)\sim N(0,1)$, so $E[(X+Y)^2/2]=1, E(X+Y)^2=2$.
$$
\text{Cov} (X,Y)=E(XY)-E(X)E(Y)\\
=\frac12 E[(X+Y)^2-X^2-Y^2]-0\times 0\\
=\frac{1}{2}\times (2-1-1)=0.
$$
So, they are independent.
Note I have used the fact that $E(X)=E(Y)=0, E(X^2)=E(Y^2)=1$.
A: I add little to Jethro's answer (which I do not understand exactly why was downvoted), but I want to put some order to the discussion in comments.
From the hypothesis we can conclude the following.


*

*Taking $\beta = 0$ and $\beta = \frac{\pi}2$ leads to $X\sim N(0,1)$, and $Y\sim N(0,1)$, respectively.

*For any $a,b\in \Bbb R$, $aX+bY \sim N(0,a^2+b^2)$. In fact we have $$aX+
bY=\sqrt{a^2+b^2}(X\cos\beta + Y\sin \beta)=\sqrt{a^2+b^2}Z,$$where $$\beta=\arctan \left(\frac ba\right).$$Therefore $(X,Y)$ have a bivariate normal distribution.

*In particular $W=X+Y\sim N(0,2)$, implying $\mbox{E}\left[W^2\right]=2$. Hence the result, already shown, that $$\mbox{Cov}(X,Y)=\mbox{E}[XY]=\frac12\mbox{E}\left[W^2-X^2-Y^2\right]=0.$$Thus $X$ and $Y$ are uncorrelated.

*Uncorrelation and 2. guarantee independence. $\blacksquare$
