$X\cos\beta+Y\sin\beta\sim N(0,1)$, show $X$ and $Y$ are independent $N(0,1)$ random variable

normal distributionprobability

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

  1. Is my method correct?
  2. Can I conclude $X\sim N(0,1), Y\sim N(0,1)$,
    since $\beta \in \mathbb{R}$?

Best Answer

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.

  1. Taking $\beta = 0$ and $\beta = \frac{\pi}2$ leads to $X\sim N(0,1)$, and $Y\sim N(0,1)$, respectively.
  2. 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.
  3. 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.
  4. Uncorrelation and 2. guarantee independence. $\blacksquare$
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