Question:
If $X\sim \mathcal{N}(0,1)$ and we define $Y$ such that $$ Y = \begin{cases} X& \text{ if }|X|<a \\ -X& \text{ if }|X|\geq a. \end{cases} $$ Show that $Y\sim \mathcal{N}(0,1)$ but $(X,Y)$ is not bivariate normal.
I'm mainly unsure on how to demonstrate the first part of the problem. I think that all $Y$ does is flip the tails of $X$ but my attempts in rigorously proving that $Y\sim \mathcal{N}(0,1)$ have been fruitless. I've tried computing the moment generating function of $Y$ but I don't know how deal with the limits of the resultant integral; where I wish to split it up in such a way as to recover the mgf of $X$.
The second part of the problem I think I've solved. I considered the distribution $X-Y$ which is clearly defined by $$ \begin{cases} 0 &\text{if }|X|<a, \\ 2X &\text{if }|X|\geq a. \end{cases} $$ Since $\mathbb{P}(|X|<a)\neq 0$, then we have that $X-Y$ cannot be continuous. Hence, they're not bivariate normal.
Best Answer
$P(Y \leq y)=P(X \leq y, |X| <a)+P(-X \leq y, |X| \geq a)$. Since $X$ and $-X$ have the same distribution we can change $X$ to $-X$ in the second term. So $P(Y \leq y)=P(X \leq y, |X| <a)+P(X \leq y, |X| \geq a)=P(X \leq y)$.