Let $X$ have the $N(0,1)$ distribution and let $a>0$, show that the random variable $Y$ given by
$$Y=\begin{cases}
X & \text{if }|X|<a\\[5pt]
-X &\text{if }|X|\geq a\;
\end{cases}$$
has the $N(0,1)$ distribution. What is cov$(X,Y)$?
Comments: Does that mean $Y$ has pdf:
$$f_Y(y)=\begin{cases}
f_X(y) & \text{if }|y|<a\\[6pt]
f_X(-y) & \text{if }|y|\geq a\;
\end{cases}$$
Best Answer
The pair $(X,Y)$ is not bivariate normal since $X+Y$ has positive probability of being equal to $0$, but is not always equal to $0$, so it's not normally distributed. (It's equal to $0$ whenever $|X|>a$. Besides, just look at the graph of $Y$ as a function of $X$ and you see that it's constrained to lie on an odd-shaped one-dimensional figure that is not a straight line.
Notice that $f_X(-y)$ is the same as $f_X(y)$ because of the symmetry of the distribution of $X$. That is indeed the density function of $Y$, and that tells you that the distribution of $Y$ is $N(0,1)$. So they're separately normal but not jointly normal.
The covariance depends on $a$ in a way that becomes clearer if you think about what happens to the correlation when $a$ is very close to $0$ and when $a$ is very very big. You see that in one case you get positive correlation and in the other you get negative correlation, so somewhere in between you get $0$ correlation. This shows that a pair of random variables can be separately normally distributied with covariance $0$ without being independent. (If they were jointly normally distributed and uncorrelated, they'd be independent.)