# The correlation between two normal distribution

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: $$f_Y(y)=\begin{cases} f_X(y) & \text{if }|y|<a\\[6pt] f_X(-y) & \text{if }|y|\geq a\; \end{cases}$$

• What the definition of $Y$ gives you more or less directly is the cumulative distribution function of $Y$. – André Nicolas Mar 14 '13 at 18:00
• $Y$ does indeed have a $N(0,1)$ distribution: this is a standard example (see e.g. Wikipedia) and the answer to the binormality of $(X,Y)$ can also be found there. – Dilip Sarwate Mar 14 '13 at 18:01
• Were (X,Y) binormal, any linear combination of X and Y would be normal. What about X+Y ? – Jean-Claude Arbaut Mar 14 '13 at 18:02
• I just posted an answer to this same question here a couple of days ago. – Michael Hardy Mar 18 '13 at 18:40

## 1 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.)

• Can the covariance be calculated explicitly? – JFK Mar 14 '13 at 20:13
• @JFK : I haven't actually examined that question. It can be done numerically, but I haven't worked out the details of that either. – Michael Hardy Mar 15 '13 at 17:22