# Thread: Show standard normal variables

1. ## Show standard normal variables

Hi

If $\displaystyle X$ and $\displaystyle Y$ has joint density function

$\displaystyle f_{X,Y}(x,y)=\begin{cases} \frac{2|x|+y}{\sqrt{2\pi}}e^{-\frac{(2|x|+y)^2}{2}}, y \geq -|x| \\ 0 , y < -|x| \end{cases}$

,how do I prove X and Y are standard normal variables?

Also show that they are not independent, but uncorrelated.

For the first part, can I calculate the marginal density functions and they would be the density function of a standard normal?

Or should I use moment generating functions?

Thanks

2. Just find the marginals to show that $\displaystyle X, Y$ are standard normals. I did it for $\displaystyle X$ and it really isn't that bad.

To see that they aren't independent, you just have to look at the support. Compare $\displaystyle f_{X|Y} (0|-3)$ to $\displaystyle f_X (0)$.

Given that you know $\displaystyle EX = EY = 0$, all you have to do to show they are uncorrelated is show $\displaystyle EXY = 0$.

3. Where did I go wrong?

$\displaystyle f_{X}(x)=\int_{\mathbb{R}}f_{X,Y}(x,y) \, dy = \int_{-|x|}^{\infty} \frac{2|x|+y}{\sqrt{2\pi}}e^{-\frac{(2|x|+y)^{2}}{2}} \, dy =$
$\displaystyle =\left[y+2|x|=t\right] = \int_{|x|}^{\infty}te^{-t^2/2} \frac{dt}{\sqrt{2\pi}} = \left[-\frac{2e^{-t^2/2}}{\sqrt{2\pi}}\right]_{|x|}^{\infty}= 2 \cdot \varphi(x)$

, where $\displaystyle \varphi(x)$ is the density of a standard normal random variable.

4. $\displaystyle \int_{a} ^ b te^{-t^2 / 2} \ dt = -e^{-t^2 / 2}\big| _a ^ b$

5. Ah yes, how careless of me. =)

For finding $\displaystyle f_{Y}(y)$, I´m having a bit of trouble with the integration boundaries.

6. Draw a picture. Graph the support of $\displaystyle f_{XY}$ as a region on the plane. That should make it easier to figure out the boundaries.

If $\displaystyle y \le 0$ then you need to integrate over $\displaystyle (-\infty, y) \cup (-y, \infty)$. Make sure you see why, and figure out what interval to integrate over for $\displaystyle y > 0$.