I need a help to solve the question regarding normal distribution. The question is :

If X and Y are standard normal distribution and are independent then how to find the pdf and mgf of XY?

Results 1 to 3 of 3

- Nov 13th 2012, 07:21 AM #1

- Joined
- Nov 2012
- From
- ottawa, canada
- Posts
- 4

- Nov 13th 2012, 08:17 AM #2

- Joined
- Mar 2008
- From
- Pennsylvania, USA
- Posts
- 339
- Thanks
- 46

## Re: Suman Thapa

The (univariate) normal PDF is given by $\displaystyle f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^-{\tfrac{(x-\mu)^2} {2\sigma^2}} $.

When $\displaystyle X$ is standard normal, $\displaystyle \mu=0 \text{ and } \sigma=1$, so that the standard normal PDF is $\displaystyle f_X(x) = \frac{1}{\sqrt{2\pi}}e^{-{\tfrac{x^2} {2}}} $.

$\displaystyle Y$ is also a standard normal random variable, so its PDF is similarly $\displaystyle f_Y(y) = \frac{1}{\sqrt{2\pi}}e^{-{\tfrac{y^2} {2}}} $.

Now, If $\displaystyle X $ and $\displaystyle Y$ are INDEPENDENT, then $\displaystyle f_X(x) \cdot f_Y(y) = f_{X,Y}(x,y)$.

So, $\displaystyle f_{X,Y}(x,y) = f_X(x) f_Y(y) = \left(\frac{1}{\sqrt{2\pi}}e^{-{\tfrac{x^2} {2}}}\right)\left(\frac{1}{\sqrt{2\pi}}e^{-{\tfrac{y^2} {2}}}\right) = \frac{1}{2\pi}e^{-{\tfrac{x^2+y^2} {2}}} $.

- Nov 13th 2012, 08:42 AM #3

- Joined
- Nov 2012
- From
- ottawa, canada
- Posts
- 4