# Dividing probability density functions?

• Jan 27th 2010, 02:50 PM
Volcanicrain
Dividing probability density functions?
I have a problem and I am at a loss:

Suppose that Z is a standard normal random variable and Y1 and Y2 are χ2 distributed random variable with υ1 and υ2 degrees of freedom, respectively. Further, assume that Z and Y1 and Y2 are independent.
a) Define W = (Z/SQRT(Y1)). Find E(W) and V(W). What assumptions do you need about the value of υ1?
b) Define U= Y1/Y2 . Find E(U) and V(U). What assumptions do you need about the value of υ1 and υ2?

I know the moment generating functions of Z, Y1 & Y2, but I don't know how I am supposed to find the distribution of W considering we are dividing densities... Any help would be vastly appreciated!
• Jan 27th 2010, 11:36 PM
Moo
• Jan 28th 2010, 07:19 AM
matheagle
These are just multiples of T and F random variables.
Just multiply and divide by the dfs and you will get your answer in seconds.
Clearly the mean of the first is zero.
• Jan 28th 2010, 07:30 PM
Volcanicrain
Quote:

Originally Posted by matheagle
These are just multiples of T and F random variables.
Just multiply and divide by the dfs and you will get your answer in seconds.
Clearly the mean of the first is zero.

Could you please elaborate? I am not sure I know what you mean by T & F random variables... or dfs. :S
• Jan 28th 2010, 08:49 PM
matheagle
df=degrees of freedom
I figured you were given this problem because you're studying the F and T distributions.
There are three ways to attack this.
1 Derive the densities, which is messy and unnecessary given that you're only asked to obtain the moments.
2 Find the moments via the Y's, which isn't that hard
3 Use the F and T moments and obtain the moments of these two in a minute.

Using the last one...

$\displaystyle U= Y_1/Y_2= {u_1\over u_2}\left({ Y_1/u_1\over Y_2/u_2}\right)= {u_1\over u_2}F_{u_1,u_2}$

So $\displaystyle E(U)={u_1\over u_2}E(F_{u_1,u_2})$

and $\displaystyle V(U)=\left({u_1\over u_2}\right)^2V(F_{u_1,u_2})$

Now look up the mean and variance of an F, which isn't that hard to derive...
(thats my second suggested technique)
http://en.wikipedia.org/wiki/F-distribution

$\displaystyle E(U)= E(Y_1)E(Y_2^{-1})$

where $\displaystyle E(Y_1)=u_1$ and $\displaystyle E(Y_2^{-1})$ isn't that hard to obtain either.
• Feb 2nd 2010, 07:06 PM
pgl1990
How do you obtain E(Y^-1)?