1. ## Density Function

Let x and y be identically distributed variables have density function $\displaystyle f(u)=\frac{e^{\frac{-u^2}{2}}}{(2\pi)^\frac{1}{2}}$

find the density function of z=x^2+y^2

i have an answer given as $\displaystyle g(z)=.5e^{\frac{-z}{2}}$ for $\displaystyle z\geq 0$

2. you're saying that X and Y are st normals
However the constant is square root of 2 pi.
not 2 pi squared.
Then $\displaystyle X^2$ is a chi-square with one degree of freedom
and so is $\displaystyle Y^2$ and their sum by independence is a chi-square with two degrees of freedom

3. yeah, i am well aware of that, but when I tried to get an answer i ended up with a root z in there that shouldnt be

4. $\displaystyle \frac{e^{\frac{-u^2}{2}}}{(2\pi)^2}$ is not a density

However $\displaystyle \frac{e^{\frac{-u^2}{2}}}{(2\pi)^{1/2}}$ is

5. yes, that was a typo on my part sorry, but can you get this problem started for me, I was trying to use the distribution function method

6. as I said

Then $\displaystyle X^2$ is a chi-square with one degree of freedom
and so is $\displaystyle Y^2$ and their sum by independence is a chi-square with two degrees of freedom
Use MGFs to add the two rvs.
Chi-squares are gammas.

7. okay, thanks

is it invalid to say z=2y^2, and then try to use the distribution function method?

8. Originally Posted by emathinstruction
okay, thanks

is it invalid to say z=2y^2, and then try to use the distribution function method?
there is no 2 infront
let W=Y^2
Squaring a st normal is very simple.
That can be found in most books, probably yours.
and it's been done here a dozen times...