Density Function

**emathinstruction** · February 16th 2010, 04:07 AM

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

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

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

but I cannot seem to get that answer, thanks in advance

**matheagle** · February 16th 2010, 06:02 AM

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

**emathinstruction** · February 16th 2010, 06:03 AM

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

**matheagle** · February 16th 2010, 06:07 AM

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

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

**emathinstruction** · February 16th 2010, 06:09 AM

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

**matheagle** · February 16th 2010, 06:14 AM

as I said

Then $X^2$ is a chi-square with one degree of freedom
and so is $Y^2$ 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.

**emathinstruction** · February 16th 2010, 06:16 AM

okay, thanks

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

**matheagle** · February 16th 2010, 06:18 AM

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...

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