# Thread: [SOLVED] normal random variable

1. ## [SOLVED] normal random variable

if

$\displaystyle f_{XY}(x,y) = f_{X}(x)f_{Y}(y) = \frac{1}{2\pi \sigma^2} e^{-\frac{x^2+y^2}{2\sigma^2}}$

I have to show that $\displaystyle Z$ is also a normal random variable

$\displaystyle Z = \frac{(X-Y)^2-2Y^2}{\sqrt{X^2+Y^2}}$

maybe using these substitions

$\displaystyle A = \sqrt {X^2+Y^2}$

$\displaystyle \tan B = \frac{Y}{X}$

2. I don't recognize any tricks here, like using the MGF.

One thing for sure is that $\displaystyle X\sim N(0,\sigma^2)$ and $\displaystyle Y\sim N(0,\sigma^2)$ are independent

So $\displaystyle \left({X\over \sigma}\right)^2 + \left({Y\over \sigma}\right)^2\sim\chi^2_2$

IF you're going to transform from X,Y to something you should USE Z and something basic like A equal to X or Y OR maybe your A or maybe A squared? BUT not A and B. Try Z and A.

3. Originally Posted by matheagle
I don't recognize any tricks here, like using the MGF.
what does MGF stand for?

Originally Posted by matheagle
So $\displaystyle \left({X\over \sigma}\right)^2 + \left({Y\over \sigma}\right)^2\sim\chi^2_2$
Originally Posted by matheagle
IF you're going to transform from X,Y to something you should USE Z and something basic like A equal to X or Y OR maybe your A or maybe A squared? BUT not A and B. Try Z and A.
this is really hard... I don't have the idea how to do it...

thank you very much

4. MGF stands for moment generating function.
YOU may have to transform from X,Y to Z,A and then integrate out the A
to obtain the density of Z.

5. Originally Posted by matheagle
MGF stands for moment generating function.
YOU may have to transform from X,Y to Z,A and then integrate out the A
to obtain the density of Z.
What would be A?

6. Originally Posted by akane
What would be A?
You can try any possibility for A.
I was suggesting the one you had already selected, that square root of X and Y squared

Though, I must say, this is a weird problem
I don't see any tricks, are you sure you copied it correctly?

7. Originally Posted by matheagle
You can try any possibility for A.
I was suggesting the one you had already selected, that square root of X and Y squared

Though, I must say, this is a weird problem
I don't see any tricks, are you sure you copied it correctly?
unfortunately I copied it correctly
that substitution won't work because I have to rewrite Z as a function of these two new variables, in the case you mentioned I would have Z = Z (X,Y,A)