# [SOLVED] joint probability density function

• May 23rd 2010, 09:32 PM
akane
[SOLVED] joint probability density function
Given two iid Gaussian Random Variables $\displaystyle X,Y$ with zero mean and variance $\displaystyle \sigma^2$

$\displaystyle U = \frac{X^2-Y^2}{\sqrt{X^2+Y^2}}$

$\displaystyle V = \frac{2XY}{\sqrt{X^2+Y^2}}$

Derive the joint pdf $\displaystyle f_{UV} (u,v)$

Can someone help me with this?
• May 24th 2010, 01:13 AM
mr fantastic
Quote:

Originally Posted by akane
Given two iid Gaussian Random Variables $\displaystyle X,Y$ with zero mean and variance $\displaystyle \sigma^2$

$\displaystyle U = \frac{X^2-Y^2}{\sqrt{X^2+Y^2}}$

$\displaystyle V = \frac{2XY}{\sqrt{X^2+Y^2}}$

Derive the joint pdf $\displaystyle f_{UV} (u,v)$

Can someone help me with this?

You (should) know the joint pdf of X and Y. Now apply the Change of Variable Theorem:

If the random variables $\displaystyle X_1$ and $\displaystyle X_2$ have joint pdf function $\displaystyle f(x_1, x_2)$ then the pair of random variables $\displaystyle Y_1 = Y_1(X_1, X_2)$ and $\displaystyle Y_2 = Y_2(X_1, X_2)$ have joint pdf given by $\displaystyle g(y_1, y_2) = f(x_1(y_1, y_2), x_2(y_1, y_2)) |J(y_1, y_2)|$ if $\displaystyle (y_1, y_2)$ is in the range of $\displaystyle (Y_1, Y_2)$ and zero otherwise, where $\displaystyle J(y_1, y_2) = \left| \begin{array}{ccc} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_2}{\partial y_1} \\ & \\ \frac{\partial x_1}{\partial y_2} & \frac{\partial x_2}{\partial y_2} \end{array} \right|$.
• May 24th 2010, 02:36 AM
akane

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

$\displaystyle f_{UV} (u,v)= |J(u,v)| f_{X,Y} (x(u,v),y(u,v))$

$\displaystyle J (u,v)= \frac {\partial(x,y)}{\partial(u,v)} = \left| \begin{array}{cc} \frac{\partial x}{\partial u}& \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{array} \right|$

To get $\displaystyle J$ I should know $\displaystyle x = x(u,v)$ and $\displaystyle y = y(u,v)$.
Since I couldn't derive these functions from expressions for $\displaystyle U$ and $\displaystyle V$, I did the opposite thing, that is

$\displaystyle \frac {\partial(u,v)}{\partial(x,y)} = \left| \begin{array}{cc} \frac{\partial u}{\partial x}& \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\ \end{array} \right|$ and obtained 2.

So, I have this now:

$\displaystyle f_{UV}(u,v) \frac{\partial(u,v)}{\partial(x,y)} = f_{XY}(x(u,v),y(u,v))$

Hopefully this way is correct. Now I have

$\displaystyle f_{UV}(u,v) = \frac{1}{2} \frac{1}{2 \pi \sigma^2} e^{-{\frac{x^2+y^2}{2 \sigma^2}}}$

I got $\displaystyle f_{UV}$ as a function of $\displaystyle x$ and $\displaystyle y$, and don't know how to substitute $\displaystyle x$ and $\displaystyle y$ to get $\displaystyle f_{UV}$ as a function of $\displaystyle u$ and $\displaystyle v$.

Assistant mentioned that we should use this substitution

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

I have tried everything I could think of, but I'm still not sure how does that help.