[SOLVED] bivariate RVs, joint and marginal densities, and transformations

**joop** · October 26th 2008, 11:56 PM

Hello, I'd like some help on this tough problem. Thanks in advance!

Given the joint density:
$f_{XY}(x,y) = {1 \over 2\pi} e^{-{1 \over 2}((x-\mu)^2 + (y-\mu)^2)}, \text{where } \mu \in\mathbb{R}$
find the joint density function
$f_{UV}(u,v) \text{ of } U = X - Y \text{ and } V = X + Y$
and find out if U, V are independent.

My work for this part of the problem:
I got a joint density function
$f_{UV}(u,v) = {1 \over 4\pi} e^{-{1 \over 2}(({U+V \over 2}-\mu)^2+({V-U \over 2}-\mu)^2)}$

is that correct? and also, how do I integrate that with the crazy exponents? or is there a simpler way to test for independence other that checking
$f_{UV}(u,v) = f_U(u)f_V(v)$
?

The second part of the problem:
Let the transformation from (X,Y) -> (R,θ) be defined as
$R = \sqrt{(X-\mu)^2+(Y-\mu)^2}$
$\theta=$
$arctan \left ({(Y-\mu) \over (X-\mu)} \right ), \text{if }X \ge \mu$
$-\pi + arctan \left ({(Y-\mu) \over (X-\mu)} \right ), \text{if }X < \mu; Y < \mu$
$\pi + arctan \left ({(Y-\mu) \over (X-\mu)} \right ), \text{if }X < \mu; Y \ge \mu$

find the joint density and marginal densities.

So far on this part I have it solved for X and Y:
$X = \sqrt{R^2 - (Y-\mu)^2} + \mu$
$Y = tan(\theta)(X-\mu) + \mu$

I tried plugging in the equations into each other but I don't know how to get rid of the tan(). Any help is appreciated.

**Laurent** · October 27th 2008, 01:14 AM

Originally Posted by joop

Hello, I'd like some help on this tough problem. Thanks in advance!

Given the joint density:
$f_{XY}(x,y) = {1 \over 2\pi} e^{-{1 \over 2}((x-\mu)^2 + (y-\mu)^2)}, \text{where } \mu \in\mathbb{R}$
find the joint density function
$f_{UV}(u,v) \text{ of } U = X - Y \text{ and } V = X + Y$
and find out if U, V are independent.

My work for this part of the problem:
I got a joint density function
$f_{UV}(u,v) = {1 \over 4\pi} e^{-{1 \over 2}(({u+v \over 2}-\mu)^2+({v-u \over 2}-\mu)^2)}$

is that correct? and also, how do I integrate that with the crazy exponents? or is there a simpler way to test for independence other that checking
$f_{UV}(u,v) = f_U(u)f_V(v)$
?

I think this is correct. As for checking the independence, it suffices to show that $f_{U,V}(u,v)$ writes as a product of a function of $u$ times a function of $v$ (not necessarily probability density functions). For that, you have to expand the exponent and see that the terms $uv$ vanish. Once you know that the joint density functions writes $f(u)g(v)$ , you deduce by Fubini that $\int f \int g =1$ , hence dividing $f$ by its integral and $g$ by its own, you end up with density functions. But you don't have to compute these integrals to show the independence.

The second part of the problem:
Let the transformation from (X,Y) -> (R,θ) be defined as
$R = \sqrt{(X-\mu)^2+(Y-\mu)^2}$
$\theta=$
$arctan \left ({(Y-\mu) \over (X-\mu)} \right ), \text{if }X \ge \mu$
$-\pi + arctan \left ({(Y-\mu) \over (X-\mu)} \right ), \text{if }X < \mu; Y < \mu$
$\pi + arctan \left ({(Y-\mu) \over (X-\mu)} \right ), \text{if }X < \mu; Y \ge \mu$

find the joint density and marginal densities.

So far on this part I have it solved for X and Y:
$X = \sqrt{R^2 - (Y-\mu)^2} + \mu$
$Y = tan(\theta)(X-\mu) + \mu$

I tried plugging in the equations into each other but I don't know how to get rid of the tan(). Any help is appreciated.

As you can guess, this is a polar change of coordinates. You can show that $X=\mu+R\cos\theta$ and $Y=\mu+R\sin\theta$ .
For instance, you can write $X=\sqrt{R^2-\tan^2\theta(X-\mu)^2}+\mu$ , square this equation and deduce the expression for $X$ (using $X\geq \mu$ at the end, and $\frac{1}{1+\tan^2\theta}=\cos^2\theta$ ).

**joop** · October 27th 2008, 02:08 AM

My book says that the joint pdf = product of the two marginal pdfs if they are independent. And as for how to integrate I was not trying to integrate the marginal pdfs but the joint pdf (to get the marginal pdf) as in:
$<br /> f_{U}(u) = \int_{-\infty}^{\infty} {1 \over 4\pi} e^{-{1 \over 2}(({u+v \over 2}-\mu)^2+({v-u \over 2}-\mu)^2)} \, dv<br />$

As for the second part, I do not see anywhere where I can use the identity you gave me. Basically I get something like
${1 \over tan(\theta)}x^2 + (X-\mu)^2 = R^2 + \mu^2$
(above eqn might not be exact eqn for this problem but is basically what I get)

Thank you for your efforts though but please advise some more, can't quite get my head around this.

**Laurent** · October 27th 2008, 11:18 AM

Originally Posted by joop

My book says that the joint pdf = product of the two marginal pdfs if they are independent. And as for how to integrate I was not trying to integrate the marginal pdfs but the joint pdf (to get the marginal pdf) as in:
$<br /> f_{U}(u) = \int_{-\infty}^{\infty} {1 \over 4\pi} e^{-{1 \over 2}(({u+v \over 2}-\mu)^2+({v-u \over 2}-\mu)^2)} \, dv<br />$

Ok, let's do it the way your book says. If you want to compute this integral, a first thing is to do is to expand the exponent (as I said before) to separate the terms with $u$ and $v$ : $-{1 \over 2}\left(\left({u+v \over 2}-\mu\right)^2+\left({v-u \over 2}-\mu\right)^2\right)=\frac{1}{2}\left(\frac{u^2}{2} +\frac{1}{2}(v-2\mu)^2\right)$ . Then $e^{-{1 \over 2}(({u+v \over 2}-\mu)^2+({v-u \over 2}-\mu)^2)}=e^{-{1\over 2}\cdot\frac{u^2}{2}}e^{-\frac{1}{2}\cdot\frac{(v-2\mu)^2}{2}}$ and the integral you want reduces to the integration of the density function of a normal distribution, which you know: $\int e^{-\frac{1}{2}\cdot\frac{(v-2\mu)^2}{2}}\frac{dv}{\sqrt{4\pi}}=1$ (mean $2\mu$ , variance $2$ ).

As for the second part, I do not see anywhere where I can use the identity you gave me. Basically I get something like
${1 \over tan(\theta)}x^2 + (X-\mu)^2 = R^2 + \mu^2$
(above eqn might not be exact eqn for this problem but is basically what I get)

Thank you for your efforts though but please advise some more, can't quite get my head around this.

Did you get $X=\sqrt{R^2-\tan^2\theta(X-\mu)^2}+\mu$ ? This is just plugging $Y$ into the expression for $X$ that you found by yourself. My equation writes also $X-\mu=\sqrt{R^2-\tan^2\theta(X-\mu)^2}$ . Square this, and solve for $X-\mu$ , so that you readily find $\left(1+\frac{1}{\tan^2\theta}\right)(X-\mu)^2=R^2$ and finally $X=\mu+R\cos\theta$ (with the identity on $\tan$ that I gave before and that results from $\tan=\frac{\sin}{\cos}$ and $\cos^2+\sin^2=1$ ).
Once you have $X=\mu+R\cos\theta$ and $Y=\mu+R\sin\theta$ , you can do the change of variable: you integrate on $0<\theta<2\pi$ and $r>0$ , the measure $dx\,dy$ becomes $r\,dr\,d\theta$ (by computing the Jacobian determinant) and $(x-\mu)^2+(y-\mu)^2=r^2$ .

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