- #1
fluidistic
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Homework Statement
The random variables X and Y have a joint probability distribution of [tex]f_{XY}(x,y)= \frac{1}{2\pi \sigma _X \sigma _Y \sqrt{1-\rho ^2}} \exp \{ \left [ -\frac{1}{2(1-\rho ^2)} \right ] \left [ \left ( \frac{x-\mu _X}{\sigma _X} \right )^2 + \left ( \frac{y-\mu _Y}{\sigma _Y} \right )^2 -\frac{2\rho (x-\mu _X)(y-\mu _Y)}{\sigma _X \sigma _Y} \right ] \}[/tex].
Where rho, sigma's and mu's are constants.
Show that the marginal distributions are also Gaussians. Calculate their mean value and their variance.
Homework Equations
a)The marginal distribution of the variable X is [itex]f_ X (x)=\int _{-\infty}^{\infty} f_{XY}(x,y)dy[/itex].
b)The marginal distribution of the variable Y is [itex]f_Y(y)=\int _{-\infty}^{\infty} f_{XY}(x,y)dx[/itex].
c)[tex]\int _{-\infty}^{\infty} e^{-ax^2+bx }dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}}[/tex].
d)For the mean values I need to calculate [itex]\int f_X(x)xdx[/itex] and [itex]\int f_Y(y)ydy[/itex] which in my case is so horrible that I've lost some hope in it.
The Attempt at a Solution
So my problem resides in calculating that huge integral. I've done it 3 times, each time reaching a different result, but I've spotted an error in my first 2 attempts and didn't in my last attempt though the result is so ugly I can't believe I've done it right.
So in order to perform [itex]\int _{-\infty}^{\infty} f_{XY}(x,y)dy[/itex], I factorize outside the integral the constant [itex]\frac{1}{2\pi \sigma _X \sigma _Y \sqrt{1-\rho ^2}}\exp \{ \left [ - \frac{1}{2(1-\rho ^2)} \right ] \left ( \frac{x-\mu _X}{\sigma _X} \right ) \}[/itex].
Therefore I am left to calculate the integral [tex]\int _{-\infty}^{\infty} \exp \{ \left [ - \frac{1}{2(1-\rho ^2)} \right ] \left [ \left ( \frac{y-\mu _Y}{\sigma _Y} \right ) ^2 - \frac{2\rho (x-\mu _X )(y- \mu _Y)}{\sigma _X \sigma _Y} \right ] \} dy[/tex].
I skip many steps (latex is very long to write!) but what I did was rewrote the argument of the exponential into a form [itex]-ax^2+bx+c[/itex] and where [itex]a = \frac{1}{2(1-\rho ^2) \sigma _Y ^2}[/itex], [tex]b= \frac{\mu _Y}{\sigma _Y ^2(1-\rho ^2)} - \frac{\rho \mu _X}{\sigma _X \sigma _Y (1-\rho ^2)} + \frac{2\rho x}{\sigma _X \sigma Y}[/tex] and [tex]c= \left [ \frac{1}{2(1-\rho ^2)} \right ] \left [ \frac{2\rho \mu _X \mu _Y}{\sigma _X \sigma _Y}-\frac{2\rho x \mu _Y}{\sigma _X \sigma _Y} -\frac{\mu _Y ^2}{\sigma _Y ^2} \right ][/tex].
This gave me as a final result, [tex]f_X(x)= \frac{1}{\sqrt {2\pi} \sigma _X} \exp \{ x^2 \left [ \frac{1}{2(1-\rho ^2)} \right ] \left ( \frac{4 \rho ^2 -1 }{\sigma _X ^2} \right ) +x \left [ \frac{1}{2(1-\rho ^2)} \left ( \frac{2\rho \mu _Y }{\sigma _X \sigma _Y}-\frac{4 \rho ^2 \mu _X}{\sigma _X ^2}+\frac{2\mu _X}{\sigma _X ^2} \right ) \right ] +\left [ \frac{1}{2(1-\rho ^2)} \right ] \left ( \frac{\rho ^2 \mu _X ^2}{\sigma _X ^2}- \frac{\mu _X ^2}{\sigma _X ^2} \right ) \}[/tex]. This would be my answer to the first question. For the marginal distribution of the random variable Y, I'd have to change all x's by y's and all X's by Y's in that result, because the joint distribution is totally symmetric with respect to x-y/X-Y.
As you can see my answer is under the form [itex]Ae^{ax^2+bx+c}[/itex] so I think this implies it's a Gaussian?
I would like to know whether my result is good and how can I calculate the mean of that function; I mean there must be a trick or something because the integral to perform looks much uglier than the first!
Thanks for any help!