Finding the marginal distribution

• Nov 22nd 2010, 05:20 AM
noob mathematician
Finding the marginal distribution
With $\displaystyle f_X(x)\propto \frac{f_{(X|Y)}(x|y_o)}{f_{(Y|X)}(y_o|x)}$

I am given that
$\displaystyle f_{(X|Y)}(x|y)\propto (y)exp(-yx)$ at interval $\displaystyle 0\le x<b<+\infty$

and

$\displaystyle f_{Y|X}(y|x)\propto \frac{(x)exp(-xy)}{\int_0^b (x)exp(-xy)dy}$ at interval $\displaystyle 0\le y<b<+\infty$

Let $\displaystyle y_o=b/2$ and $\displaystyle \int_0^b (x)exp(-xy)dy=1-exp(-bx)$

$\displaystyle \frac{f_{(X|Y)}(x|y_o)}{f_{Y|X}(y_o|x)}=(b)exp(-bx)/\frac{(x)exp(-bx)}{1-exp(-bx)}=\frac{b}{x}(1-exp(-bx))$

Now In order to find the normalising constant c such that $\displaystyle f_X(x)=(c)\frac{f_{(X|Y)}(x|y_o)}{f_{Y|X}(y_o|x)}$, I want to integrate the right hand side.. i.e $\displaystyle \displaystyle\int^b_0\frac{1}{x}(1-exp(-bx))$ but it seems can't be done? For example: $\displaystyle \int_0^b\frac{1}{x}$ undefined. May I know where the problems lie at?
• Nov 22nd 2010, 05:43 AM
Moo
Hello,

Are you told to let $\displaystyle y_0=\frac b2$ ? because if so, your substitution into the two pdfs is wrong.

You can prove that the integral exists, by taking the limit of the integrand as x tends to 0, and noting that the function is continuous on (0,b). Then in order to solve for the integral, find the power series of $\displaystyle \exp(-bx)$ and hence simplify $\displaystyle \frac 1x (1-\exp(-bx))$

Under certain circumstances (which I suck at stating), you can reverse the integral and the sum signs and it becomes easy to compute.
• Nov 22nd 2010, 06:39 AM
noob mathematician
Quote:

Originally Posted by Moo
Hello,

Are you told to let $\displaystyle y_0=\frac b2$ ? because if so, your substitution into the two pdfs is wrong.

You can prove that the integral exists, by taking the limit of the integrand as x tends to 0, and noting that the function is continuous on (0,b). Then in order to solve for the integral, find the power series of $\displaystyle \exp(-bx)$ and hence simplify $\displaystyle \frac 1x (1-\exp(-bx))$

Under certain circumstances (which I suck at stating), you can reverse the integral and the sum signs and it becomes easy to compute.

I was told to subst it by b/2.

Think it should be $\displaystyle \frac{b}{2x}[1-exp(-bx)]$. Nevertheless ignoring the b/2 I will still need to integrate $\displaystyle \frac 1x (1-\exp(-bx))$

Using power series seems like a good idea.. will try and post again