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Math Help - Finding the marginal distribution

  1. #1
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    Finding the marginal distribution

    With f_X(x)\propto \frac{f_{(X|Y)}(x|y_o)}{f_{(Y|X)}(y_o|x)}

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

    and

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


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

    \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 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\int^b_0\frac{1}{x}(1-exp(-bx)) but it seems can't be done? For example: \int_0^b\frac{1}{x} undefined. May I know where the problems lie at?
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  2. #2
    Moo
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    Hello,

    Are you told to let 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 \exp(-bx) and hence simplify \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.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    Are you told to let 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 \exp(-bx) and hence simplify \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 \frac{b}{2x}[1-exp(-bx)]. Nevertheless ignoring the b/2 I will still need to integrate \frac 1x (1-\exp(-bx))

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