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I have the following equation

$$ f(x,y) = \frac{2}{x^2 (x-1) y^{(2 x-1)/(x-1)}} \quad \forall x>1, y>1 $$

I am trying to find that marginal pdf w.r.t. $x$, $f_X(x)$. Normally I would just take $\int f(x,y) dy$ using $x$'s domain as limits, aka over $[1, \infty)$. But that integral won't converge here. What am I doing wrong? This problem should be solve-able.

SheerSt
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  • tried my best to interpret your equation correctly, please verify – gt6989b Feb 13 '14 at 20:04
  • If you integrate over $y$, you will obtain the marginal pdf w.r.t. $x$ i.e. $f_X(x)$, rather than $f_Y(y)$. To obtain $f_Y(y)$ you would need to evaluate $\int f(x,y)dx$ which is far harder than calculating $\int f(x,y)dy$. – Alijah Ahmed Feb 13 '14 at 20:37
  • Yes, I meant $f_X(x)$. Sorry about that. – SheerSt Feb 13 '14 at 20:47
  • No problem. It's an error that is quite easy to make - I used to fall into that trap when calculating marginals, as a student. – Alijah Ahmed Feb 13 '14 at 20:49

1 Answers1

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If you marginalise with respect to $y$ you will obtain $f_X(x)$, which for the joint pdf in question is a far easier task than marginalising with respect to $x$ to obtain $f_Y(y)$

Thus

$\large\int f(x,y)dy=\frac{2}{x^2(x-1)} \int_1^\infty \frac{1}{y^{(2x-1)/(x-1)}}dy$

This simplifies to a straightforward integral of the form $\int t^ndt=[\frac{t^{n+1}}{n+1}]$

$\large\int f(x,y)dy=\frac{2}{x^2(x-1)} \int_1^\infty y^{(1-2x)/(x-1)}dy=\frac{2}{x^2(x-1)}(\frac{(1-x)}{x})[y^{-x/(x-1)}]_1^\infty$

As $x>1$, $-x/(x-1)$ is negative so that as $y \rightarrow \infty$, $y^{-x/(x-1)} \rightarrow 0$.

Thus the integral is

$\large\int f(x,y)dy=\frac{2}{x^2(x-1)}(\frac{(1-x)}{x})(0-1)=\frac{2}{x^3}$

Resulting in

$\large f_X(x)=\frac{2}{x^3}$

Integrating this marginal from $1$ to $\infty$ results in unity, which is a good sanity check.

Alijah Ahmed
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  • Thanks for the help. I guess it didn't end up being so bad - just a matter of recognizing the limit of y as it approaches infinity. – SheerSt Feb 13 '14 at 21:11
  • Yes, that's quite true. Once you find the limiting value of a parameter as it approaches infinity, it tends to zero in many cases, and the expressions simplify quite a bit. – Alijah Ahmed Feb 13 '14 at 21:17