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Let $\displaystyle{{\rm f}\left(x, y\right) = \left\{x\quad \mbox{if}\quad 0< y < {1 \over x}\right\}\ \mbox{and}\ 0}$ otherwise.

I need to calculate the marginal pdf of $Y$. I know I need to integrate out $x$, but I'm having a hard time seeing what to integrate to. Please help I've been stuck on this problem for $3$ days !.

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Can you write down the "marginal pdf of $\large Y$" definition ?. – Felix Marin Nov 19 '13 at 20:22
I have no idea how to write it on here, but the marginal pdf of Y can by found by integrating the joint probability density function from negative infinity to positive infinity, with respect to X – user110033 Nov 19 '13 at 20:31
up vote 0 down vote accepted

Hint: Graphical Hint: Try drawing the region you're integrating over. Draw the curve $y = 1/x$ for $0 < x < 1$ and realize that the region you're integrating over will be the region bounded below this curve (in the first quadrant between $0<x<1$). Now if you are going to integrate out the $x$ variable, fix some positive $y$ value and ask when you enter and exit the region as you travel parallel to the $x$-axis at the height of $y$. If you chose $y$ with $0 < y \leq 1$, the you enter the region at $x=0$ and exit at $x=1$, otherwise, if you chose $y$ with $y > 1$ then you enter the region when $x =0$ and exit when $x=1/y$. Therefore you find your marginal PDF for $Y$ will be calculated as $$ \begin{cases} \int_0^1 f(x,y)dx & 0 < y \leq 1 \\ \int_0^{1/y} f(x,y) dx & 1 < y \end{cases} $$

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I want to say that you integrate from 0 to 1/y but this pdf is supposed to have two different formulas, one when 0< y < 1, and one when y < 1. – user110033 Nov 19 '13 at 20:29
There must be more restrictions on the PDF you gave, since if $R = \{(x,y) : 0 < y < 1/x\}$ is the region you're integrating over, then $\iint_{R} f(x,y) dx\,dy = \infty$ but should be $1$. Are you sure that $R$ is not supposed to have more restrictions? – Tom Nov 19 '13 at 20:38
@user110033 Perhaps $R = \{(x,y) : 0 < y< 1/x < 1\}$?? – Tom Nov 19 '13 at 20:40
I'm sorry it got left out somehow, 0< x < 1 should be there as well. (Also note that all < should be ≤, sorry). – user110033 Nov 19 '13 at 20:41
@user110033 see the edits to the solution. – Tom Nov 19 '13 at 21:00

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