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I am having trouble understanding how to solve this when the variables are not discrete.

Let the simultaneous density of the non-discrete stochastic variables (X,Y) be Formula

I am then supposed to find marginal densities g(x) and h(y), conditional density f(y|x) and P(Y > 1|X=1/2)

I understand i should integrate the joint probability density function, but with what boundaries?


marked as duplicate by whuber Jan 26 '16 at 23:07

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  • $\begingroup$ You have to figure out the marginal and the conditional boundaries. For instance, $X$ and $Y$ vary over $(0,2)$, while conditionally $Y$ varies on $(x,2)$ and $X$ on $(0,y)$. $\endgroup$ – Xi'an Jan 25 '16 at 17:17
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    $\begingroup$ @Xi'an You really provided the answer to the question. Is there a point in elaborating with a formal answer? $\endgroup$ – Antoni Parellada Jan 25 '16 at 18:42
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    $\begingroup$ When in doubt, draw. You have that $y$ is greater than $x$ but smaller than $2$ which tells you that your region is triangular. Can you see why? $\endgroup$ – JohnK Jan 25 '16 at 21:37
  • $\begingroup$ @Antoni The fact that "and conditional density" is part of the title of this question suggests there's some merit to this request. If this question is not going to get an answer, but rather linger as a duplicate for SEO purposes, then I'd rather that part of the title were edited out, or this is going to be a source of frustration for later searchers. $\endgroup$ – Silverfish Jan 27 '16 at 3:35
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    $\begingroup$ @Antoni One of the reasons that duplicate questions are often left, rather than deleted, is search engine optimization - it can make it easier for people to find what they're looking for, if the way they search for a question matches the way a duplicate was written rather than how the original was written. But in this case it would have a downside: someone searching for how to find the marginal density from the joint density could end up here, then redirected to the "duplicate", and find no such discussion. $\endgroup$ – Silverfish Jan 27 '16 at 15:14

The marginal pdf will be calculated over the area defined by a triangle as mentioned in the comments. The reason for it lies in the boundary constraints $0 < x < y < 2$, where the bivariate joint pdf is defined. To see this, we can mentally slide along the $x$ axis from $0$ to $2$, and see how at any given point, the $y$ axis will be past the bisecting line ($y = x$) on the $xy$ plane by the inequality $y > x$, and with a maximum of $2.$

I tried illustrating this with the following plot, which looks at the $z= \frac{1}{2}\,x\,y$ surface slightly from the top and front. The area we are going to integrate over when obtaining the marginal pdf's will be the light blue (crayon) triangle, which will cut through the overhead surface "in front of" the yellow plane:

view from the front

And here is the view of the region where the pdf is defined (in blue) seen from above with the $z = 1/2\,x\,y$ surface in gray:

enter image description here

Therefore the marginal pdf of $Y$ will be found by integrating the joint pdf, $f(x,y)$ over the support of $Y$, corresponding to the triangle in the plot:

$f_Y(y) =\displaystyle\int_{x\,=\,0}^{x\,=\,y} f(x,y)\, dx = \displaystyle\int_0^y \frac{1}{2}\,x\,y\, dx = \frac{x^2}{4}\,y\,\Big|_{x\,=\,0}^{x\,=\,y}=\frac{y^3}{4}$, for $0<y<2.$

And the marginal of $X$ will be obtained by integrating away the $y$, keeping in mind, the the lower boundary of $Y$ is $X$:

$f_X(x) =\displaystyle\int_{y\,=\,x}^{y\,=\,2} f(x,y)\, dy = \displaystyle\int_x^2 \frac{1}{2}\,x\,y\, dy =x\, \frac{y^2}{4}\,\Big|_{y\,=\,x}^{y\,=\,2}=x-\frac{x^3}{4}$, for $0 < x < 2.$

The conditional pdf of $Y$ given $X=x$, $f_{Y|X}$ is given by

$\large \frac{f(x,y)}{f_X(x)}=\frac{2\,y}{4-x^2}$.

We know that the marginal of $X$ and the joint pdf. The conditional will have the same support, $0<x<y$ and $x<y<2.$

  • $\begingroup$ It is difficult to interpret these figures. Perhaps you could explain how they work? What problem are they intended to solve? In what sense are the shaded areas "limits," as the arrows appear to claim? $\endgroup$ – whuber Jan 27 '16 at 16:21
  • $\begingroup$ @whuber OK... I redid the whole thing. I hope it is now accurate, and helpful in addressing the issue. $\endgroup$ – Antoni Parellada Jan 28 '16 at 6:06
  • $\begingroup$ I tried to complete the answer because it didn't sound like the integrals were the issue, but assuming there are no other issues, I can modify the answer according to guidelines. $\endgroup$ – Antoni Parellada Jan 28 '16 at 6:53
  • $\begingroup$ (+1) for the detailed explanation. Now, I also find the picture hard to understand as x and y do not appear to be at the right places... $\endgroup$ – Xi'an Jan 28 '16 at 14:31
  • $\begingroup$ @Xi'an I marked the x and y axis more conspicuously - the region where the pdf is defined is the blue triangle that has an apex at x=0, y=0, and fans out to x=2, y=2. $\endgroup$ – Antoni Parellada Jan 28 '16 at 14:42

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