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One of the problems in my textbook is posed as follows. A two-dimensional stochastic continuous vector has the following density function:

$$ f_{X,Y}(x,y)= \begin{cases} 15xy^2 & \text{if 0 < x < 1 and 0 < y < x}\\ 0 & \text{otherwise}\\ \end{cases} $$

Show that the marginal density functions $f_X$ and $f_Y$ are:

$$ f_{X}(x)= \begin{cases} 5x^4 & \text{if 0 < x < 1}\\ 0 & \text{otherwise}\\ \end{cases} $$

$$ f_{Y}(y)= \begin{cases} \scriptsize{\frac{15}{2}}\normalsize y^2(1-y^2) & \text{if 0 < y < 1}\\ 0 & \text{otherwise}\\ \end{cases} $$

I understand how the density function $f_X$ is calculated, by integrating $f_{X,Y}$ from $0$ to $x$ with respect to $y$. I'm however totally lost on $f_Y$, where is the $(1-y^2)$ coming from? If I integrate from $0$ to $1$ with respect to $x$ then I only get $\scriptsize{\frac{15}{2}}\normalsize y^2$, and why is the range $0 < y < 1$?

I've graphed the support for $X,Y$, all values where $f_{X,Y}>0$ are colored blue:

The support for $X,Y$

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    $\begingroup$ It might help you to draw a picture of the support of $(X,Y)$ (which is the set of $(x,y)$ for which $f(x,y)\ne 0$). That should immediately answer some of your questions. $\endgroup$ – whuber Jan 22 '13 at 23:07
  • $\begingroup$ @whuber Okay so I've graphed the support and I think I understand why it's 0 < y < 1, it's because x is only defined in 0 < x < 1 and since 0 < y < x we then naturally have that y is only defined from 0 to 1, correct? But I still don't understand the (1-y^2) part. $\endgroup$ – soren.qvist Jan 22 '13 at 23:21
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    $\begingroup$ Hint: The marginal density of $f_Y(y)$ is the integral of $f_{X,Y}(x,y)$ which, for a fixed value of $y$, $0 < y < 1$, is nonzero only for those $x$ satisfying $y < x < 1$. That is, $$f_Y(y) = \int_{-\infty}^\infty f_{X,Y}(x,y)dx = \int_y^1 15xy^2 dx$$ and that is where the $(1-y^2)$ part comes from. $\endgroup$ – Dilip Sarwate Jan 23 '13 at 3:23
  • $\begingroup$ Thanks for the hint Dilip, I'm afraid I don't understand it fully though. ".. for a fixed value of $y$, $0 < y < 1$, is nonzero only for those $x$ satisfying $y<x<1$." Are you referring to the blue area on the chart? $\endgroup$ – soren.qvist Jan 23 '13 at 10:51
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    $\begingroup$ @soren.qvist Yes. I am referring to the blue area on the chart. $f_Y(0.4)$ is the integral (area under the curve) of a function of $x$ which has value $(15(0.4)^2)x=2.4x$ if $x$ is between $0.4$ and $1$ (the blue area) and $0$ otherwise. Repeat for other fixed values of $y$, and notice that each time the numerical value of $f_Y(y)$ works out to be the same number as obtained by "plugging in" the chosen value of $y$ into the expression $f_Y(y)$ as given in your answer sheet. Then, comes the "Hey Ma, I think I see a pattern!" moment and you realize that $f_Y(y)$ equals the integral shown. $\endgroup$ – Dilip Sarwate Jan 23 '13 at 15:43
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As you correctly pointed out in your question $f_{Y}(y)$ is calculated by integrating the joint density, $f_{X,Y}(x,y)$ with respect to X. The critical part here is identifying the area on which you integrate. You have already clearly showed graphically the support of the joint distribution function $f_{X,Y}(x,y)$. So, now, you can note that the range of $X$ in the shaded region is from $X=y$ to $X=1$ (i.e. graphically, you can visualize horizontal lines, parallel to the x-axis, going from the diagonal line $Y=X$ to the vertical line at $X=1$).

Thus, the lower and upper limits of the integration are going to be $X=y$ and $X=1$. Thus, the solution to the problem is as follows: $$f_{Y}(y)= \int_{y}^{1} f_{X,Y}(x,y) dx= \int_{y}^{1} 15xy^{2} dx=15y^{2}\int_{y}^{1} x dx=15y^{2}\left(\frac{1}{2}x^2\Big|_y^1\right)\\=\frac{15}{2}y^2(1-y^2). $$

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