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The joint PDF of $X$ and $Y$ are given by $$f(x,y)=3y, for 0<y<1, 0<x<y$$

As usual, I tried to find the marginal PDF of X by $$\int_{0}^{1}f(x,y) dy $$ However, this gives me 1.5 which is greater than 1.

How can I approach the problem?

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    $\begingroup$ You missed the $x<y$ part. Also, a pdf may take values greater than $1$. $\endgroup$
    – user251257
    Dec 2, 2016 at 1:33
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    $\begingroup$ @user251257 how to include $x<y$? $\endgroup$ Dec 2, 2016 at 1:44
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    $\begingroup$ you obviously computed $\int_0^1 f(x,y) dy = \int_0^1 3y\, dy$. But what is, say $f(1/2, 1/4)$? $\endgroup$
    – user251257
    Dec 2, 2016 at 1:49
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    $\begingroup$ @user251257 should be 0? $\endgroup$ Dec 2, 2016 at 2:14
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    $\begingroup$ Yes. So what are the correct integral bounds? $\endgroup$
    – user251257
    Dec 2, 2016 at 2:15

2 Answers 2

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To fully formulate this problem, it is helpful to introduce the indicator function defined in the following way:
$$ I_A(x) = \begin{cases} 1 \quad \text{if $x \in A$}\\ 0 \quad \text{if $x \notin A$} \end{cases} $$ Thus, the joint PDF of $X$ and $Y$ can be written as: $$ f(x, y) = 3 y I_{(0, y)}(x) I_{(0, 1)}(y) = 3 y I_{(x, 1)}(y) I_{(0, 1)}(x) $$ The marginal PDF of $X$ is given by: $$ f(x) = \int_{-\infty}^{+\infty} f(x, y) \; d y = \int_{-\infty}^{+\infty} 3 y I_{(x, 1)}(y) I_{(0, 1)}(x) \; d y = I_{(0, 1)}(x) \int_{-\infty}^{+\infty} 3 y I_{(x, 1)}(y) \; d y = I_{(0, 1)}(x) \int_{x}^{1} 3 y \; d y = I_{(0, 1)}(x) \bigg ( \frac{3}{2}y^2 \bigg ) \Bigg |_x^1 = \frac{3}{2} (1 - x^2) I_{(0, 1)}(x) $$ Write this in the usual way, we have: $$ f(x) = \begin{cases} \frac{3}{2} (1 - x^2) \quad \text{if $0 < x < 1$}\\ 0 \quad \text{otherwise} \end{cases} $$ It is not hard to verify that $f(x)$ is a PDF.


Remark: Always remember to analyze the range of random variables first. A general way to do this is using the indicator function to extend the range of random variables to the entire real line(for a certain real random variable) and compute the integral with the infinite upper and lower bound. Then, consider the "true" bound(range) of the integral indicated by those indicator functions.

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We have to integrate the joint distribution $f_{X,Y}(x,y)$ with respect to $Y$ to get the marginal $f_X(x)$. The limits of integration are:

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So it amounts to getting rid of the $y$'s by integrating:

$$f_X(x)=\int_{y=x}^1 3y\,dy=\frac{3}{2}y^2\Big|_x^1=\frac{3}{2}(1 - x^2).$$

This is a valid pdf, because if we integrate it over the domain of $X: 0<x<1:$

$$\int_0^1 \frac{3}{2}(1-x^2)\,dx=1$$

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