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A pair of random variable $(X, Y)$ is uniformly distributed in the quadrilateral region with $(0,0),(a,0),(a,b),(2a,b)$, where $a,b$ are positive real numbers.

What is the joint pdf $f(X,Y.)$

Find the marginal probability density functions $f_X (x)$ and $f_Y (y)$.

Find $E(X)$and $E(Y)$.

Find $E(X,Y)$.

My understanding is the uniformly distributed pdf $=\frac1{\text{area}}=\frac1{ab}$. It seems that $X$ and $Y$ are independent R.V. because it seems the joint pdf can be factored as $\frac1a \frac1b$ if the pdf is correct. However, after I found the marginal pdf $f_X=\frac2{a^2}; f_Y=\frac2{b}-\frac{2}{b^2}$, which shows $X, Y$ are not independent. If I can not get correct marginal pdf, I can not finish the question d and e. Could anyone help me out? Thank you!

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    $\begingroup$ $X$ and $Y$ are not independent, and the expressions of $f_X(x)$ and $f_Y(y)$ are incorrect! $\endgroup$ Commented Nov 3, 2017 at 3:59
  • $\begingroup$ Factorization (if possible) should take account of allowable region of variables. $\endgroup$ Commented Jul 19, 2024 at 10:10

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Note that \begin{align}f_X(x) &= \int_{0}^b f_{X,Y}(x,y) \,dy \\ &= \begin{cases} \int_0^{\frac{b}{a}x} f_{X,Y}(x,y) \, dy & , 0 \leq x \leq a \\ \int_{\frac{b}{a}x-b}^b f_{X,Y}(x,y) \, dy &, a < x \leq 2a\end{cases}\end{align}

Recheck your value value for $f_Y$ as well. You should get a simpler expression. Notice that pdf should integrate to $1$ of which both of your proposed density functions have failed the test.

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  • $\begingroup$ Thank you very much for the clear explanation. Could you also check whether I got the joint pdf correctly? Then I can complete the solution. $\endgroup$ Commented Nov 3, 2017 at 4:30
  • $\begingroup$ you should state the support of the pdf, that is the region where the density is positive. $\frac1{ab}$ is fine but over which region. $\endgroup$ Commented Nov 3, 2017 at 4:33
  • $\begingroup$ I think I have trouble with limits when I integrate joint density function for the marginal functions. I also integrate the joint pdf, which is 2. It fails the test. Would you help me further? Thanks a lot $\endgroup$ Commented Nov 3, 2017 at 4:48
  • $\begingroup$ I think support can not be x is between 0 and 2a, y is between 0 and b. I know there are two congruent triangles, but how do I choose the limits to support the joint pdf? $\endgroup$ Commented Nov 3, 2017 at 4:51
  • $\begingroup$ The region can be described by $\{ (x,y) : 0 \leq x \le a, 0 \leq y \leq bx \} \cup \{ (x,y) : a \leq x \le 2a, \frac{b}{a}x-b \leq y \leq b \} $ $\endgroup$ Commented Nov 3, 2017 at 4:57

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