Given the following PDF of continuous 2 random variables:

$$ f_{X,Y}(x,y)=\begin{cases} y^2 & 0\le y\le x\le 1;\newline 0 & \text{otherwise}. \end{cases} $$

Graph showing the region of integration with x and y random variables:

limit of 0<y<x<1

Question is to find the marginal CDF $F_{Y}(y)$ and $P[Y > x]$

My attempt to find joint CDF: $F_{X,Y}(x,y)$ $$ \begin{alignedat}{0} F_{X,Y}(x,y)&=\int_{-\infty}^{x}\int_{-\infty}^{x}y^2 dydx \newline &=\int_{0}^{x}\int_{0}^{x}y^2 dy dx\newline &=\int_{0}^{x}\frac{y^3}{3}\Bigg|_{0}^{x} dx=\int_{0}^{x}\frac{x^3}{3}dx\newline &=\frac{x^4}{12}\Bigg|_{0}^{x}=\frac{x^4}{12} \end{alignedat} $$

And then to find marginal CDF of: $F_{Y}(y)=F_{X,Y}(x,y)$

$$ F_{Y}(y)=F_{X,Y}(\infty,y)= $$

But where do I go from there? What does setting x to $\infty$ even mean?

And the question to find $P[Y>X]$, shouldn't it be 0 since $0\leq y\leq x\leq 1$?

UPDATE 12/17/2022

Based on @Xi'an's tip, $$ \int_{0}^{1}\int_{0}^{1}cy^2dydx=1\\ \text{ implies } c = 3\\ $$ The original PDF must have been: $$f_{X,Y}(x,y)=\begin{cases} 3y^2 & 0\le y\le x\le1 \newline 0 & otherwise \end{cases} $$

Find PDF between $0\le y\le x\le1$

$$ \begin{alignedat}{0} F_{X,Y}(x,y)&=\int_{0}^{x}\int_{0}^{y}3y^2dy_0dx_0\\ &= y^3x \end{alignedat} $$ Therefore, the joint CDF is $$ F_{X,Y}(x,y)= \begin{cases} 0 & y\le0,x\le0,\\ y^3x & 0\le y\le x\le1,\\ 1 & y\ge1, x\ge1. \end{cases} $$ How do I compute for other conditions, for example $y>=1, 0<=x<=1$ $$ F_{X,Y}(x,y)=\int_0^x\int_1^\infty 3y^2 dydx\\ $$ y goes to infinity?

  • 2
    $\begingroup$ What have you tried? $\endgroup$
    – Zhanxiong
    Dec 14, 2022 at 23:47
  • 3
    $\begingroup$ Instead of reposting the same, you can edit your previous post. $\endgroup$ Dec 15, 2022 at 8:03
  • 3
    $\begingroup$ And also please add the self-study tag to tour post. $\endgroup$ Dec 15, 2022 at 8:03
  • 2
    $\begingroup$ And indeed the joint pdf cannot be $y^2$ on $0 \le y \le x \le 1$ since $\int_{x=0}^1\int_{y=0}^x y^2 \, dy \, dx= \frac1{12} \not=1$ $\endgroup$
    – Henry
    Dec 15, 2022 at 9:27
  • 3
    $\begingroup$ There is an option to reopen the previous question but it's an option offered to the users of the site who have enough privilege to vote to reopen the question if it meets our standards (which occasionally happens via the original poster editing the question) $\endgroup$
    – Glen_b
    Dec 15, 2022 at 10:22

1 Answer 1


Warning: The density $f_{X,Y}$ does not integrate to one (1). This error can be easily corrected and the proper multiplying constant $$c=1\Big/\iint y^2 \mathbb I_{0\le y\le x\le 1}\text dy\text dx$$ derived.

Hint: Writing \begin{alignedat}{0} F_{X,Y}(x,y)&=\int_{-\infty}^{x}\int_{-\infty}^{x} cy^2 \text dy\text dx \end{alignedat} is not correct, because the symbol $x$ takes two different meanings in this expression, namely upper bound value and integrand symbol. The upper bound $y$ is furthermore missing.

The correct entry is, for $x_0,y_0\ge 0$, \begin{alignedat}{0} F_{X,Y}(x_0,y_0)&=\int_{-\infty}^{x_0}\int_{-\infty}^{y_0} cy^2 \mathbb I_{0\le y\le x\le 1}\text dy\text dx\\ &= \int_{0}^{x_0\wedge 1}\left\{\int_{0}^{y_0\wedge x} cy^2 \text dy\right\}\text dx \end{alignedat} where $a\wedge b=\min\{a,b\}$. Once the computation of $F_{X,Y}(x,y)$ is correctly completed, $$F_Y(y)=F_{X,Y}(1,y)$$ since $1$ is the maximal possible value of $X$.

Note: The same care must be taken when computing $P(Y>X)$ [if I assume there is a typo in the question about $P(Y>x)$].


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