Random variables $X$ and $Y$ are uniformly distributed on the set: $$D=\{(x,y): 1\leqslant x\leqslant 2,~ 0\leqslant y\leqslant 2,~ x \leqslant y\}$$ I want to determine the joint PDF of $X$ and $Y$ but I'm not sure how to do that. EDIT: I've drawn the figure, and the area that respects all the conditions given by the set D is a rectangle with an area of 2. This would mean that the PDF is 1/2. Is this correct?
4
- 1$\begingroup$ The PDF is $\frac 1 A$ on $D$ where $A$ is the area of $D$. Draw a picture and find the area. $\endgroup$Kavi Rama Murthy– Kavi Rama Murthy2021-06-21 10:05:03 +00:00Commented Jun 21, 2021 at 10:05
- 1$\begingroup$ If ever a question was ambiguous, this one is. Your title and your initial sentence and your statement about $f_X$ seem to imply that you mean $X$ is uniformly distributed and $Y$ is uniformly distributed, but then your definition of $D$ seems to suggest you mean that the pair $(X,Y)$ is uniformly distributed (and that would imply that $X$ is not uniformly distributed and $Y$ is not uniformly distributed). $\endgroup$Michael Hardy– Michael Hardy2021-06-21 10:07:49 +00:00Commented Jun 21, 2021 at 10:07
- $\begingroup$ @KaviRamaMurthy I've drawn the figure, and the area that respects all the conditions given by the set D is a rectangle with an area of 2. This would mean that the PDF is 1/2. Is this correct? $\endgroup$Andrei0408– Andrei04082021-06-21 19:57:11 +00:00Commented Jun 21, 2021 at 19:57
- $\begingroup$ @MichaelHardy The statement is about pairs of random variables, I wrote the formula about $f_{X}$ because I wanted to make a connection between the case of a single uniform r.v., thinking it would be somewhat relevant $\endgroup$Andrei0408– Andrei04082021-06-21 19:59:26 +00:00Commented Jun 21, 2021 at 19:59
Add a comment |
1 Answer
$\begingroup$ $\endgroup$
\begin{align} D &= \{(x,y) \in \mathbb{R}^2:1 \le x \le 2, 0 \le y \le 2, x \le y \}\\ &=\{(x,y) \in \mathbb{R}^2: 1 \le x \le 2, x \le y \le 2 \} \end{align}
The area of the triangle is $\frac12 \cdot (2-1)\cdot (2-1)=\frac12$.
Hence the joint pdf is
$$f(x,y) = \begin{cases} 2 &, 1 \le x \le 2, x \le y \le 2 \\ 0 &, \text{otherwise}\end{cases}$$
