# Interpreting the Joint pdf of Two Random Variables

Suppose I have the following joint pdf of $X$ and $Y$.

$$f_{X,Y}(x,y) = \begin{cases}c\cdot x \cdot \max(x,y) \text{ if } 0\leq x\leq 1, 0\leq y\leq 1\text{ and }x+y\leq 1;\\ 0 \text{ otherwise}\end{cases}$$

How do I

(a) find the constant $c$,

(b) compute $P(X>2Y)$, and

(c) compute the marginal pdfs of $X$ and $Y$?

Currently, I have the following. For part (a), I calculated this. Is this right?

I apologize for not being able to write in the proper notation on this site. For part (b), I calculated this. Is this right?

What is part (c)? How should I set up the integrals? I am having trouble getting them to equal 1.

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Hint: Draw a reasonably large sketch (say 10 cm by 10 cm) of the plane with coordinate axes $x, y$ and mark on it the lines $x = y$ and $x+y = 1$ (make the point $(1,0)$ be $5$ cm from the point $(0,0)$). You should be able to see a large right triangle divided into two congruent right triangles in your sketch. $f_{X,Y}(x,y)$ is a surface above the large triangle whose height above the plane at $(x,y)$ is given by two different expressions $cx^2$ and $cxy$ depending on which smaller triangle the point $(x,y)$ lies in. Mark the triangles with the appropriate expressions. Now can you proceed? – Dilip Sarwate Sep 17 '12 at 10:35
Yes, I've made significant progress. What do you think? – idealistikz Sep 17 '12 at 19:14
Did you draw the sketch? I have no idea what your WA page is trying to calculate. – Dilip Sarwate Sep 17 '12 at 20:09
Yes, there are four distinct regions which are represented by the four double integrals. – idealistikz Sep 18 '12 at 3:20
Got something from my answer? – Did Sep 30 '12 at 10:12

(a) Integrate the function $f_{X,Y}$ on the set $[0,1]^2$ and choose $c$ such that the result is $1$.
(b) Integrate the function $(x,y)\mapsto f_{X,Y}(x,y)$ on the set $(x,y)\in[0,1]^2$, $x\gt2y$.
(c) Integrate the functions $f_{X,Y}(x,\cdot)$ and $f_{X,Y}(\cdot,y)$.
You seem to want to integrate $cxy$ on $0\lt x\lt y\lt1-x$ and $cx^2$ on $0\lt y\lt x\lt1-y$ (and now, your job is to understand and to check this proposition, maybe it is wrong...). – Did Sep 16 '12 at 19:17