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If i have the region: $$W=\left\{ \left( x,y\right) \in \mathbb{R} ^{2}:0\leq y\leq x\leq 1\right\} $$

And the random vector $(X,Y)$ with the following joint PDF:

$$ \begin{equation} f_{(X,Y)}(x,y)= \left\{ \begin{array}{@{}ll@{}} c(x^4 + y^2), & \text{if}\ \left( x,y\right) \in W \\ 0, & \text{otherwise} \end{array}\right. \end{equation} $$

How can i find the probability distribution of the variable $S=(X-Y)$?

Gotey
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2 Answers2

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Note first that $S$ ranges from $0$ to $1$. Now given any $0\leq s\leq 1$, if we want to write $s=x-y$, then we must have $s\leq x\leq 1$ and $y=x-s$. Thus we should have, for $0\leq s\leq 1$, $$ f_S(s)=\int_s^1f_{X,Y}(x,x-s)dx=\int_s^1c(x^4+(x-s)^2)dx $$

TomGrubb
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Hint: First find c using the below

$$\int_{0}^{1}\int_{y}^{1} c(x^4+y^4)dxdy = 1$$

Then follow the procedure outlined by the other responder, this is not an answer. Just an aid.

Satish Ramanathan
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  • How do you know from which regions do the integration (from $y$ to $1$ and from $0$ to $1$)? – Gotey Jan 02 '16 at 01:08
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    You should get that information from the region where the joint probability function is defined,i.e $0\le y\le x\le1$. For every y from o to 1, x ranges from y to 1. That is how I formulate my limits to the integral. – Satish Ramanathan Jan 02 '16 at 01:14