Given two random variables $X$ and $Y$ with joint probability density function $$ f_{X,Y}(x,y) = \begin{cases} 2, & \text{if x $\ge$ 0, y $\ge$ 0 , $x + y$ $\le$ 1} \\ 0, & \text{elsewhere} \end{cases} $$

let $W = X + Y$

My question is how I calculate the probabilty density function $f_W(w)$ from here?

If calculated the marginal probabilty density functions $$f_x(x) = 2-2x \text{ for } 0 \le x \le 1$$ $$f_Y(y) = 2-2y \text{ for } 0 \le y \le 1$$

But i have no clue how to go from here to the probability density function $f_W(w)$

Any pointers or solutions are highly appreciated!

Thanks in advance!

  • $\begingroup$ The marginals you calculated look "suspicious". They can take positive values outside $[0,1]$ while $(X,Y)$ has a subset of $[0,1]^2$ as support. There is definitely something wrong with your calculation. Fortunately you do not need them for calculating $f_W(w)$. $\endgroup$ – drhab Aug 12 at 15:09
  • $\begingroup$ Made a typo in the marginals, they should be correct now. $\endgroup$ – xbys Aug 12 at 15:13
  • $\begingroup$ Not completely. Also $0\leq x\leq 2$ must be changed into $0\leq x\leq 1$. Same story of $y$. $\endgroup$ – drhab Aug 12 at 15:15
  • $\begingroup$ You're right, they should be complete now . $\endgroup$ – xbys Aug 12 at 15:17

For $w\geq0$ define triangle: $$\Delta_w=\{(x,y)\in[0,1]^2\mid x+y\leq w\}$$

Then $(X,Y)$ has uniform distribution on triangle $\Delta_1$

For $w\in[0,1]$ we find: $$F_W(w)=P(X+Y\leq w)=P((X,Y)\in\Delta_w)=2\lambda(\Delta_w)=w^2$$ where $\lambda$ denotes the Lebesgue measure.

The PDF of $W$ can be found as derivative of the CDF, i.e. for $w\in[0,1]$ we have:$$f_W(w)=2w$$ Further we can take $f_W(w)=0$ if $w\notin[0,1]$.


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