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Find the PDF of $W=X+Y$,when X and Y have the joint probability density function

$$f_{X,Y}(x,y)=\begin{cases} 1, &0\le x \le 1,0 \le y \le 1 \\[0.5ex] 0, & \text{otherwise} \end{cases}$$

My solution is let $W=X+Y,X_1=X$,then

$f_{X_1,W}(x ,w)=f_{X,Y}(x,w-x)=1\times |J|=1\times 1=1 , 0\le x \le w \le 2$

so now i just $f_W(w)=\int^{w}_{0}1 dx=w$ ,for $0\le w \le 2$

Am i wrong?

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    Your proposed answer does not integrate to 1. You should give the inequality $x_1\le w$ a harder look. Does it guarantee that $x\le 1$? – 2017-02-22

1 Answers 1

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Rather than $0\le x \le w \le 2$ you require $0\le x\le 1, 0\le w-x\le 1$, which is:$$0\le w\le 2~,~ \max\{0, \underline\quad\}\le x\le \min\{1,\underline\quad\}$$

So $$f_W(w) =\begin{cases}\displaystyle \int_0^\underline\quad f_{X,Y}(x,w-x)\operatorname d x &:& 0\le w\lt\underline\qquad\\[1ex]\displaystyle \int_\underline\quad^\underline\quad f_{X,Y}(x,w-x)\operatorname d x &:& \underline\qquad\le w\lt \underline\qquad \\[1ex]\displaystyle \int_\underline\quad^1 f_{X,Y}(x,w-x)\operatorname d x &:& \underline\qquad\le w\le 2 \\[2ex] 0 &:& w\lt 0~\vee~2\lt w \end{cases}$$

Fill in the blanks as appropriate.

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    Well,i do not understanding here, max{0,−−}≤x≤min{1,−−},so you mean i have to consider the range of w,that is,$0 \le w \le 1$ ,and $1\le w \le 2$? – 2017-02-23