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RV $X$ and $Y$ have joint PDF: $$f_{X,Y}(x,y)= \begin{cases} 8xy & 0\le y \le x \le1 \\ 0 & \text{otherwise} \end{cases}$$

Find PDF of W=X+Y

I know that I need to use : $f_W(w)=\int_{-\infty}^\infty f_{X,Y}(x,w-y) \, dx$. But im confused about limits. I saw some examples but couldnt figure out how they choose limits. (obviously not $-\infty$ to $\infty$) and it seems not $(0$ to $y$ and $y$ to $x)$. Can anyone explain it in simple way?

sorry for my bad english but i think i hit the point

Andy
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    Functional duplicate. – Did Jan 08 '15 at 15:37
  • @Did There says taking integal in two different region. $0\le w \le 1$ and $1\le w \le 2$ in your solution there is only $0\le w \le 1$ part whats the difference between that question and this question ? why u didnt compute $1\le w \le 2$ part? – Andy Jan 08 '15 at 16:13
  • In the other solution one computes a PDF on $w\gt0$, not on $0\lt w\lt1$. Anyway, the point is that the method fully applies to your case. – Did Jan 08 '15 at 16:40

2 Answers2

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You wrote $$0\le y\le x\le 1.\tag1$$ That is how you choose limits. Suppose you want to integrate the density over its whole domain and get $1$. You have $$ \int_0^1 \cdots\cdots \,dx $$ and inside that you have an integral with respect to $y$. According to $(1)$, $y$ must remain between $0$ and $x$, so you have $$ \int_0^1 \left( \int_0^x\cdots\cdots\,dy \right) \,dx. $$ Alternatively, you can put the integral with respect to $y$ on the outside, so you have $$ \int_0^1 \cdots\cdots\,dy. $$ Inside that you integrate with respect to $x$. According to $(1)$, $x$ runs from $y$ to $1$, so you have $$ \int_0^1\left( \int_y^1 \cdots\cdots\,dx \right)\,dy. $$

Michael Hardy
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  • ohh my bad i dont need to take double integal, i just need to single intergral with respect to one of the variables. But i tried to compute with that double integral and its ended up with just single $1$ as an answer. – Andy Jan 08 '15 at 15:50
  • Then your single integral should look like one of the two inner integrals in what is done above, not like the outer ones. – Michael Hardy Jan 08 '15 at 15:57
  • @Micheal Hardy But as in the title stated. If i just take an integral with respect to one of the variables by making other one like $x=w-y$ and taking integral looks enough – Andy Jan 08 '15 at 16:10
  • @Andy : Are you saying $w$ would be a constant so that $x$ would decrease as $y$ increases? If so, that might effect the bounds of integration. – Michael Hardy Jan 08 '15 at 17:03
  • PS: Note the spelling of my first name. That is the usual spelling; the way you spelled it is unusual. – Michael Hardy Jan 08 '15 at 17:03
  • No $w$ is not constant but like your expression in inner integral we assuming that $x$ is constant and taking integral with respect to $y$ right ? like that we assuming that $w$ is constant and eliminate $y$ from variables. and taking integral with respect to $x$. what u have write is right so i gave vote, but was not the answer for the question, @MichaelHardy – Andy Jan 08 '15 at 19:39
  • Anyways i got the answer from Did`s comment under the question – Andy Jan 08 '15 at 19:40
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Express the joint density function using indicators, substitute $y=w-x$, rearrange the indicators to give the bounds on $x$ relative to $w$, and then integrate with respect to $x$ over those bounds to give the density function for $W = X+Y$.

$$\begin{align} f(x, y) & = 8 xy \;\mathbf 1_{0\le y\le x\le 1} \\[2ex] f(x, w-x) & = 8 x(w-x) \;\mathbf 1_{0\le (w-x)\le x\le 1} & y = w- x \\ & = 8 (wx-x^2) \;\mathbf 1_{0\le x\le w\le 2 x \le 2} \\ & = 8 (wx-x^2) \;\mathbf 1_{0\le w\le 2}\;\mathbf 1_{w/2\le x \le \min(w,1)} \\[2ex] f_{\small W}(w) & = \mathbf 1_{0\le w\le 1}\int_{w/2}^w 8(wx-x^2)\operatorname d x + \mathbf 1_{1 < w\le 2}\int_{w/2}^1 8(wx-x^2)\operatorname d x \\[1ex] & = \frac{2w^3}{3}\mathbf 1_{0\le w\le 1} + \frac{12w -8-2w^3}{3} \mathbf 1_{1\le w\le 2} \end{align}$$

Graham Kemp
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