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Find the density function of $Z=X+Y$, $X,Y$ have the joint density function $f(x,y) = \frac{1}{2} (x+y) e^{-(x+y)},\, x,y \geq 0$.

My initial idea is to calculate the distribution function of Z like this:

$P(Z < z) = P(X+Y < z) = P(X < z-Y)$

$F_{z} = \frac{1}{2}\int\limits^{z}_{0}\int\limits^{z-y}_{0} (x+y) e^{-(x+y)} dx\, dy$

and then calculate its derivative $F_{z}'$

Is this the way to go ?

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Yes, that looks right. But notice that you are computing a double integral and then differentiating the result to get the density. Since "differentiation undoes integration" it is possible to combine the two steps and compute the density using just one integration. The resulting single integral is shown in this answer to a related problem. –  Dilip Sarwate Apr 26 '13 at 11:03
    
I noticed that, but I am unable get those densities $f_x, f_y$ such that $f(x,y) = f_x(x) \cdot f_y(y)$ –  jed Apr 26 '13 at 11:29
    
Err... you should have stopped reading my answer at the point where it said $$\begin{align}f_Z(z) &= \cdots\\&=\cdots\\&\vdots\\&= \int_{-\infty}^{\infty} f_{X,Y}(x,z-x)\,\mathrm dx\end{align}$$ which applies to all random variables, not just to independent random variables. –  Dilip Sarwate Apr 26 '13 at 11:34
    
ok, with the hint it becomes \begin{array}[rl]{l} f_{X+Y}(z) &= \frac{1}{2} \int\limits_{0}^{z} (x+z-x)\, e^{-(x+z-x)}\, dx \\ &= \frac{1}{2} \int\limits_{0}^{z} z\, e^{-z}\, dx \\ &= \frac{1}{2} z\cdot e^{-z} \int\limits_{0}^{z} \, dx \\ &= \frac{1}{2} z^2\,e^{-z} \\ \end{array} –  jed Apr 26 '13 at 12:50
1  
See? Wasn't that a lot easier? I recommend that you write up your solution, including the way you got to the simple calculation, and post it as an answer to your own problem. After a while, you could even accept your own answer as the best of all the answers if you like. By the way, such actions are not just acceptable but also an approved way of proceeding in such cases. –  Dilip Sarwate Apr 26 '13 at 14:22

1 Answer 1

Joint density is $f(x,y) = \frac{1}{2} (x+y) e^{-(x+y)},\, x,y \geq 0$ and the distribution function is

$F_Z(z) = P(Z < z) = P (X+Y < z) = P ( X < z, Y < X -z)$

then we have for the density of Z

\begin{array}[rl]{l} f_{Z}(z) &= \frac{\partial}{\partial z} F_z(z) = \frac{\partial}{\partial z} \int\limits_{0}^{z}\int\limits_{0}^{x-z} f(x,y)\,dy\,dx \\ &= \frac{\partial}{\partial z} F_z(z) = \int\limits_{0}^{z}\frac{\partial}{\partial z} \int\limits_{0}^{x-z} f(x,y)\,dy\,dx \\ &= \int\limits_{0}^{z} f(x,z-x)\cdot 1 - f(x,0)\cdot 0 \,dx \\ &= \frac{1}{2} \int\limits_{0}^{z} z\, e^{-z}\, dx \\ &= \frac{1}{2} z\cdot e^{-z} \int\limits_{0}^{z} \, dx \\ f_{Z}(z) &= \frac{1}{2} z^2\,e^{-z} \\ \end{array}

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I think there may be some typos such as $\frac{\partial}{\partial x}$ instead of $\frac{\partial}{\partial z}$ and the third line really should resolve to the integral of $f(x,z-x)$ and not of $f(x,x-z)$ the way you have it. –  Dilip Sarwate May 15 '13 at 14:17

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