Find distribution of $Z=\frac{X+Y}{2}$ given $f_{X,Y}(x,y)=e^{-(x+y)}$

Question

Excercise

Let $X, Y$ be random variables such that their joint density function is defined by:

$f_{X,Y}(x,y)=e^{-(x+y)}, \enspace x,y>0$.

Find the distribution of Z defined as:

$Z=\frac{X+Y}{2}$.

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Attempt of solution

I am using the fact that:

$F_Z(z)=P\{\frac{X+Y}{2}<z\}$, which implies that:

$F_Z(z)=\int_{0}^{\infty}\int_{0}^{2z-y}e^{-(x+y)}dxdy$.

My problem is that this leads to:

$F_Z(z)=\int_{0}^{\infty}e^{-y}-e^{-2z}dy$,

and this integral does not converge.

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What am I getting wrong here? Thanks in advance.

Answer 1

Without doing any calculation, just observe that $f_{XY}(x,y)=f_X(x)f_Y(y)$ where $X,Y$ are iid $Exp(1)=Gamma(1;1)$ distributed.

Thus $(X+Y)\sim Gamma(2;1)$ and $Z=\frac{1}{2}(X+Y)\sim Gamma(2;2)$ that is

$f_Z(z)=4ze^{-2z}$

$z\geq 0$

Answer 2

HINT

The support for $f_{X,Y}$ is only $x,y > 0$. So the area you're integrating over is a triangle with vertices $(0,0), (2z,0), (0,2z)$ in the $(x,y)$ plane. So the outer integral should not be $\int^\infty_0$...

Answer 3

The integral domain should not go to infinity.

Since $X+Y\leqslant 2z$ and $X>0$, therefore $Y$ cannot be greater than $2z$.

Answer 4

Note that marginal densities of $X$ and $Y$ are $f_X(x)=e^{-x}$ and $f_Y(y)=e^{-y}$. Then, $Z=X+Y$ has density according to the convolution: $$f_Z(z)=\int_0^\infty f_X(x)f_Y(z-x)dx = \int_0^z e^{-x}e^{-(z-x)}dx=ze^{-z}.$$

Density of $W=Z/2$ is

$$f_W(w)=f_Z(2w)|dz/dw| = 4we^{-2w}.$$