Find the PDF, and the conditional PDFs of $Y$ when $Y = X + Z$, where $X$ and $Z$ are exponential functions.

Question

$Y = X + Z$
$X$ and $Z$ are independant, and are exponentially distributed with parameters: $\lambda_X=5$ and $\lambda_Z=1$
a) Find the PDF of $Y$.
b) Find the conditional pdf of $Y$ when $X = 2$, and also the conditional PDF of $Y$ when X = x, where $x\in\mathbb{R}$.
c) Find the minimum mean square estimate of $X$ when $Y = y$, where $y > 0$

I have found that $$f_X(x)= \begin{cases} 5e^{-5x}&x \geq 0 \\ 0 & otherwise\end{cases}$$ and $$f_Z(z)= \begin{cases} e^{-z}&z \geq 0 \\ 0 & otherwise\end{cases}$$ but I do not know where to go from there for the first part.

Answer 1

Use a double integral to generate the cdf of $Y=X+Z$ from the pdfs of $X$ and $Z$:

$$F_Y(y_0)=\int\int_{x+z<y_0} f_X(x)f_Z(z)\ dz\ dx$$

In writing down the integral explicitly it will help to sketch the region in the $xz$-plane where $Y\le y_0$ for a fixed but arbitrary $y_0$. Once you have the cdf, differentiate to obtain the pdf. That should get you started.

Answer 2

EDIT:

To calculate $\mathbb{E}[X|Y=y]$

First of all we calculate the joint distribution $f_{XY}(x,y)$ using an auxiliary variable $W=X$

$$\begin{cases} w=x \\ y=x+z \end{cases} $$

$$\begin{cases} x=w \\ z=y-w \end{cases} $$

the Jacobian is $|J|=1$ so the joint distribution is

$f_{WY}(w,y)=5e^{-4w}e^{-y}\mathbb{1}_{(0;+\infty)}(y)\mathbb{1}_{(0;y)}(w)$

This because $0<y-w<\infty$ then $0<w<y<\infty$

now the conditional distribution is ($f_Y(y)$ is immediate via convolution)

$f_{X|Y}(x|y)=\frac{f_{XY}(x,y)}{ f_Y(y)}=\frac{4e^{-4x}}{ 1-e^{-4y}}\mathbb{1}_{(0;y)}(x)$

Concluding the expected value

$$\mathbb{E}[X|Y=y]=\frac{1}{1-e^{-4y}}\int_0^y 4xe^{-4x}dx$$