$Y = X + Z$
$X$ 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.
Best Answer
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.