Question: I have two independent random variables (say $X$ and $Y$) such that $X \sim U[0,1]$ and $Y \sim$ Exp$(1)$, and I want to find the PDF of $Z=X+Y$.
My attempt:
I know $f_X(x)=1$ for $x \in[0,1]$, and $f_Y(y)=e^{-y}$ for $y \in [0, \infty)$.
I also know that, due to their independence, $f_Z(z)=(f_X * f_Y)(z)$ where $(f_X * f_Y)(z)$ is the convolution of $f_X$ and $f_Y$.
Furthermore, $(f_X * f_Y)(z)=\int^{\infty}_{-\infty} f_X(z-y)f_Y(y) dy = \int^{\infty}_{-\infty} f_Y(z-x)f_X(x) dx$.
However, I am unsure of a few things:
- Can I use a convolution approach even though $f_X$ and $f_Y$ are not defined for all real numbers?
- If I can, how would I determine the bounds of the integral given $f_X$ and $f_Y$ are defined for different subsets of the real numbers ($x \in[0,1]$ and $y \in [0, \infty)$ respectively)?
Context: Ultimately, I need to compute $P(Z>z)$ for two different cases (when $z\in[0,1]$ and when $z>1$), so I planned on integrating $f_Z(z)$ to get the CDF for $Z$.
Any help would be greatly appreciated.
Best Answer
When you say $f_X(x)=1$ for $0<x<1$ what you really mean is $f_X(x)=1$ for $0<x<1$ an d $f(x)=0$ for all other $x$. All density functions are defined on the entire real line. So there is no problem in using the convolution formula.
In this case $(f_X*f_Y)(z)=\int_{-\infty}^{\infty} f_X(z-y)f_Y(y) dy$. [This is the general formula for convolution]. Let $z >0$. Note that $f_Y(y)=0$ if $y <0$ and $f_X(z-y)=0$ if $z-y \notin (0,1)$ i.e., if $y \notin (z-1,z)$. Hence integration is over all positive $y$ satisfying $z-1<y<z$. In order to carry out this integration you have to consider two cases: $z >1$ and $z <1$. In the first case the integration is from $z-1$ to $z$. In the second case it is from $0$ to $z$.