Assuming $X$ and $Y$ are i.i.d. random variables let $Z = X + Y$
$$
f_X(x) =
\begin{cases}
\lambda e^{- \lambda x} & x \gt 0 \\
0 & \text{else}
\end{cases}
$$
$$
f_Y(y) =
\begin{cases}
\mu e^{- \mu x} & x \gt 0 \\
0 & \text{else}
\end{cases}
$$
Find the density $f_z$ of $Z$.
I have attempted to use convolution with indicator functions, but am yet to get the right answer.
$$
f_X(x) * f_Y(y) = \int_{-\infty}^{\infty} f_X(x)f_Y(z-x) dx \\
= \int_{0}^{\infty} \lambda e^{- \lambda x} 1_{[x \gt 0]} \mu e^{- \mu (z-x)} 1_{[z-x \gt 0]} dx \\
$$
since $z-x \gt 0$ can be written as $z \gt x$ I believe the product of the two indicator functions should be $\text{max}\{0, z\}$, but this is where I become uncertain as I think that would make the integral:
$$
= \lambda \mu e^{-\mu z} \int_{\text{max}\{0,z\}}^{\infty} e^{- \lambda x} e^{\mu x} dx \\
$$
Intuitively, I feel I have messed up as the integral from z to infinity is contained in both the solutions for $z \gt 0$ and $z \lt 0$. More concretely I know I have messed up as evaluating the integral leaves me with:
$$
\begin{cases}
\mu \lambda e^{- \mu z } \left[ e^{-x(\lambda – \mu)} \right]_z^\infty & z \gt 0 \\
\mu \lambda e^{- \mu z } \left[ e^{-x(\lambda – \mu)} \right]_0^\infty & z \lt 0
\end {cases}
$$
$$
=
\begin{cases}
\frac{\mu \lambda}{\lambda – \mu} e^{- \lambda z } & z \gt 0 \\
\frac{\mu \lambda}{\lambda – \mu} e^{- \mu z } & z \lt 0
\end {cases}
$$
which does not agree with the solution listed in the book I am studying with:
$$
f_Z(x) =
\begin{cases}
\frac{\mu \lambda}{\mu + \lambda} e^{\lambda x} & x \lt 0 \\
\frac{\mu \lambda}{\mu + \lambda} e^{- \mu x} & x \ge 0
\end{cases}
$$
This solution really confuses me, as I do not understand how the convolution of two functions that are strictly equal to 0 when $x \lt 0$ could be a function that has non zero values for $x \lt 0$.
Does anyone have any suggestions on how to fix the integral or intuition on why this convolution works the way it does?
Best Answer
Obviously, the pdf of $Z$ is zero for $Z<0$. Also, since $Z=X+Y$, for any given value of $Z=z$, $X\le z$. Thus, the upper limit of the integral is $z$, not $\infty$. So the pdf of $Z$ for $Z\geq 0$ is
$f_Z(z)=fX(x)∗fY(y)=\int_0^\infty f_X(x)f_Y(z−x)dx=\int_0^z λe^{−λx}μe^{−μ(z−x)}dx$,
that is,
$f_Z(z)=λμe^{-μz}\int_0^z e^{-(λ-μ)x} dx$.
The rest derivation is straightforward. Simply remember to deal with two cases separately, that is, Case I: if $λ = μ$, and Case II: if $λ \neq μ$.