I have a problem where, in order to solve it, I need to find the distribution of $Z$. Say that $X\sim\text{exp}(\lambda)$ and $Y\sim\text{exp}(\mu)$. I don't want to use the convolution formula but instead the mgf's. I have that
$$M_X(t)=\frac{\lambda}{\lambda-t},\quad \quad M_Y(t)=\frac{\mu}{\mu-t}.$$
Thus,
$$M_Z(t)=M_{X+Y}(t)=M_X(t)M_Y(t)=\frac{\lambda\mu}{\lambda\mu-\lambda t-\mu t+t^2}.$$
I'm trying to find what distribution this is the mgf for. If $X$ and $Y$ were I.I.D, then it would just be an mgf for the gamma distribution. But this is not the case since I can't rewrite it in the correct form.
Is there any way to proceeed here?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\expo{-y/\mu} \over \mu} \braces{\bracks{z - y > 0}{\expo{-\pars{z - y}/\lambda} \over \lambda}}\dd y} \\[5mm] = &\ \bracks{z > 0}{\expo{-z/\lambda} \over \lambda\mu} \int_{0}^{z}\exp\pars{{\mu - \lambda \over \lambda\mu}\,y}\dd y \\[5mm] = &\ \bracks{z > 0}{\expo{-z/\lambda} \over \lambda\mu} {\expo{\pars{\mu - \lambda}z/\pars{\lambda\mu}} - 1 \over \pars{\mu - \lambda}/\pars{\lambda\mu}}\dd y = \bbx{\bracks{z > 0}{\expo{-z/\mu} - \expo{-z/\lambda} \over \mu - \lambda}} \end{align}