Determine CDF of $S=X+Y$ using partial fraction decomposition. $X\sim \mathrm{Exp}(\alpha)$ independent of $Y\sim \mathrm{Exp}(\beta)$

Question

Let $X\sim \mathrm{Exp}(\alpha)$ and $Y\sim \mathrm{Exp}(\beta)$ be independent and assume $0<\alpha<\beta$.

I'm asked to determine the CDF of $S=X+Y$ using the following method: Determine the moment generating function of $S$ and perform a partial fraction decomposition. Explain how you can find out the PDF and CDF of $S$ by 'simply looking' at this partial fraciton decomposition.

As $X$ and $Y$ are independent, we have $M_S(t)=\frac{\alpha}{\alpha-t}\frac{\beta}{\beta-t}$ for $t<\alpha$. How can I use partial fraction decomposition here?

Okay, so we have $$M_S(t)=\frac{\frac{\alpha\beta}{\beta-\alpha}}{\alpha-t}-\frac{\frac{\alpha\beta}{\beta-\alpha}}{\beta-t}$$

Answer 1

As an alternative method we can use convolution to determine the density of the sum; for $t>0$ we have:

\begin{align} f_S(t) &= (f_X\star f_Y)(t)\\ &= \int_{\mathbb R} f_X(s)f_Y(t-s)\ \mathsf ds\\ &= \int_0^t \alpha e^{-\alpha s}\beta e^{-\beta (t-s)}\ \mathsf ds\\ &= \frac{\alpha \beta }{\beta-\alpha }\left(e^{-\alpha t}-e^{-\beta t}\right). \end{align} The distribution function is found by integrating the density; for $t>0$ we have \begin{align} F_S(t) &= \int_0^t f_S(s)\ \mathsf ds\\ &= \int_0^t \frac{\alpha \beta }{\beta-\alpha }\left(e^{-\alpha s}-e^{-\beta s}\right)\ \mathsf ds\\ &= \frac{\alpha +\alpha \left(-e^{\beta (-t)}\right)+\beta \left(e^{\alpha (-t)}-1\right)}{\alpha -\beta }. \end{align}