Answer

Note that the required probability distribution is given by the following formula

\[ \begin{aligned} p_{X+Y}(x) = \int_{-\infty}^{\infty} p_{X}(x-y) \times p_{Y}(y) dy\end{aligned} \]

However, note that the exponential distribution is not positive everywhere. For values \(< 0\), the probability density is 0. Hence, we break the integral into three parts as follows

\[ \begin{aligned} p_{X+Y}(x) = \int_{-\infty}^{0} p_{X}(x-y) \times p_{Y}(y) dy + \int_{0}^{x} p_{X}(x-y) \times p_{Y}(y) dy + \int_{x}^{\infty} p_{X}(x-y) \times p_{Y}(y) dy\end{aligned} \]

Carefully note that for \(y\) in range \((-\infty,0]\), \(p_{Y}(y) = 0\), and in the range \([x,\infty)\), \(x-y < 0\), which implies \(p_{X}(x) = 0\). Hence,

\[ \begin{aligned} p_{X+Y}(x) &= \int_{0}^{x} p_{X}(x-y) \times p_{Y}(y) dy\newline &= \lambda \mu \exp(-\lambda x) \int_{0}^{x} \exp((\lambda - \mu)y) dy\newline &= \frac{\lambda \mu}{\lambda - \mu} \exp(-\lambda x) (\exp((\lambda - \mu)x) - 1)\newline &= \frac{\lambda \mu}{\lambda - \mu}(\exp(\mu x) - \exp(-\lambda x))\end{aligned} \]