Expectation of geometric summation of exponential random variables

We have $\{X_i\}_{i\in\mathbb N}$ as a sequence of independent exponentially distributed rv's with parameter $\lambda$. We also have, $Y_N =\sum_{i=1}^{N} X_i$.

I need to prove that, $Y_N$ has the exponential distribution with parameter $p\lambda$ where $N$ is a geometrically distributed rv with parameter $p$. Also $p$ here is independent of $\{X_i\}_{i\in\mathbb N}$.

Any help is highly appreciated. thanks.

Solution 1:

$$\begin{align*} F_Y(y) &= \Pr[Y \le y] = \sum_{n=1}^\infty \Pr[Y \le y \mid N = n]\Pr[N = n] \\ &= \sum_{n=1}^\infty \int_{t=0}^y \frac{\lambda^n t^{n-1} e^{-\lambda t}}{\Gamma(n)} \, dt \cdot (1-p)^{n-1} p \\ &= \int_{t=0}^y \lambda p e^{-\lambda t} \sum_{n=1}^\infty \frac{(\lambda t (1-p))^{n-1}}{(n-1)!} \, dt \\ &= \int_{t=0}^y \lambda p e^{-\lambda t} e^{\lambda t (1-p)} \, dt \\ &= \int_{t=0}^y \lambda p e^{-\lambda p t} \, dt \\ &= 1 - e^{-\lambda p y}, \end{align*}$$ which is clearly exponential with parameter $\lambda p$.