$Z = \sum_{i=1}^T X_i$, $T$ ~ $Geo(p)$, $X_i$ ~ $exp(\lambda)$. Then, $Z$ ~ $exp(p\lambda)$

Question

Let $X_1, X_2, \dots$ be independent random variables with distribution exp($\lambda$). Let $T$ be a random variable taking values in $\{1, 2, \dots\}$ with distribution $Geo(p)$, independent of $X_i$.

Show that $Z = \sum_{i=1}^T X_i$ has distribution exp$(p\lambda)$.

So I know that T is a random stopping time with distribution $Geo(p)$, that is, $\mathbb{P}(T = n) = (1 -p)^np$.

Then, $Z = \sum_{i=1}^T X_i = \sum_{i=1}^n 1_{\{T =n\}}X_i$.

Now I would like to use charactheristic functions to conclude that $Z$ has distribution exp($p\lambda), that is, \psi_Z(t) = \frac{p\lambda}{p\lambda - it}$.

I also know that as $X_1, X_2, \dots$ are independent, $\psi_{X_1 + X_2 + \dots + X_n}(t) = \prod_{i=1}^{n} \psi_{X_i}(t) = \prod_{i=1}^n \frac{\lambda}{\lambda - it}$.

Answer 1

Using the law of total expectation, $$ \mathbb{E}[e^{itZ}]=\sum_{n=1}^{\infty}\mathbb{E}[e^{itZ}\mid T=n]\mathbb{P}(T=n)$$ Next, $$ \mathbb{E}[e^{itZ}\mid T=n]=\mathbb{E}[\exp(it(X_1+\dots+X_n))]=\Big(\frac{\lambda}{\lambda-it}\Big)^n$$ since the $X_i$ are independent of $T$, so $$ \sum_{n=1}^{\infty}\mathbb{E}[e^{itZ}\mid T=n]\mathbb{P}(T=n)=\sum_{n=1}^{\infty}\Big(\frac{\lambda}{\lambda-it}\Big)^n(1-p)^{n-1}p$$ $$ =\frac{p\lambda}{\lambda-it}\sum_{m=0}^{\infty}\Big[\frac{(1-p)\lambda}{\lambda-it}\Big]^m=\frac{p\lambda}{\lambda-it}\frac{1}{1-\frac{(1-p)\lambda}{\lambda-it}}=\frac{p\lambda}{\lambda-it-\lambda+p\lambda}=\frac{p\lambda}{p\lambda-it}$$