Geometric sum of exponential random variables

Here’s an interesting result: Let {N\sim\mathrm{Geo}(p)}, i.e. {\mathbb P(N=n) = p(1-p)^{n-1}, n=1,2,3,\ldots} and assume that {X} is a continuous random variable such that {X\mid N} follows {\mathrm{Erlang}(N,\lambda)} distribution. What is the distribution of {X}?

For {x>0} we have

\displaystyle  \begin{aligned} f_X(x) &= \sum_{n=1}^\infty f_{X\mid N=n}(x)\\ &= \sum_{n=1}^\infty p(1-p)^{n-1}\frac{(\lambda x)^{n-1}}{(n-1)!}\lambda e^{-\lambda x}\\ &= \lambda p e^{-\lambda x} \sum_{n=1}^\infty \frac{((1-p)\lambda x)^{n-1}}{(n-1)!}\\ &= \lambda p e^{-\lambda x}e^{(1-p)\lambda x}\\ &= \lambda p e^{-\lambda p x}, \end{aligned}

so that {X\sim\mathrm{Exp}(\lambda p)}.

Note that if {\{N(t):t\geqslant 0\}} is a Poisson process with intensity {\lambda} and {\{X_n : n=1,2,3,\ldots \}} is a Bernoulli process with parameter {p}, then we can consider the split processes {\{N_p(t):t\geqslant0\}}, {\{N_{1-p}(t):t\geqslant 0 \}}, where {\{T_n\}} are the arrival times of the original process, and {N_p(t)} is the number of arrivals in {(0,t]} with {B_n=1}, {N_{1-p}(t)} the number of arrivals in {(0,t]} with {B_n=0}. It is well-known that {N_p(t)} and {N_{1-p}(t)} are independent Poisson processes with intensity {\lambda p} and {\lambda(1-p)}, respectively. So {X} is the interarrival distribution of {N_p(t)}.

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