Poisson Process: distribution for time until next arrival

Question

The arrival of a subway follows a Poisson Process with rate $\lambda$ and each subway that arrives is full independently with probability $1-p$ such that the arrival of the next (non-full) subway you can take is distributed like $T =\sum_{i=1}^{N} X_{i}$, where each $X_{i}$ ~ Exp($\lambda$) and $N$ ~ Geo($p$). Conceptually, why does $T$ ∼ Exp($p\lambda$)?

Answer 1

Arrival of Subway(s)

Arrival of subway follows a Poisson process with rate $\lambda$, means:

In the next arbitrary $t$ time units, the expected number of subway arrivals is $\lambda\cdot t$. Hence the term “rate”.

Arrival of Non-Full Subway(s)

Each subway only has a probability $p$ of being non-full, means:

In the next arbitrary $t$ time units, the expected number of non-full subway arrivals is only $p\lambda\cdot t$.

When you compare with the first section, notice that arrival of non-empty subway(s) follow a Poisson process with rate $p\lambda$

Answer 2

The arrivals of non-full trains follow the Poisson process with rate $p\lambda$. Also, the interarrival time in Poisson process with rate $\lambda$ is exponential r.v. with parameter $\lambda$. So the interarrival time for non-full trains is $$ T \sim Exp(p\lambda)$$