I have a poisson process with 0.1 observations per minute. What is the expected time of the 2nd arrival, given that the 2nd arrival occurs in the first 2.5 minutes?
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1 Answer
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Without conditioning on the 2nd arrival occurring in the first 2.5 minutes, we know the second arrival is distributed as
$$ X\sim\text{Erlang}(2,\lambda), $$
where you have rate $\lambda=0.1$ for arrivals. In what follows I'll assume you want the second arrival; if you were looking for the nth arrival, it would be $\text{Erlang}(n,\lambda)$ and the algebra that follows would need to be updated accordingly.
If $p_X(x)$ is the erlang pdf, then the pdf of our distribution conditioned on the second arrival happening by time $t$ is
$$ p_{X|X\leq t}(x) = \frac{p_X(x)}{Pr(X\leq t)}\mathbb{I}_{x\leq t} $$
Using the Erlang pdf $p_X(x) = \lambda^2xe^{-\lambda x}$ and cdf $Pr(X\leq t) = 1-e^{-\lambda t}(1+\lambda t)$, it is an exercise in integration by parts to obtain
$$ \mathbb{E}[X|X\leq t] = \frac{1}{\lambda}\cdot\frac{2-e^{-\lambda t}(\lambda^2t^2+2\lambda t+2)}{1-e^{-\lambda t}(1+\lambda t)} $$
For your parameters of $\lambda=0.1$ and $t=2.5$, $\mathbb{E}[X|X\leq t]\approx 1.632$. As $t\rightarrow\infty$, $E[X|X\leq t] \rightarrow 2/\lambda$, which is the unconditional expectation of the second arrival.
