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?


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.

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