The random incidence paradox explains why the expected inter-arrival time observed at a randomly chosen point in time is longer than the average inter-arrival time suggested by the process itself. This phenomenon is commonly studied in the context of the [[Poisson Process]] **Intuition:** The paradox arises because, during the selection of a random point in time, we are more likely to observe a longer inter-arrival time, since they make up more time and thus they are more likely to be selected. **Setup:** - Let $X \sim \mathrm{Pois}(\lambda)$ be a Poisson process with rate $\lambda$. - The inter-arrival times consequentially follow an [[Exponential Distribution]], $T_i \sim \mathrm{Exp}(\lambda)$. At a random observation point $t^\star$, the inter-arrival time containing $t^\star$ can be expressed as $V-U$, where: - $U$ is the arrival time before $t^\star$ - $V$ is the arrival time after $t^\star$ ![[random-incidence-paradox.png|center|400]] The inter-arrival time can be decomposed into two independent exponential [[Random Variable|r.v's.]] $ V-U = \underbrace{(V- t^\star)}_{\mathrm{Exp}(\lambda)}+ \underbrace{(t^\star -U)}_{\mathrm{Exp}(\lambda)} $ Applying [[Linearity of Expectations]] we can compute the expected length of the interval between $U$ and $V$. $ \begin{align} \mathbb{E}[V-U]&=\mathbb E[V-t^\star] + \mathbb E[t^\star - U]\\[2pt] \mathbb{E}[V-U]&=\frac{1}{\lambda}+\frac{1}{\lambda} = \frac{2}{\lambda} \end{align} $ The observed inter-arrival time at a random point is twice the regular inter-arrival time of the process. > [!note:] > The “random-incidence-paradox” is not special to the Poisson process. It is only the magnitude of increase in the inter-arrival time that changes. So, the sampling method matters!