Consider a [[Poisson Process]] $X_1 \sim \mathrm{Pois}(\lambda)$, but instead of observing a fixed time interval, the interval length is a [[Random Variable]] from [[Exponential Distribution]], $T \sim \mathrm{Exp}(\mu)$. We want to compute the probability of observing $k$ successes until $T$. ![[poisson-process-random-interval-1.png|center|400]] ## Mathematical Approach 1. *Challenge:* The interval length $T$ is not fixed but is itself a random variable. Hence, we cannot directly use the [[Probability Mass Function|PMF]] of the [[Poisson Distribution]] (which is for a fixed interval $\tau$). 2. *Conditional probability:* However, we can condition on the interval length $(T=t)$ and calculate the conditional probability of observing $k$ successes $\mathbf P(N_T=k \vert T=t)$. $ \mathbf P(N_t=k \vert T=t)=\frac{\lambda^k e^{-\lambda t}}{k!}$ 3. *Total probability theorem:* Using [[Total Probability Theorem]] we compute the unconditional probability, by integrating over all possible values of $T$, weighted by the density $f_T(t)$. $ \mathbf P(N_T=k)= \int_0^\infty \mathbf P(N_t=k \vert T=t) * f_T(t) * dt $ 4. *Substitute the exponential:* Replace $f_T(t)$ with the actual [[Probability Density Function|PDF]]. $\mathbf P(N_T=k) =\int_0^\infty \frac{\lambda^k e^{-\lambda t}}{k!}*\mu e^{-\mu t}*dt $ ## Intuitive Approach Since $T$ is an exponential r.v. we can view it as the first arrival in a Poisson process. We denote it as $X_2 \sim \mathrm{Pois}(\mu)$. By mixing $X_2$ with the original process $X_1$ we create a [[Poisson Process#Merged Poisson Process|merged Poisson process]] with parameter $(\lambda+\mu)$. ![[poisson-process-random-interval-2.png|center|400]] 1. In this merged process each success has probability $\frac{\mu}{\lambda + \mu}$ to originate from $X_2$. 2. In the binomial perspective we call each success a trial, that either comes from $X_1$ (no success) or from $X_2$ (success). We want to know the number of trials needed, until the first success. 3. This is the definition of a r.v. from a [[Geometric Distribution]] with parameter $(\frac{\mu}{\mu+\lambda})$. $ \mathbf P(N_T=k)=\text{Geom}\Big (\frac{\mu}{\lambda + \mu}\Big) $