Arrival Counts and Waiting Times

Definition Let $N(t)$, for $t \ge 0$, be a homogeneous Poisson process with rate $\lambda > 0$. So $N(0) = 0$, the counts in disjoint time intervals are independent, and the count in any interval of length $t$ is $\operatorname{Pois}(\lambda t)$. Here $N(t)$ is the number of arrivals up to and including time $t$. The first arrival time $T_1$ is the time at which the count first reaches $1$. For every $t \ge 0$,

because the first arrival comes after time $t$ exactly when no arrival has happened by time $t$. Taking complements gives $\{T_1 \le t\} = \{N(t) \ge 1\}$.

Definition In a homogeneous Poisson process with rate $\lambda > 0$, the first arrival time has the exponential distribution with rate $\lambda$, written $T_1 \sim \operatorname{Exp}(\lambda)$. For $t \ge 0$ its survival function, CDF and density are

and $F(t) = f(t) = 0$ for $t < 0$. The survival function is $P(N(t) = 0)$ from the previous card, and the density is the derivative of $F$ for $t > 0$. Since $e^{-\lambda t} \to 0$ as $t \to \infty$, the first arrival happens at some finite time with probability $1$. The mean is $E[T_1] = 1/\lambda$, the variance is $\operatorname{Var}(T_1) = 1/\lambda^2$ and the standard deviation is $\operatorname{SD}(T_1) = 1/\lambda$. With $\lambda$ in arrivals per unit of time and $t$ in the same unit, the product $\lambda t$ is a pure number, the density is measured per unit of time, and the mean $1/\lambda$ is a time.

Definition For all $s \ge 0$ and $t \ge 0$, a waiting time $T_1 \sim \operatorname{Exp}(\lambda)$ satisfies

The conditioning event has probability $e^{-\lambda s} > 0$, and the event $\{T_1 > s + t\}$ lies inside $\{T_1 > s\}$, so the conditional probability is the ratio of the two survival probabilities. Given that no arrival has happened by time $s$, the remaining wait $T_1 - s$ is again $\operatorname{Exp}(\lambda)$. In a Poisson process the same holds from any fixed time $s$. The count $N(s + t) - N(s)$ is $\operatorname{Pois}(\lambda t)$ and is independent of the arrivals up to time $s$, so the wait from time $s$ until the next arrival is $\operatorname{Exp}(\lambda)$ whatever happened before $s$.

Definition Let $T_k$ be the time of the $k$th arrival for $k = 1, 2, \dots$, and set $T_0 = 0$. The $k$th interarrival time is the gap

so $G_1 = T_1$ is the first wait. In a homogeneous Poisson process with rate $\lambda > 0$, the gaps $G_1, G_2, \dots$ are independent and each has the $\operatorname{Exp}(\lambda)$ distribution. Conversely, if independent $\operatorname{Exp}(\lambda)$ gaps are added up to give arrival times, counting the arrivals up to each time $t$ produces a homogeneous Poisson process with rate $\lambda$.

Definition For $k \ge 1$, the $k$th arrival time is the sum of the first $k$ gaps,

By linearity of expectation and the independence of the gaps,

For $k = 1$ this is the exponential first wait. For $k \ge 2$ the arrival time $T_k$ is not exponential, because an exponential variable with mean $k/\lambda$ has variance $k^2/\lambda^2$, which is larger than $k/\lambda^2$.

Definition For every $k \ge 1$ and $t \ge 0$,

since the $k$th arrival has happened by time $t$ exactly when at least $k$ arrivals have happened by then. Taking complements and using $N(t) \sim \operatorname{Pois}(\lambda t)$,

For $k = 1$ the sum has the single term $e^{-\lambda t}$, the exponential survival function again.