From a Counting Martingale to an Expected Arrival Time
A Poisson process counts events as they occur, but its arrival times are random. To calculate the mean time of the \(k\)th event, one can relate the count accumulated by that time to the linear rate of the process. The key is to use the compensated counting process \(N_t-\lambda t\), a martingale, and to stop it first at a bounded time. This avoids assuming in advance that the unbounded arrival time has a finite expectation—the very fact we want to establish.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a Poisson process of rate \(\lambda>0\), with its usual right-continuous natural filtration \((\mathcal{F}_t)_{t\ge0}\), where \(\mathcal{F}_t\) records the counts observed up to time \(t\). Thus \(N_0=0\), its increments over disjoint time intervals are independent, and \(N_t-N_s\) has a Poisson distribution with mean \(\lambda(t-s)\) for \(0\le s<t\). In particular, its jumps are of size one.
The process \(M_t\) is an integrable martingale: for \(s\le t\), independent increments give \(\mathbb{E}[N_t-N_s\mid\mathcal{F}_s]=\lambda(t-s)\), and hence \(\mathbb{E}[M_t\mid\mathcal{F}_s]=N_s-\lambda s=M_s\). We use the bounded optional stopping theorem: for a right-continuous martingale and a bounded stopping time \(\sigma\), \(\mathbb{E}[M_\sigma]=\mathbb{E}[M_0]\). The theorem applies to \(M\) and \(T_k\wedge t\), which is bounded by the deterministic time \(t\).
The Expected-Time Result and Its Proof
Proof. First show that \(T_k\) is finite almost surely. For every \(t\ge0\), the event \(\{T_k>t\}\) is the event that fewer than \(k\) arrivals have occurred by time \(t\), so $$ \mathbb{P}(T_k>t)=\mathbb{P}(N_t<k)=e^{-\lambda t}\sum_{j=0}^{k-1}\frac{(\lambda t)^j}{j!}. $$ This expression tends to zero as \(t\to\infty\): it is an exponential \(e^{-\lambda t}\) multiplied by a polynomial in \(t\), and \(\lambda>0\). Therefore \(\mathbb{P}(T_k=\infty)=0\).
Fix \(t\ge0\) and set \(\sigma_t=T_k\wedge t\). Since the process has unit jumps, \(N_{\sigma_t}=\min(N_t,k)\): if the \(k\)th arrival has occurred by \(t\), the count at \(\sigma_t=T_k\) is \(k\); otherwise \(\sigma_t=t\) and \(N_t<k\). Bounded optional stopping and \(M_0=0\) now give $$ 0=\mathbb{E}[M_{\sigma_t}] =\mathbb{E}[N_{\sigma_t}]-\lambda\mathbb{E}[\sigma_t] =\mathbb{E}[\min(N_t,k)]-\lambda\mathbb{E}[T_k\wedge t]. $$
As \(t\to\infty\), \(\min(N_t,k)\to k\) almost surely because \(T_k<\infty\) almost surely. These variables are bounded by \(k\), so bounded convergence gives \(\mathbb{E}[\min(N_t,k)]\to k\). Also \(T_k\wedge t\) increases to \(T_k\), so the monotone convergence theorem gives \(\mathbb{E}[T_k\wedge t]\to\mathbb{E}[T_k]\), allowing this expectation initially to be infinite. Taking limits in the displayed identity yields \(k=\lambda\mathbb{E}[T_k]\). Since \(k\) and \(\lambda\) are positive and finite, \(\mathbb{E}[T_k]=k/\lambda<\infty\).
It remains to check the unbounded stopping condition rather than simply assume optional stopping at \(T_k\). For every \(t\), \(N_{t\wedge T_k}\le k\) and \(t\wedge T_k\le T_k\), so $$ |M_{t\wedge T_k}|\le k+\lambda T_k. $$ The right side is integrable because \(\mathbb{E}[T_k]=k/\lambda<\infty\). This domination also proves uniform integrability: if \(Z=k+\lambda T_k\), then for any \(K>0\), $$ \sup_{t\ge0}\mathbb{E}\!\left[|M_{t\wedge T_k}|\mathbf{1}_{\{|M_{t\wedge T_k}|>K\}}\right] \le \mathbb{E}[Z\mathbf{1}_{\{Z>K\}}]\longrightarrow0 $$ as \(K\to\infty\), by integrability of \(Z\). Almost surely, \(M_{t\wedge T_k}\to M_{T_k}\), and the same integrable bound permits dominated convergence of expectations. Since \(\mathbb{E}[M_{t\wedge T_k}]=0\) for every \(t\), it follows that \(\mathbb{E}[M_{T_k}]=0\). At the arrival time \(N_{T_k}=k\), so this last identity gives \(k-\lambda\mathbb{E}[T_k]=0\), in agreement with the value already established. \(\square\)
The truncation is important to the logic. Bounded optional stopping gives an identity at \(T_k\wedge t\) without needing any unproved integrability of \(T_k\). Passing to the limit proves that integrability, after which the domination argument verifies optional stopping at \(T_k\) itself.
Worked Example: The Third Arrival at Rate Two
Let \((N_t)_{t\ge0}\) be a Poisson process with rate \(\lambda=2\), and find the expected time \(T_3\) of its third arrival. The rate is positive and \(k=3\) is a positive integer, so the theorem applies directly: $$ \mathbb{E}[T_3]=\frac{3}{2}. $$
Check by integrating the tail. For \(t\ge0\), the third arrival has not occurred by \(t\) exactly when \(N_t\) is \(0\), \(1\), or \(2\). Thus $$ \mathbb{P}(T_3>t)=e^{-2t}\left(1+2t+\frac{(2t)^2}{2}\right) =e^{-2t}(1+2t+2t^2). $$ Since \(T_3\) is nonnegative, its expectation is the integral of this tail. Evaluating the three terms gives $$ \int_0^\infty e^{-2t}\,dt=\frac12,\qquad \int_0^\infty 2t e^{-2t}\,dt=\frac12,\qquad \int_0^\infty 2t^2 e^{-2t}\,dt=\frac12. $$ For example, the last integral is \(2\cdot 2!/2^3=1/2\), using \(\int_0^\infty t^j e^{-at}\,dt=j!/a^{j+1}\) for \(a>0\) and nonnegative integer \(j\). Their sum is \(1/2+1/2+1/2=3/2\), as required.
Worked Example: The Fourth Arrival at Rate One-Half
Suppose the arrival rate is \(\lambda=\tfrac12\), and find the mean time to the fourth arrival. The theorem gives $$ \mathbb{E}[T_4]=\frac{4}{1/2}=8. $$
Check using interarrival times. For a Poisson process of rate \(\lambda\), the successive interarrival times are independent exponential random variables with mean \(1/\lambda\). Here each has mean \(1/(1/2)=2\). If these four interarrival times are denoted by \(\tau_1,\tau_2,\tau_3,\tau_4\), then \(T_4=\tau_1+\tau_2+\tau_3+\tau_4\). Linearity of expectation gives $$ \mathbb{E}[T_4]=2+2+2+2=8. $$ This agrees with the martingale calculation. The check also confirms the scale: halving the arrival rate doubles the expected wait for each arrival.
Proof. Since \(T_r\le T_k\) almost surely, the difference is nonnegative. Both arrival times have finite expectation by the theorem, so linearity of expectation applies: $$ \mathbb{E}[T_k-T_r]=\mathbb{E}[T_k]-\mathbb{E}[T_r] =\frac{k}{\lambda}-\frac{r}{\lambda} =\frac{k-r}{\lambda}. $$ This proves the stated formula. \(\square\)
Worked Example: The Time from the Second to the Fifth Arrival
For a Poisson process with rate \(\lambda=\tfrac32\), find the expected time from the second arrival to the fifth. The expected time of the fifth arrival is $$ \mathbb{E}[T_5]=\frac{5}{3/2}=\frac{10}{3}, $$ and the expected time of the second arrival is $$ \mathbb{E}[T_2]=\frac{2}{3/2}=\frac{4}{3}. $$ Therefore the expected elapsed time is $$ \mathbb{E}[T_5-T_2]=\frac{10}{3}-\frac{4}{3}=\frac{6}{3}=2. $$
Check by counting the interarrival intervals. There are three intervals between arrivals two and five: the third, fourth, and fifth interarrival times. Each has mean \(1/(3/2)=2/3\), so the sum of their means is \(2/3+2/3+2/3=2\). This verifies the elapsed-time calculation.
Why the Stopping Condition Matters
A common mistake is to take the fixed-time identity \(\mathbb{E}[N_t]=\lambda t\) and substitute the random time \(T_k\), writing \(\mathbb{E}[N_{T_k}]=\lambda\mathbb{E}[T_k]\) without justification. For example, at rate \(2\), the third arrival gives \(N_{T_3}=3\); the tempting calculation \(3=2\mathbb{E}[T_3]\) produces \(3/2\), which happens to be the correct answer, but the substitution alone is not a valid argument. The fixed-time identity is not a stopping theorem. At a random time, an optional-stopping condition is needed, and the finiteness of \(\mathbb{E}[T_3]\) cannot be presumed when the calculation is intended to prove it.
The rigorous correction is to stop at \(T_k\wedge t\), where bounded optional stopping applies. The resulting identity both establishes the mean and, through the integrable bound \(k+\lambda T_k\), verifies uniform integrability for passage to the unbounded stopping time. The distinction matters in other stopping problems: an algebraic identity at a random time does not by itself show that the stopped martingale is integrable or that its expectation is preserved.
Check Your Understanding
Use the theorem and its hypotheses to answer the following questions.
- For a Poisson process of rate \(4\), what is the expected time of the sixth arrival?
- Why is \(T_k\wedge t\) a bounded stopping time, and what identity follows from bounded optional stopping?
- Explain how \(\mathbb{P}(T_k>t)=\mathbb{P}(N_t<k)\) shows that \(T_k\) is almost surely finite when \(\lambda>0\).
- What integrable random variable dominates \(|M_{t\wedge T_k}|\), and how does this establish uniform integrability?
- For rate \(3\), compute the expected elapsed time between the first and fourth arrivals, and check it by counting the interarrival intervals.