Count-Arrival-Time Duality

One Event, Two Ways to Describe It

In a Poisson process, a question about when the kth event occurs can often be answered by asking how many events have occurred by a specified time. These are not merely events with equal probabilities: with consistent endpoint conventions, they are the same event on each sample path. The equivalence lets us choose whichever description makes a calculation simpler.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(N(t))_{t\ge0}$ be a Poisson counting process with state space $\mathbb{N}_0=\{0,1,2,\ldots\}$, time index $t\ge0$, and rate $\lambda>0$. Thus $N(0)=0$, and for each $t\ge0$, $N(t)$ has the Poisson distribution with mean $\lambda t$. We count arrivals in $(0,t]$: an arrival at time $t$ is included in $N(t)$. For each positive integer $k$, let $S_k$ be the time of the kth arrival, with $S_k=\infty$ if fewer than $k$ arrivals ever occur. Equivalently, $S_k=\inf\{u\ge0:N(u)\ge k\}$.

Theorem (Count-arrival-time duality): For every positive integer $k$ and every $t\ge0$, $$ \{S_k\le t\}=\{N(t)\ge k\}, \qquad \{S_k>t\}=\{N(t)\le k-1\}. $$ Consequently, $\mathbb{P}(S_k\le t)=\mathbb{P}(N(t)\ge k)$ and $\mathbb{P}(S_k>t)=\mathbb{P}(N(t)\le k-1)$.

Proof. Fix a sample path and a time $t$. If $S_k\le t$, the kth arrival has occurred by time $t$, so at least $k$ arrivals have been counted by then: $N(t)\ge k$. Conversely, if $N(t)\ge k$, the count has reached $k$ by time $t$, so the kth arrival time satisfies $S_k\le t$. This proves the first equality of events. The events $\{S_k\le t\}$ and $\{S_k>t\}$ are complements, as are $\{N(t)\ge k\}$ and $\{N(t)\le k-1\}$, since $N(t)$ is integer-valued. Taking probabilities proves both probability identities. $\square$

The endpoint matters in these statements. Because $N(t)$ counts arrivals in $(0,t]$, an arrival exactly at $t$ contributes to $N(t)$ and makes $S_k\le t$, not $S_k>t$. At $t=0$, $N(0)=0$ and $S_k>0$ for $k\ge1$, so the stated identities also hold at the left endpoint.

Poisson Formulas for Arrival-Time Probabilities

The Poisson count law gives an explicit version of the duality. For $j\in\mathbb{N}_0$, $\mathbb{P}(N(t)=j)=e^{-\lambda t}(\lambda t)^j/j!$. Summing the appropriate probabilities gives the distribution function and survival probability of $S_k$.

Theorem (Equivalent Poisson tail formulas): For $\lambda>0$, $t\ge0$, and positive integer $k$, $$ \mathbb{P}(S_k\le t) =\mathbb{P}(N(t)\ge k) =1-e^{-\lambda t}\sum_{j=0}^{k-1}\frac{(\lambda t)^j}{j!}, $$ and $$ \mathbb{P}(S_k>t) =\mathbb{P}(N(t)\le k-1) =e^{-\lambda t}\sum_{j=0}^{k-1}\frac{(\lambda t)^j}{j!}. $$

Proof. By the count-arrival-time duality, $\mathbb{P}(S_k\le t)=\mathbb{P}(N(t)\ge k)$. The complement of $\{N(t)\ge k\}$ is $\{N(t)\le k-1\}$. Since $N(t)$ has the Poisson distribution with parameter $\lambda t$, its probability on that finite set is $$ \mathbb{P}(N(t)\le k-1) =\sum_{j=0}^{k-1}e^{-\lambda t}\frac{(\lambda t)^j}{j!} =e^{-\lambda t}\sum_{j=0}^{k-1}\frac{(\lambda t)^j}{j!}. $$ Subtracting this probability from $1$ gives the formula for $\mathbb{P}(S_k\le t)$. The formula for $\mathbb{P}(S_k>t)$ follows directly from the second identity in the duality theorem. These calculations use the Poisson marginal law at the specified time $t$; no statement about independent increments is needed. $\square$

These formulas are the count-based form of the distribution of the kth arrival. The lower tail of the count, from zero through $k-1$, gives the probability that the kth arrival is still in the future. The complementary upper tail gives the probability it has already occurred. The event equality lets us move between these descriptions without changing the event or its probability.

Worked Example: The Third Arrival by Time One

A Poisson process has rate $\lambda=2$. Find the probability that its third arrival occurs by time $1$, first using the count and then checking with the arrival-time formula.

Set $t=1$. The Poisson parameter for the count is $\lambda t=2\cdot1=2$, so $N(1)$ has probabilities $\mathbb{P}(N(1)=j)=e^{-2}2^j/j!$. By duality, the requested probability is $\mathbb{P}(N(1)\ge3)$. Its complement consists of counts $0,1,2$. Compute each term:

\begin{aligned} \mathbb{P}(N(1)=0)&=e^{-2}\frac{2^0}{0!}=e^{-2},\\ \mathbb{P}(N(1)=1)&=e^{-2}\frac{2^1}{1!}=2e^{-2},\\ \mathbb{P}(N(1)=2)&=e^{-2}\frac{2^2}{2!}=2e^{-2}. \end{aligned}

Thus $\mathbb{P}(N(1)\le2)=5e^{-2}$, and $$ \mathbb{P}(S_3\le1)=\mathbb{P}(N(1)\ge3)=1-5e^{-2}. $$ Check: The arrival-time formula uses $k-1=2$ and $\lambda t=2$, giving $1-e^{-2}(1+2+2)=1-5e^{-2}$, the same answer.

Worked Example: The Second Arrival Is Still Pending

A Poisson process has rate $\lambda=\tfrac32$. Find the probability that its second arrival occurs strictly after time $t=\tfrac23$. Then express the complementary event as a count probability.

The count parameter is $\lambda t=\tfrac32\cdot\tfrac23=1$. By the duality, $S_2>t$ exactly when $N(t)\le1$. The two possible counts are zero and one: $$ \begin{aligned} \mathbb{P}(N(\tfrac23)=0)&=e^{-1}\frac{1^0}{0!}=e^{-1},\\ \mathbb{P}(N(\tfrac23)=1)&=e^{-1}\frac{1^1}{1!}=e^{-1}. \end{aligned} $$ Therefore $\mathbb{P}(S_2>\tfrac23)=2e^{-1}$. The complementary event has probability $1-2e^{-1}$, and the duality identifies it as $\mathbb{P}(N(\tfrac23)\ge2)$, or equivalently $\mathbb{P}(S_2\le\tfrac23)$.

Check: The survival formula gives $e^{-1}(1+1)=2e^{-1}$. Adding the probabilities of $S_2>\tfrac23$ and $S_2\le\tfrac23$ gives $2e^{-1}+(1-2e^{-1})=1$.

Worked Example: The Fourth Arrival by Time Two

A Poisson process has rate $\lambda=\tfrac12$. Find the probability that at least four arrivals have occurred by time $2$, and rewrite the answer as a probability about the fourth arrival.

Here $\lambda t=\tfrac12\cdot2=1$. The requested count event is $N(2)\ge4$. Its complement consists of counts $0,1,2,3$, whose probabilities are $$ \begin{aligned} \mathbb{P}(N(2)=0)&=e^{-1}\frac{1^0}{0!}=e^{-1},\\ \mathbb{P}(N(2)=1)&=e^{-1}\frac{1^1}{1!}=e^{-1},\\ \mathbb{P}(N(2)=2)&=e^{-1}\frac{1^2}{2!}=\tfrac12e^{-1},\\ \mathbb{P}(N(2)=3)&=e^{-1}\frac{1^3}{3!}=\tfrac16e^{-1}. \end{aligned} $$ The sum of the coefficients is $1+1+\tfrac12+\tfrac16=\tfrac83$. Hence $$ \mathbb{P}(N(2)\ge4)=1-\tfrac83e^{-1} =\mathbb{P}(S_4\le2). $$

Check: The arrival-time formula for $k=4$ has terms $j=0,1,2,3$, with $\lambda t=1$, so it gives $1-e^{-1}(1+1+\tfrac12+\tfrac16)=1-\tfrac83e^{-1}$, as required.

Choosing the Useful Description

Count-arrival-time duality is useful because the same question can be framed in two ways. If the count distribution is directly available, compute the relevant Poisson tail and translate the result into an arrival-time statement. If the question starts with an arrival time, translate it into a count event before summing probabilities. The integer threshold is the key: “the kth arrival has occurred” means “at least $k$ arrivals,” while “the kth arrival has not occurred” means “at most $k-1$ arrivals.”

The equality is pathwise, not an independence claim. In particular, it does not say that an arrival-time event and a count event are separate or independent events; they describe the same outcome set. Nor does it require deriving a new arrival-time distribution from scratch. Once the Poisson count probabilities are known, the finite sum through $k-1$ supplies the lower tail, and its complement supplies the probability of reaching count $k$.

Common mistake. For the third arrival by time $1$ in the first example, one might incorrectly calculate only $\mathbb{P}(N(1)=3)$. Substituting the parameters gives $$ \mathbb{P}(N(1)=3)=e^{-2}\frac{2^3}{3!}=\tfrac43e^{-2}. $$ That is the probability of exactly three arrivals, not at least three. The desired event includes counts $3,4,5,\ldots$, so its probability is $\mathbb{P}(N(1)\ge3)=1-5e^{-2}$. The difference between the correct probability and the exactly-three probability is $\mathbb{P}(N(1)\ge4)$, which is positive because, for example, $\mathbb{P}(N(1)=4)=e^{-2}2^4/4!=\tfrac23e^{-2}>0$.

A second frequent error is shifting the cutoff by one. For $S_k>t$, the count must be at most $k-1$, not at most $k$: a count equal to $k$ means the kth arrival has already occurred by $t$. Keeping the event statements beside the calculation prevents both errors:

S_k\le t\ \Longleftrightarrow\ N(t)\ge k, \qquad S_k>t\ \Longleftrightarrow\ N(t)\le k-1.

Check Your Understanding

For each question, identify the equivalent event before calculating or explaining the probability.

Write the event $\{S_5\le t\}$ as an event involving $N(t)$.
For a Poisson process with rate $\lambda$, express $\mathbb{P}(S_3>t)$ as a finite sum involving $\lambda t$.
A count is observed at time $t$. Which count values correspond to the event that the second arrival has occurred by $t$?
Explain why an arrival exactly at $t$ belongs to the event $\{S_k\le t\}$ under the convention used here.
For the fourth arrival by time $t$, explain why using only $\mathbb{P}(N(t)=4)$ does not give the desired probability.

What You'll Learn