Conditioning a Poisson Process on Its Endpoint
A Poisson process is usually described by its rate: over an interval of length \(u\), its count has a Poisson distribution with mean proportional to \(u\). Conditioning on the count at a fixed terminal time changes that picture. Once the total number of arrivals is fixed, the count at an earlier time cannot exceed that total, and the arrivals remaining after any time are constrained by the endpoint. The resulting process is called a Poisson bridge.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N(t))_{t\ge0}\) be a Poisson process of rate \(\lambda>0\), with \(N(0)=0\). Thus counts on disjoint intervals are independent, and \(N(t)-N(s)\) has a Poisson distribution with mean \(\lambda(t-s)\) for \(0\le s<t\). Fix a terminal time \(T>0\) and an integer \(n\ge0\). Since \(\mathbb{P}(N(T)=n)=e^{-\lambda T}(\lambda T)^n/n!>0\), conditioning on \(N(T)=n\) is well-defined. Write \(\mathbb{P}^{(n,T)}(\,\cdot\,)=\mathbb{P}(\,\cdot\mid N(T)=n)\). The conditioned process is considered on \(0\le t\le T\), with its natural filtration \(\mathcal{F}_t=\sigma(N(u):0\le u\le t)\).
Proof. For \(0<t<T\), independent increments give, for \(k=0,\ldots,n\), $$ \mathbb{P}(N(t)=k,\ N(T)=n) =e^{-\lambda t}\frac{(\lambda t)^k}{k!}\, e^{-\lambda(T-t)}\frac{(\lambda(T-t))^{n-k}}{(n-k)!}. $$ Dividing by \(\mathbb{P}(N(T)=n)=e^{-\lambda T}(\lambda T)^n/n!\) cancels the exponential factors and gives $$ \frac{n!}{k!(n-k)!}\left(\frac{t}{T}\right)^k \left(\frac{T-t}{T}\right)^{n-k}, $$ which is the stated binomial probability. If \(k>n\), the joint event is impossible, so its conditional probability is zero. At \(t=0\), \(N(0)=0\); at \(t=T\), the conditioning fixes \(N(T)=n\). These also agree with the degenerate endpoint versions of the binomial law. \(\square\)
The parameter \(\lambda\) disappears from this conditional distribution. The rate determines how likely the conditioning event is, but, once the total \(n\) is fixed, the fraction \(t/T\) determines how many of those arrivals have occurred by time \(t\). In particular, the conditional mean and variance are \(nt/T\) and \(n(t/T)(1-t/T)\), respectively, by the binomial mean and variance formulas.
Transitions Between Intermediate Times
The one-time law does not by itself describe how the bridge moves. To obtain a transition, suppose \(0\le s<t<T\), and that \(N(s)=k\) under the conditioning, where \(k\in\{0,\ldots,n\}\) is a possible count. There are \(n-k\) arrivals still to occur by time \(T\). Of those, the interval \((s,t]\) occupies a fraction \((t-s)/(T-s)\) of the remaining time.
Proof. Set \(r=j-k\), so \(0\le r\le n-k\). By independent increments, the counts on \((s,t]\) and \((t,T]\) are independent Poisson variables with means \(\lambda(t-s)\) and \(\lambda(T-t)\). Given \(N(s)=k\), the endpoint condition requires their sum to be \(n-k\). The conditional probability that the first of these counts equals \(r\) is therefore the ratio $$ \frac{e^{-\lambda(t-s)}\dfrac{(\lambda(t-s))^r}{r!}\, e^{-\lambda(T-t)}\dfrac{(\lambda(T-t))^{n-k-r}}{(n-k-r)!}} {e^{-\lambda(T-s)}\dfrac{(\lambda(T-s))^{n-k}}{(n-k)!}}. $$ Canceling the exponential and rate factors gives $$ \binom{n-k}{r} \left(\frac{t-s}{T-s}\right)^r \left(\frac{T-t}{T-s}\right)^{n-k-r}. $$ Since \(r=j-k\) and \(n-k-r=n-j\), this is the displayed formula. Counts cannot decrease or exceed the fixed total, which gives the zero probabilities outside \(k\le j\le n\). At \(t=T\), the endpoint condition forces the count to equal \(n\). \(\square\)
These transitions also establish the Markov property under the conditioned law. Given the past \(\mathcal{F}_s\), the transition probabilities to time \(t\) depend only on the current count \(N(s)\), not on earlier counts. Indeed, on each event where \(N(s)=k\), independent increments make the probability of a specified increment and the required remaining count depend on the past only through \(k\). The transition probabilities depend explicitly on \(s\) and \(t\), however, so this is a time-inhomogeneous Markov process, not a time-homogeneous one.
Worked Example: A Midpoint Count
Let the rate be \(\lambda=3\), the terminal time be \(T=4\), and condition on \(N(4)=5\). Find the full distribution of \(N(2)\).
The conditioning event has positive probability because \(3\cdot4=12>0\). The bridge count law gives \(N(2)\sim\operatorname{Bin}(5,2/4)=\operatorname{Bin}(5,1/2)\). For \(k=0,\ldots,5\), the probabilities are $$ \begin{aligned} \mathbb{P}^{(5,4)}(N(2)=0)&=\binom50(1/2)^5=1/32,\\ \mathbb{P}^{(5,4)}(N(2)=1)&=\binom51(1/2)^5=5/32,\\ \mathbb{P}^{(5,4)}(N(2)=2)&=\binom52(1/2)^5=10/32=5/16,\\ \mathbb{P}^{(5,4)}(N(2)=3)&=\binom53(1/2)^5=10/32=5/16,\\ \mathbb{P}^{(5,4)}(N(2)=4)&=\binom54(1/2)^5=5/32,\\ \mathbb{P}^{(5,4)}(N(2)=5)&=\binom55(1/2)^5=1/32. \end{aligned} $$ Check: the probabilities sum to \((1+5+10+10+5+1)/32=32/32=1\). The mean is \(5(1/2)=5/2\), consistent with five conditioned arrivals distributed over a time interval whose first half is half of the total interval.
Worked Example: A Bridge Transition
Let \(T=5\), condition on \(N(5)=6\), and suppose \(N(1)=2\). Find the distribution of \(N(3)\), including the probability that \(N(3)=4\).
There are \(6-2=4\) arrivals left after time \(1\). The interval from \(1\) to \(3\) is \(2\) units long, and the remaining time from \(1\) to \(5\) is \(4\) units long. Thus the increment has binomial parameters \(4\) and \(2/4=1/2\): $$ N(3)-N(1)\mid\{N(1)=2,N(5)=6\}\sim\operatorname{Bin}(4,1/2). $$ Consequently \(N(3)\) can equal \(2,3,4,5,6\), with respective probabilities \(1/16,4/16,6/16,4/16,1/16\). In particular, $$ \mathbb{P}^{(6,5)}(N(3)=4\mid N(1)=2) =\binom{4}{2}(1/2)^2(1/2)^2 =6/16=3/8. $$ Check: the five probabilities sum to \((1+4+6+4+1)/16=16/16=1\), and \(N(3)=4\) requires exactly two of the four remaining arrivals to occur between times \(1\) and \(3\), as used in the calculation.
What the Conditioned Path Looks Like
Under the event \(N(T)=n\), the path starts at \(0\), is nondecreasing, and ends at \(n\). A Poisson process has unit jumps and finitely many jumps on a finite interval almost surely, so the conditioned path has exactly \(n\) unit jumps on \((0,T]\), with no further jumps before \(T\). When \(n=0\), it is the constant zero path. For \(n>0\), the earlier uniform-order-statistics result for Poisson arrivals given a fixed count says that the \(n\) jump times have the distribution of the ordered values of \(n\) independent Uniform\((0,T)\) random variables. This gives a path-level description consistent with the binomial count and transition formulas: the number of those \(n\) locations falling in an interval is binomial with success probability equal to the interval's fraction of the total length.
The transition formula also quantifies how the bridge is pulled toward its endpoint. If the current time is \(s<T\) and the count is \(k\), then for a short interval of length \(h>0\), the conditional probability of at least one arrival is $$ 1-\left(1-\frac{h}{T-s}\right)^{n-k}, \qquad 0<h\le T-s. $$ For \(n-k>0\), dividing by \(h\) and taking \(h\) down to zero gives the instantaneous arrival rate \((n-k)/(T-s)\). If \(n-k=0\), the probability is zero for every such \(h\), and the rate is zero. Thus, when arrivals remain, the conditional rate increases as the terminal time approaches; the process must use its remaining time to complete the remaining \(n-k\) jumps.
Worked Example: Two Successive Time Intervals
Let \(T=6\), condition on \(N(6)=4\), and calculate the probability of both \(N(2)=1\) and \(N(5)=3\). Verify it by a direct conditional allocation across the three intervals.
First, \(N(2)\sim\operatorname{Bin}(4,2/6)=\operatorname{Bin}(4,1/3)\), so $$ \mathbb{P}^{(4,6)}(N(2)=1) =\binom41(1/3)(2/3)^3 =4\cdot(1/3)\cdot(8/27)=32/81. $$ Given \(N(2)=1\), there are three arrivals remaining, and the interval \((2,5]\) is \(3\) of the \(4\) remaining time units. To have \(N(5)=3\), exactly two of those three arrivals must occur in \((2,5]\). Hence $$ \mathbb{P}^{(4,6)}(N(5)=3\mid N(2)=1) =\binom32(3/4)^2(1/4) =3\cdot(9/16)\cdot(1/4)=27/64. $$ Multiplying gives $$ \mathbb{P}^{(4,6)}(N(2)=1,N(5)=3) =(32/81)(27/64)=1/6. $$ Check by interval allocation: the three intervals \((0,2],(2,5],(5,6]\) must contain \(1,2,1\) arrivals. Their length proportions are \(1/3,1/2,1/6\). Conditional on four total arrivals, the allocation probability is $$ \frac{4!}{1!2!1!}(1/3)(1/2)^2(1/6) =12\cdot(1/72)=1/6, $$ which agrees with the successive-transition calculation.
A Common Conditioning Error
Pitfall: conditioning on the terminal count does not leave the count at time \(t\) Poisson with mean \(\lambda t\). For example, let \(\lambda=2\), \(T=3\), and condition on \(N(3)=4\). At \(t=1\), the bridge law is \(\operatorname{Bin}(4,1/3)\), so $$ \mathbb{P}^{(4,3)}(N(1)=0)=(1-1/3)^4=(2/3)^4=16/81. $$ The incorrect unconditioned calculation would give \(\mathbb{P}(N(1)=0)=e^{-2}\), which is not the conditional probability. More fundamentally, the unconditioned Poisson law assigns positive probability to counts larger than four, whereas under the bridge every count is at most four. The correction is to condition using the joint probability with \(N(T)=n\), or to use the binomial bridge law directly.
For any bridge calculation, first identify the total count \(n\), the terminal time \(T\), and the interval fraction. For a single count at \(t\), that fraction is \(t/T\); for a transition from \(s\) to \(t\), it is \((t-s)/(T-s)\), and the number of remaining arrivals is \(n-N(s)\). Using the correct remaining time in the transition formula is essential: the interval fraction is not generally \((t-s)/T\).
Check Your Understanding
Use the binomial count law and bridge transition formula to answer each question. State the relevant parameters before computing.
- Condition on \(N(8)=6\). Find the distribution and mean of \(N(2)\).
- Condition on \(N(10)=5\), and suppose \(N(4)=2\). What is the probability that \(N(7)=4\)?
- For a bridge with terminal count \(n=0\), describe the count at every time \(0\le t\le T\).
- Explain why the bridge is Markov but time-inhomogeneous.
- For a bridge with \(n-k>0\) arrivals remaining at time \(s<T\), give its instantaneous arrival rate and describe its behavior as \(s\) approaches \(T\).