Why a Fixed Count Gives a Beta Law
Suppose a Poisson process has exactly \(n\) arrivals by a fixed time \(T\). The arrival times are then dependent: once their total number is fixed, knowing that one arrival occurred early affects the possible locations of the others. Yet their ordered locations have a simple description. Conditional on the total count, they have the same distribution as the ordered values of \(n\) independent uniform random variables on \((0,T)\). This gives an exact law for the \(k\)th arrival, not just an approximation.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((N_t)_{t\ge0}\) be a homogeneous Poisson process with rate \(\lambda>0\), with state space \(\{0,1,2,\ldots\}\) and continuous time \(t\ge0\). Let \(S_k\) denote its \(k\)th arrival time, for \(k\ge1\). Fix \(T>0\). The conditioning event \(\{N_T=n\}\) has positive probability for every integer \(n\ge0\), since \(\mathbb{P}(N_T=n)=e^{-\lambda T}(\lambda T)^n/n!>0\). The conditional law of an arrival within \([0,T]\) requires \(1\le k\le n\).
The beta parameters are \(k\) and \(n-k+1\). The rate \(\lambda\) determines how likely it is to observe \(n\) arrivals, but after conditioning on that count, it does not affect their relative locations in \([0,T]\).
Deriving the Density
We use the conditional uniform order-statistic result: given \(N_T=n\), the \(n\) arrival locations have the distribution of \(n\) independent uniform random variables on \((0,T)\), arranged in increasing order. Rescaling locations by \(T\), write the ordered values as \(U_{(1)}<\cdots<U_{(n)}\), where the unordered variables are independent uniform random variables on \((0,1)\). Their joint density on \(0<u_1<\cdots<u_n<1\) is \(n!\): the \(n!\) possible orderings of the independent variables have equal density \(1\), and ties have probability zero.
Proof. Fix \(k\in\{1,\ldots,n\}\). To obtain the density of \(U_{(k)}\) at \(u\in(0,1)\), place the first \(k-1\) ordered values below \(u\) and the last \(n-k\) above \(u\). The volume of the ordered region for \(k-1\) points in an interval of length \(u\) is \(u^{k-1}/(k-1)!\): the cube has volume \(u^{k-1}\), and its \((k-1)!\) strict orderings have equal volume. This also holds when \(k=1\), since the empty region has volume \(1\). The corresponding volume above \(u\) is \((1-u)^{n-k}/(n-k)!\), including the case \(k=n\). Multiplying these volumes by the joint density \(n!\) gives the marginal density $$ f_{U_{(k)}}(u)=\frac{n!}{(k-1)!(n-k)!}u^{k-1}(1-u)^{n-k}. $$ This density integrates to \(1\), because it was obtained by integrating the joint probability density over all ordered values other than \(u\). Since \(S_k=TU_{(k)}\), a change of variable \(u=t/T\), with derivative \(du/dt=1/T\), gives the stated density for \(S_k\). It has no mass outside \((0,T)\) under the condition \(n\ge k\). \(\square\)
The factor \(1/T\) in the time density is essential: it converts a density per unit of scaled time into a density per unit of clock time. The density also shows how the position of the \(k\)th arrival changes with \(k\). Small \(k\) emphasizes earlier times, while \(k\) near \(n\) emphasizes later times.
Deriving the Distribution Function
For \(0\le t\le T\), the event \(S_k\le t\) means that at least \(k\) of the \(n\) conditional uniform arrival locations fall in \((0,t]\). Each location falls there with probability \(t/T\), independently of the others under the conditional uniform representation. Thus the number in that interval has a binomial distribution with parameters \(n\) and \(t/T\).
Proof. Let \(U_1,\ldots,U_n\) be the unordered conditional uniform locations on \((0,1)\), and put \(u=t/T\). For \(0\le t\le T\), each indicator \(\mathbf{1}_{\{TU_i\le t\}}\) equals \(1\) with probability \(u\), independently across \(i\). Their sum therefore has probability \(\binom{n}{j}u^j(1-u)^{n-j}\) of equalling \(j\). The \(k\)th ordered location is at most \(t\) exactly when this sum is at least \(k\), so summing these probabilities for \(j=k,\ldots,n\) proves the formula. If \(t\le0\), no location is at or before \(t\), up to events of probability zero at the endpoint; if \(t\ge T\), all \(n\) locations are at or before \(t\). This proves the endpoint cases as well. \(\square\)
The density and distribution function answer different practical questions. Integrating the density is convenient for a short time interval or a simple polynomial. The finite sum is often convenient for a threshold probability because it is a binomial tail. In either calculation, use the ratio \(t/T\), not \(t\) by itself.
Worked Example: The Second Arrival Before Half the Horizon
Suppose \(N_t\) is a Poisson process with rate \(\lambda=2\), and condition on \(N_4=5\). Find \(\mathbb{P}(S_2\le2\mid N_4=5)\).
Here \(T=4\), \(n=5\), \(k=2\), and \(t/T=2/4=1/2\). The distribution-function formula gives $$ \mathbb{P}(S_2\le2\mid N_4=5) =\sum_{j=2}^{5}\binom{5}{j}\left(\frac12\right)^j\left(\frac12\right)^{5-j} =\frac{10+10+5+1}{32} =\frac{26}{32} =\frac{13}{16}. $$
Check by integrating the density. The scaled density is \(20u(1-u)^3\), so the time density is \(20(t/4)(1-t/4)^3/4\). Its integral from \(0\) to \(2\), using \(u=t/4\) and \(dt=4\,du\), is $$ \int_0^{1/2}20u(1-u)^3\,du =20\left[\frac{u^2}{2}-u^3+\frac{3u^4}{4}-\frac{u^5}{5}\right]_0^{1/2} =20\left(\frac18-\frac18+\frac{3}{64}-\frac{1}{160}\right) =20\cdot\frac{13}{320} =\frac{13}{16}. $$ The two calculations agree.
Worked Example: The Third Arrival in a Middle Interval
Condition on \(N_3=4\). Find the probability that the third arrival occurs after time \(1\) and no later than time \(2\).
Now \(T=3\), \(n=4\), and \(k=3\). The scaled density is $$ \frac{4!}{2!1!}u^2(1-u)=12u^2(1-u), $$ so its distribution function on \([0,1]\) is \(4u^3-3u^4\), obtained by integrating \(12u^2-12u^3\). The scaled endpoints are \(1/3\) and \(2/3\). Therefore $$ \begin{aligned} \mathbb{P}(1<S_3\le2\mid N_3=4) &=\left(4\left(\frac23\right)^3-3\left(\frac23\right)^4\right) -\left(4\left(\frac13\right)^3-3\left(\frac13\right)^4\right)\\ &=\left(\frac{32}{27}-\frac{16}{27}\right) -\left(\frac{4}{27}-\frac{1}{27}\right) =\frac{16}{27}-\frac{3}{27} =\frac{13}{27}. \end{aligned} $$
Check by the finite sum. At time \(2\), the probability is the chance that at least three of four locations are in the first \(2/3\) of the horizon: $$ \binom43\left(\frac23\right)^3\left(\frac13\right)+\left(\frac23\right)^4 =\frac{32}{81}+\frac{16}{81} =\frac{16}{27}. $$ At time \(1\), it is $$ \binom43\left(\frac13\right)^3\left(\frac23\right)+\left(\frac13\right)^4 =\frac{8}{81}+\frac{1}{81} =\frac19. $$ Subtracting gives \(16/27-1/9=16/27-3/27=13/27\), as before. The strict lower endpoint does not change the probability because the conditional distribution has a density and hence no atom at time \(1\).
Worked Example: The Last Arrival
Condition on \(N_2=3\). Find the probability that the last arrival occurs in \((1,3/2]\).
For \(k=n=3\), the scaled density is \(3u^2\), and the time density is \(3t^2/2^3=3t^2/8\) for \(0<t<2\). Its distribution function is \((t/2)^3\). Hence $$ \mathbb{P}(1<S_3\le3/2\mid N_2=3) =\left(\frac{3/2}{2}\right)^3-\left(\frac12\right)^3 =\left(\frac34\right)^3-\frac18 =\frac{27}{64}-\frac{8}{64} =\frac{19}{64}. $$
Check by integration. Direct integration gives $$ \int_1^{3/2}\frac{3t^2}{8}\,dt =\left[\frac{t^3}{8}\right]_1^{3/2} =\frac{27}{64}-\frac18 =\frac{19}{64}. $$ The density integrates to \(1\) over its full support, since \(\int_0^2 3t^2/8\,dt=8/8=1\).
Scaling and a Common Pitfall
A frequent error is to substitute \(t/T\) into the beta density but forget the change-of-variable factor \(1/T\). For example, in the first example the scaled density is \(20u(1-u)^3\). Substituting \(u=t/4\) but omitting the Jacobian would give the incorrect time density \(20(t/4)(1-t/4)^3\). Its integral over \(0<t<4\) is \(4\), not \(1\): under \(u=t/4\), that integral is \(4\int_0^1 20u(1-u)^3\,du=4\). The correction is to multiply by \(du/dt=1/4\). The correct density integrates to \(1\), and the factor has units of inverse time, as a time density should.
Conditioning on the count is also essential to interpreting the result. The beta law describes the location of \(S_k\) given \(N_T=n\), with \(k\le n\). It is not the unconditional distribution of \(S_k\), which also reflects the randomness of how many arrivals occur by \(T\). When \(n<k\), the event \(N_T=n\) still has positive probability, but \(S_k>T\) on that event; the beta density on \((0,T)\) does not apply.
The conditional law is useful whenever timing matters after the total number of events has been fixed, for example when comparing event locations within a study window. It separates two features of the process: the count by time \(T\), which depends on \(\lambda\), and the relative positions of the arrivals conditional on that count, which are uniform order statistics. For the \(k\)th location, that order-statistic structure is exactly the scaled beta law derived above.
Check Your Understanding
Use the beta density or the finite-sum distribution function, and check that any probability lies between \(0\) and \(1\).
- Given \(N_5=4\), write the scaled beta density for \(S_2/5\) and the time density for \(S_2\).
- Given \(N_6=3\), calculate \(\mathbb{P}(S_1\le2\mid N_6=3)\) using the finite sum.
- Given \(N_2=4\), calculate \(\mathbb{P}(S_4>1\mid N_2=4)\) using the distribution function.
- Why must the time density include a factor \(1/T\) when it is obtained from the density of \(S_k/T\)?
- What can be said about \(S_k\) on \(\{N_T=n\}\) when \(n<k\), and why does the beta formula not apply there?