Means and Variances of Conditional Arrival Times

Arrival-Time Moments Under a Fixed Count

Suppose a Poisson process has exactly \(n\) arrivals during a fixed interval \([0,T]\). The arrival times are then dependent: once their total number is fixed, an early arrival leaves more of the interval available for later arrivals. Nevertheless, their joint conditional law has a simple description. The ordered arrival times behave like the order statistics of \(n\) independent uniform points on \([0,T]\). This gives explicit formulas for the conditional mean and variance of the \(k\)th arrival time.

Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space carrying a homogeneous Poisson process \((N_t)_{t\ge0}\) of rate \(\lambda>0\), with state space \(S=\{0,1,2,\ldots\}\) and natural filtration \(\mathcal F_t=\sigma(N_s:0\le s\le t)\). Fix \(T>0\). Write \(S_k=\inf\{t\ge0:N_t\ge k\}\) for the time of the \(k\)th arrival. We condition on the event \(\{N(T)=n\}\), which has probability \(e^{-\lambda T}(\lambda T)^n/n!>0\). A \(k\)th arrival within the interval exists under this conditioning precisely when \(1\le k\le n\).

Here is a brief derivation of the conditional law. For ordered times \(0<s_1<\cdots<s_n<T\), the joint density of arrivals at those locations and no other arrivals in \([0,T]\) is \(e^{-\lambda T}\lambda^n\). Dividing by \(\mathbb P(N(T)=n)=e^{-\lambda T}(\lambda T)^n/n!\) gives the stated conditional density. The same density is obtained by taking \(n\) independent uniform points, whose unordered joint density is \(1/T^n\), and sorting them: each strictly ordered configuration has \(n!\) possible label orderings. Ties and arrivals exactly at the endpoints have probability zero.

The Marginal Density and Moment Formulas

For a fixed \(k\) with \(1\le k\le n\), the event that the \(k\)th ordered point is near \(s\) requires \(k-1\) points below \(s\), one point in a small interval around \(s\), and \(n-k\) points above \(s\). Counting which of the independent uniform points occupy these positions gives the marginal density below.

Proof. Set \(U_i\) to be the independent Uniform\((0,T)\) points whose order statistics are the arrival times under the conditional law. For a small interval \((s,s+ds)\), the probability that one specified point lies there is approximately \(ds/T\); the probability that a point lies below \(s\) is \(s/T\), and the probability that it lies above \(s\) is \((T-s)/T\). Choose the point in the small interval in \(n\) ways, choose the \(k-1\) points below it in \(\binom{n-1}{k-1}\) ways, and place the remaining \(n-k\) above it. The resulting density is \[ \frac{n!}{(k-1)!(n-k)!}\frac{1}{T} \left(\frac{s}{T}\right)^{k-1} \left(\frac{T-s}{T}\right)^{n-k}, \] which is the displayed marginal density.

To calculate its moments, use \(u=s/T\). For positive integers \(a,b\), the integral \(\int_0^1 u^{a-1}(1-u)^{b-1}\,du=(a-1)!(b-1)!/(a+b-1)!\). This identity follows by integration by parts repeatedly: when \(b>1\), integration by parts gives \(I(a,b)=\frac{b-1}{a}I(a+1,b-1)\), and \(I(a,1)=1/a\). Applying this identity to the density shows that \(S_k/T\) has first moment \(k/(n+1)\) and second moment \(k(k+1)/((n+1)(n+2))\). Therefore \[ \mathbb E[S_k\mid N(T)=n]=\frac{kT}{n+1}, \qquad \mathbb E[S_k^2\mid N(T)=n]=\frac{k(k+1)T^2}{(n+1)(n+2)}. \] Subtracting the square of the mean gives \[ \begin{aligned} \operatorname{Var}(S_k\mid N(T)=n) &=T^2\left(\frac{k(k+1)}{(n+1)(n+2)}-\frac{k^2}{(n+1)^2}\right)\\ &=\frac{k(n+1-k)T^2}{(n+1)^2(n+2)}. \end{aligned} \] The density is normalized because the same integral identity, with \(a=k\) and \(b=n+1-k\), makes its integral equal to \(1\). This proves the density and both moment formulas. \(\square\)

The conditional mean is the fraction \(k/(n+1)\) of the observation horizon, not \(k/n\). The \(n\) random points divide \([0,T]\) into \(n+1\) gaps, including the gaps before the first and after the last arrival; this helps explain the denominator \(n+1\). The variance is largest near the middle ranks and is symmetric under replacing \(k\) by \(n+1-k\). Both moments scale with the observation horizon: multiplying \(T\) by a factor multiplies the mean by that factor and the variance by its square. Once we condition on the count, the rate \(\lambda\) no longer appears in these formulas.

Why Conditioning Changes the Answer

The unconditional \(k\)th arrival time in a rate-\(\lambda\) Poisson process has mean \(k/\lambda\) and variance \(k/\lambda^2\). Those are not the appropriate moments after fixing the count by time \(T\). For example, when \(T=6\), \(\lambda=2\), and \(N(6)=3\), the unconditional mean of the first arrival is \(1/2\), but the conditional mean is \(6/4=3/2\). Conditioning on exactly three arrivals changes the distribution of the arrival locations; it is not valid to retain the unconditional gamma-distribution moments.

Common mistake. A frequent incorrect calculation uses \(k/n\) as the expected fraction of the interval, giving \(S_1\) mean \(T/n\). For \(n=3\) and \(T=6\), that gives \(6/3=2\), which is wrong. The correct denominator is \(n+1\): the first order statistic has mean \(6/(3+1)=3/2\). A direct check uses the first-arrival survival function above, which gives \(\mathbb E[S_1\mid N(6)=3]=\int_0^6(1-t/6)^3dt=6/4=3/2\).

There is no \(k\)th arrival among the arrivals in \([0,T]\) when \(n=0\), so the conditional formulas do not apply for \(k\ge1\) in that case. Likewise, a rank \(k>n\) is not defined as an arrival time within the fixed-count sample. The conditions \(T>0\), \(\lambda>0\), and \(1\le k\le n\) ensure that the interval and requested arrival are meaningful and that the conditioning event has positive probability.