Homogeneous Poisson Processes from Exponential Waiting Times

From Waiting Times to Event Paths

A homogeneous Poisson process models events that occur randomly over time at a constant rate. In simulation, one direct way to generate its path is to draw the waiting time to the next event, add that wait to the current time, and repeat. This gives event times themselves, so the output can answer questions about both the number of events by a horizon and when they occurred.

Fix a rate \(\lambda>0\). On a probability space \((\Omega,\mathcal{F},\mathbb{P})\), let \(X_1,X_2,\ldots\) be independent and identically distributed exponential random variables with rate \(\lambda\), so that \(\mathbb{P}(X_k>x)=e^{-\lambda x}\) for \(x\ge0\). Define the arrival times \(S_k=X_1+\cdots+X_k\), and let \(N_t\) count arrivals by time \(t\). The state space is \(S=\{0,1,2,\ldots\}\), and time is continuous, \(t\ge0\). We use the convention \(S_0=0\), and count an arrival at the horizon: \(N_t=\#\{k\ge1:S_k\le t\}\). For a fixed \(t\), equality \(S_k=t\) has probability zero, but the inclusive convention makes the algorithm precise.

To simulate a wait, draw \(U_k\) independently from the uniform distribution on \((0,1)\) and set \(X_k=-\log(U_k)/\lambda\). Indeed, for \(x\ge0\), \(\mathbb{P}(-\log(U_k)/\lambda>x)=\mathbb{P}(U_k<e^{-\lambda x})=e^{-\lambda x}\). Thus the transformed draws have the required exponential distribution, and independence is preserved because each wait is a function of a separate independent draw. In practice, a pseudorandom generator supplies the uniform draws; exact endpoint values 0 or 1 must be avoided because the logarithm is not finite there.

The loop stops almost surely after finitely many events on any finite horizon. The count-law proof below shows that \(N_T\) has a probability distribution on the nonnegative integers whose probabilities sum to one, so infinitely many arrivals before \(T\) have probability zero.

Why the Construction Gives Poisson Counts

Proof. First fix \(T>0\) and an integer \(n\ge1\). The joint density of the first \(n\) waiting times \(X_1,\ldots,X_n\) at positive values \(x_1,\ldots,x_n\) is \(\lambda^n e^{-\lambda(x_1+\cdots+x_n)}\), by independence. Changing variables from the waits to ordered arrival times \(s_k=x_1+\cdots+x_k\) has Jacobian determinant \(1\). Therefore the density of \(S_1,\ldots,S_n\) on \(0<s_1<\cdots<s_n\) is \(\lambda^n e^{-\lambda s_n}\). Given these first \(n\) arrivals, the event that there is no further arrival by \(T\) is \(X_{n+1}>T-s_n\), whose probability is \(e^{-\lambda(T-s_n)}\), by the exponential tail and independence. Multiplying gives the density-and-survival factor \(\lambda^n e^{-\lambda T}\) for exactly these \(n\) arrivals and no others before \(T\). For \(n=0\), the probability of no arrival is \(\mathbb{P}(X_1>T)=e^{-\lambda T}\), the same formula with the empty product.

Now partition \([0,T]\) at \(0=t_0<t_1<\cdots<t_m=T\), and write \(\Delta_r=t_r-t_{r-1}\). If interval \(r\) contains \(k_r\) arrivals, the ordered arrival times within that interval occupy a region of volume \(\Delta_r^{k_r}/k_r!\). This is because the cube of side length \(\Delta_r\) is divided, apart from ties of volume zero, into \(k_r!\) equally sized orderings; when \(k_r=0\), the volume is \(1\). The intervals together have total ordered-region volume equal to the product of these volumes. Integrating the factor \(\lambda^{\sum_r k_r}e^{-\lambda T}\) over that region gives

This factorization proves both the Poisson count law for each interval and independence of the counts in disjoint consecutive intervals. Finally, for \(a=\lambda t\), summing the count probabilities gives \(\mathbb{E}[N_t]=e^{-a}\sum_{n\ge1}n a^n/n!=a\), after shifting the index in the exponential series. Similarly, \(\mathbb{E}[N_t(N_t-1)]=e^{-a}\sum_{n\ge2}n(n-1)a^n/n!=a^2\). Since \(N_t^2=N_t(N_t-1)+N_t\), its variance is \(a^2+a-a^2=a\). At \(t=0\), \(N_0=0\), so the same conclusions hold. \(\square\)

Arrival-Time Summaries and Repeated Runs

Counts are only one summary of a simulated path. The event list \(S_1,\ldots,S_{N_T}\) also supports summaries such as the first observed arrival, the last arrival, or the average of all observed event times. For a fixed arrival index \(k\), the time \(S_k\) is a sum of \(k\) independent exponential waits. A rate-\(\lambda\) exponential variable has mean \(1/\lambda\) and variance \(1/\lambda^2\); these follow by integrating its density \(\lambda e^{-\lambda x}\) on \(x\ge0\). Linearity of expectation and independence therefore give the next result.

Proof. By definition, \(S_k=\sum_{r=1}^kX_r\). Each summand has mean \(1/\lambda\), so \(\mathbb{E}[S_k]=\sum_{r=1}^k\mathbb{E}[X_r]=k/\lambda\). Independence makes every covariance between distinct summands zero, so \(\operatorname{Var}(S_k)=\sum_{r=1}^k\operatorname{Var}(X_r)=k/\lambda^2\). \(\square\)

Common Pitfalls and Practical Checks

Common mistake: treating waits as event times. If the first two waiting times are \(1/4\) and \(1/2\), the arrival times are \(S_1=1/4\) and \(S_2=1/4+1/2=3/4\). A wrong calculation records the second event at \(1/2\), using the second wait itself as an absolute time. The correction is to add each new wait to the previous arrival time.

Another important check is the horizon comparison. Record an event when its accumulated time satisfies \(S_k\le T\), and stop when \(S_k>T\). Do not count a wait merely because it was drawn: it is the proposed next arrival time that must be compared with \(T\). For continuous exponential waiting times, an exact tie with a fixed \(T\) has probability zero, but implementing a consistent inclusive rule avoids ambiguity.

For estimating an expected count, simulate independent whole paths and average their counts. For estimating a rate from a fixed common horizon \(T\), total arrivals divided by total exposure, \(\sum_{r=1}^R N_T^{(r)}/(RT)\), is the corresponding pooled rate estimate. Arrival-time averages require a clearly specified target: averaging first arrival times only over runs with an arrival by \(T\) excludes runs with no such arrival, while averaging all recorded event times weights runs with more events more heavily. Report which summary was calculated and how events or runs were included.