Zero-Rate and Positive-Rate Cases

One Rate, Two Different Regimes

A rate parameter describes how quickly events occur, but the value zero is not just an especially slow positive rate. At rate zero, no event occurs at any finite time: every count is zero, and the first arrival never happens. At a positive rate, counts have Poisson distributions and waiting times are finite with probability one. Understanding this distinction prevents formulas for positive rates from being used at a parameter value where their assumptions fail.

Let \((N_t)_{t\ge0}\) be a counting process on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(S=\{0,1,2,\ldots\}\) and continuous time \(t\ge0\). Its natural filtration is \(\mathcal{F}_t=\sigma(N_s:0\le s\le t)\). A Poisson process of rate \(\lambda\ge0\) starts at \(N_0=0\), has independent increments, and has an increment \(N_{t+h}-N_t\) with a Poisson distribution of mean \(\lambda h\) for \(t,h\ge0\). The process is nondecreasing and integer-valued. When \(\lambda>0\), its count probabilities are \(\mathbb{P}(N_t=n)=e^{-\lambda t}(\lambda t)^n/n!\), for \(n=0,1,\ldots\).

The separate definition at zero matters. The positive-rate probability formula contains \((\lambda t)^n\), and at \(\lambda t=0,n=0\) a blind substitution raises the ambiguous expression \(0^0\). The point-mass definition gives the count law unambiguously, including at time \(t=0\). It also makes the increments consistent: every increment of the zero-rate process is zero, so increments over disjoint intervals are independent and each has the required Poisson distribution of mean zero.

Counts and Waiting Times

For \(k\ge1\), define the \(k\)th arrival time by \(T_k=\inf\{t\ge0:N_t\ge k\}\), with \(\inf\varnothing=\infty\). Under the usual count convention, \(T_k>t\) exactly when \(N_t<k\). For a positive-rate process, this gives the waiting-time tail directly from the count probabilities. In particular, for every \(t\ge0\),

For \(k=1\), this becomes \(\mathbb{P}(T_1>t)=e^{-\lambda t}\), the exponential waiting-time tail. For the zero-rate process, \(N_t=0\) for every finite \(t\). Therefore \(T_k=\infty\) almost surely for each \(k\ge1\): for every finite \(t\), \(\mathbb{P}(T_k>t)=1\). There is no exponential waiting-time distribution with rate zero; the first arrival does not occur at a finite time.

Proof. If \(\lambda=0\), the point-mass count law gives \(\mathbb{P}(N_t=0)=1\) for every finite \(t\). Since \(T_1\) is the first time the count reaches at least one, \(N_t=0\) at all finite times implies \(T_1=\infty\) almost surely. Conversely, if \(\mathbb{P}(N_t=0)=1\) for every finite \(t\), then for any fixed \(t>0\) the Poisson count formula rules out \(\lambda>0\), because it would give \(\mathbb{P}(N_t=0)=e^{-\lambda t}<1\). Thus \(\lambda=0\). If \(T_1=\infty\) almost surely, then \(N_t=0\) almost surely for every fixed finite \(t\), so the same argument gives \(\lambda=0\). Finally, when \(\lambda>0\), the count formula gives \(e^{-\lambda t}<1\) for \(t>0\). Also \(\mathbb{P}(T_1>t)=e^{-\lambda t}\), which tends to zero as \(t\) tends to infinity; hence \(\mathbb{P}(T_1=\infty)=0\). \(\square\)

The waiting-time formula follows from the count law and the defining relation between \(T_k\) and \(N_t\); it does not require a separate derivation of the gamma distribution. For \(k>1\), the event of still waiting at time \(t\) includes the possibilities of zero, one, up to \(k-1\) arrivals. At rate zero, all of those possibilities collapse to the single outcome of zero arrivals.

How the Positive-Rate Case Approaches Zero

Although rate zero is a distinct case, positive-rate probabilities have sensible limits as the rate decreases to zero. Fix a finite \(t\ge0\). For \(\lambda>0\), the probability of at least one arrival by \(t\) is \(1-e^{-\lambda t}\). As \(\lambda\) decreases to zero, this probability tends to zero, matching the zero-rate probability. More generally, for \(n\ge1\), the probability of exactly \(n\) arrivals tends to zero, while the probability of zero arrivals tends to one.

Proof of the limit. For fixed finite \(t\), continuity of the exponential function gives \(e^{-\lambda t}\to1\) as \(\lambda\downarrow0\), so \(1-e^{-\lambda t}\to0\). For \(n\ge1\), the positive-rate formula is \(e^{-\lambda t}(\lambda t)^n/n!\). Its first factor tends to one and its second factor tends to zero, hence the product tends to zero. For \(n=0\), the formula is \(e^{-\lambda t}\), which tends to one. These limits agree with the zero-rate point mass.

The same limit describes waiting times. For every fixed finite \(t\), \(\mathbb{P}(T_1\le t)=1-e^{-\lambda t}\to0\) as \(\lambda\downarrow0\). Thus the first arrival becomes unlikely to occur within any fixed finite observation window. This is a statement about finite-time probabilities; it does not say that the waiting time at rate zero is an ordinary exponential random variable with a very large but finite mean.

Common Mistakes and Practical Interpretation

A common error is to write the positive-rate exponential density \(f(t)=\lambda e^{-\lambda t}\) and set \(\lambda=0\), obtaining \(f(t)=0\) for every finite \(t\). That cannot be a probability density: its integral over finite \(t\ge0\) is zero, not one. The correction is that the exponential waiting-time model assumes \(\lambda>0\). At zero rate the waiting time is \(T_1=\infty\) almost surely, so its probability is concentrated at an infinite waiting time rather than described by a density on finite times.

Another error is to confuse a small positive rate with a zero rate. For any \(\lambda>0\) and finite \(t>0\), the probability of at least one arrival is \(1-e^{-\lambda t}>0\). At rate zero it is exactly zero. A small positive rate may make arrivals rare over a particular observation window, but it does not make the process identically empty.

Finally, do not use the expression \((\lambda t)^0\) to decide the zero-count probability at \(\lambda t=0\). The zero-rate law is defined directly as a point mass, which avoids the \(0^0\) ambiguity and correctly covers \(t=0\) as well as \(\lambda=0\). Whenever a waiting-time question involves zero rate, first ask whether the arrival can occur at all; here it cannot.