Thinning Rate Calculations for Inspection and Detection

From Detection Probability to an Observed Rate

An inspection system can miss events even when the underlying event process is regular. To calculate how many events it records, separate two quantities: the rate at which events occur and the probability that any one event is detected. Under independent event-by-event detection, their product is the observed rate. The same calculation gives the rate of missed events, which is often needed to quantify what the inspection system does not record.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N_t)_{t\ge 0}\) be a homogeneous Poisson process with state space \(\{0,1,2,\ldots\}\) and rate \(\lambda\ge 0\). Thus, for an interval of length \(T\ge 0\), \(N_T\) has a Poisson distribution with mean \(\lambda T\). Suppose each event is detected with probability \(p\in[0,1]\), independently of all other events and of the process. The detection decision is a Bernoulli mark attached to each event. Write \(D_T\) for the number detected and \(M_T\) for the number missed by time \(T\). Then \(N_T=D_T+M_T\).

In particular, the number of detections in an interval of length \(T\) is Poisson with mean \(\lambda pT\). Therefore the probability of no detections is \(e^{-\lambda pT}\), and the probability of at least one detection is \(1-e^{-\lambda pT}\). These probabilities concern the observed stream, not the underlying event stream: missed events may still occur when the observed count is zero.

Why the Rate Rule Holds

Proof. First suppose \(T>0\), \(\lambda>0\), and \(0<p<1\). For integers \(r,m\ge 0\), the event \(D_T=r,\ M_T=m\) requires \(N_T=r+m\), followed by exactly \(r\) detections among those \(r+m\) events. The Poisson count law and the independent Bernoulli decisions give

The second line is the product of the Poisson probability of \(r\) with mean \(\lambda pT\) and the Poisson probability of \(m\) with mean \(\lambda(1-p)T\). Thus the joint probabilities factor for every \(r,m\), which proves both the marginal distributions and independence. If \(p=0\), then \(D_T=0\) and \(M_T=N_T\); if \(p=1\), then \(D_T=N_T\) and \(M_T=0\). These are the stated Poisson distributions with a zero mean interpreted as a count that equals zero with probability one. If \(\lambda=0\) or \(T=0\), then \(N_T=0\) with probability one, so both counts are zero. This covers all allowed parameters. \(\square\)

The theorem also gives the chance of missing every event in an interval: \(M_T\) is Poisson with mean \(\lambda(1-p)T\), so \(\mathbb{P}(M_T=0)=e^{-\lambda(1-p)T}\). This is different from the chance of recording no events, \(e^{-\lambda pT}\). The two probabilities answer different questions.

Proof. By the preceding theorem, the detected count from source \(i\) in the interval is Poisson with mean \(\lambda_i p_iT\). These counts are independent because the source processes and their detection decisions are independent. For \(0\le z\le 1\), the probability generating function of a Poisson count with mean \(a\) is \(e^{a(z-1)}\), as follows by summing \(e^{-a}\sum_{k\ge0}(az)^k/k!\). Independence means the generating function of the sum is the product of the source generating functions. Hence it is

This is the generating function of a Poisson random variable with mean \(T\sum_i\lambda_i p_i\), proving the result. If there are no sources, the sum is empty, the total count is zero, and the formula gives mean zero. \(\square\)

Worked Rate Calculations

Common Pitfall: Using the Miss Probability as the Detection Probability

Suppose a process has rate \(8\) per hour and an inspection detects each event with probability \(\tfrac34\). A common error is to use the miss probability \(\tfrac14\) as the observed-rate multiplier, then report \(8\cdot\tfrac14=2\) detections per hour. That calculation gives the missed rate, not the detected rate. The correction is to use the probability corresponding to the count being calculated: detected rate \(=8\cdot\tfrac34=6\) per hour, missed rate \(=8\cdot\tfrac14=2\) per hour. Their sum is \(6+2=8\), recovering the underlying rate.

The simple rate rule also depends on the model assumptions. If detection probabilities vary by source, calculate each source's detected rate separately and add them when the sources are independent. If detection decisions for different events are dependent, or detection probability changes with time, the homogeneous independent-thinning result cannot be applied without further analysis. In particular, knowing only an average detection probability need not justify treating the detections as a homogeneous Poisson process.