Why Practical Screening Matters
A Poisson approximation can simplify a binomial count when many independent trials each have a small chance of success. The key is not merely that a Poisson model has the same mean: the trial structure must also make the approximation plausible. Practical screening helps identify good candidates before doing detailed probability calculations.
The conditions are a matter of degree. There is no single cutoff that guarantees a chosen level of accuracy for every event of interest. A rule of thumb is therefore a screening device, not a substitute for a model check or a formal error bound. The central practical question is whether the number of trials is large enough and the success probability per trial is small enough for the binomial count to behave like a count of rare events.
A Practical Screening Framework
A useful first screen is to check both \(n\) and \(p\), not just their product. As a conservative classroom rule of thumb, \(n\geq100\) and \(p\leq0.05\) indicate a plausible candidate for Poisson approximation. Some settings use less conservative cutoffs, such as \(n\geq20\) with \(p\leq0.05\), but those are weaker screens. Neither set of cutoffs is a theorem, and passing either does not guarantee that every requested probability is accurate enough.
Next calculate \(\lambda=np\). This is the expected count under both models. A modest expected count is common in rare-event applications and is the setting in which many introductory rules of thumb are used. However, \(\lambda\leq10\) is not a universal validity condition: a larger expected count does not, by itself, rule out approximation. It should prompt care about the purpose of the approximation and the precision required.
A supporting diagnostic is the difference between the two variances. For a binomial count, the variance is \(np(1-p)\); for a Poisson count with mean \(\lambda=np\), the variance is \(\lambda\). The derivation below shows why small \(p\) matters beyond matching the mean.
Thus, the Poisson variance exceeds the binomial variance by \(np^2=\lambda p\). Relative to the Poisson variance, the difference is \(p\). When \(p\) is small, the two variances are close. This is a useful diagnostic, not proof that all probabilities or tails are close: distributions can have similar means and variances yet differ in particular regions.
Check that the count is over a fixed number \(n\) of trials, with independent outcomes and a common success probability \(p\).
Record \(n\), \(p\), and \(\lambda=np\). A large \(n\) does not compensate automatically for a probability \(p\) that is not small.
Calculate \(np^2\), the absolute variance difference, and \(p\), the difference as a proportion of the Poisson variance.
Say whether the approximation is a reasonable candidate, borderline, or unsuitable as a default. If precision matters or the screen is weak, retain the exact binomial model when feasible.
Worked Screening Examples
Worked Example: A Large Portfolio with Rare Claims
A hypothetical insurer reviews a fixed group of 8,000 independently covered items. Each item has probability \(0.00125\) of producing a specified claim during the review period. Assess whether a Poisson approximation is a reasonable candidate. Do not calculate a particular claim-count probability.
Method. Check the binomial assumptions and apply the practical screen, then compare the variances.
The count is assumed to have a fixed number of trials, independent claim outcomes, and the same probability for each item. Thus \(n=8{,}000\) and \(p=0.00125\). The expected count is
The sample size is well above 100, and \(0.00125<0.05\). Both parts of the conservative screening rule are satisfied. The variance difference is
The Poisson variance is \(10\), while the binomial variance is \(10(1-0.00125)=9.9875\). Their difference is \(10-9.9875=0.0125\), or \(0.0125/10=0.00125\), which is \(0.125\%\) of the Poisson variance.
Assessment. Under the stated assumptions, the Poisson model is a reasonable candidate for this rare-claim count. The conclusion is about suitability, not a guarantee for a specific probability calculation. The common mistake avoided here is treating \(\lambda=10\) as the evidence of suitability by itself; the large \(n\) and very small \(p\) also matter.
Worked Example: A Moderate Trial Count Near the Screening Boundary
A hypothetical device test consists of 60 independent units, each with probability \(0.06\) of a specified failure. Assess whether to use a Poisson approximation as the default model for the failure count.
Method. Compare both trial-level conditions with the conservative screen and use the variance discrepancy to describe the limitation.
Here \(n=60\) and \(p=0.06\). The common mean parameter would be
Although \(3.6\) is a modest expected count, \(n=60<100\), and \(p=0.06>0.05\). The conservative screen is not met. The variance difference is
The Poisson variance is \(3.6\). The binomial variance is \(3.6(1-0.06)=3.6(0.94)=3.384\), a difference of \(3.6-3.384=0.216\). The difference is \(0.216/3.6=0.06\), or \(6\%\) of the Poisson variance.
Assessment. This is borderline rather than an automatic rejection: the values are close to, but do not meet, the conservative screen, and the variance check shows a noticeable difference. If the requested result requires dependable precision, use the exact binomial model when feasible. The common mistake avoided is accepting the Poisson model solely because the expected count is small.
Worked Example: Same Expected Count, Different Rarity
Two hypothetical screening procedures each have an expected count of five flagged records. Procedure A uses 25 independent records with flag probability \(0.20\); Procedure B uses 500 independent records with flag probability \(0.01\). Compare their suitability for a Poisson approximation.
Method. Compare \(n\) and \(p\) separately, then calculate the variance difference for each procedure. Both have \(\lambda=5\), but equal means do not make their trial structures equivalent.
For Procedure A,
The Poisson variance is \(5\); the binomial variance is \(5(1-0.20)=4\). The difference is \(1\), or \(1/5=20\%\) of the Poisson variance. Also, \(n_A=25<100\) and \(p_A=0.20>0.05\), so Procedure A fails the conservative screen.
For Procedure B,
The Poisson variance is again \(5\); the binomial variance is \(5(1-0.01)=4.95\). The difference is \(0.05\), or \(0.05/5=1\%\) of the Poisson variance. Procedure B has \(n_B=500\geq100\) and \(p_B=0.01\leq0.05\), so it passes the screen.
Assessment. Procedure B is the stronger candidate, even though both expected counts are five. The contrast shows why checking only \(\lambda\) is insufficient. The common mistake avoided is assuming that identical means imply equally suitable Poisson approximations.
When the Screening Logic Does Not Apply
The binomial assumptions are part of the screen, not optional details. If trial probabilities differ, the count is not generally \(\operatorname{Bin}(n,p)\) for a single \(p\). If outcomes are dependent, the binomial variance formula used above may not apply. In either case, the ordinary binomial-to-Poisson rule of thumb cannot be invoked without further modeling.
Likewise, passing the screening conditions should not be described as proof that an approximation meets a particular error tolerance. Accuracy can depend on which event is being evaluated and how much precision is needed. When exact binomial calculations are practical and the consequences of error matter, the exact model is a sensible reference. Formal error assessment is a separate task from this preliminary practical screen.
Common Exam Trap
A frequent error is to confuse the success probability \(p\) with the expected count \(\lambda=np\). Suppose a hypothetical count has \(n=100\) and \(p=0.04\). The incorrect calculation is
This omits the number of trials. The correction is
The count has a \(4\%\) success probability on each trial and an expected total of four successes. These are different quantities. Another related trap is to calculate \(\lambda\) correctly and stop: a plausible mean does not remove the need to assess whether \(n\) is large and \(p\) is small.
Check Your Understanding
For each hypothetical count, apply the stated practical screen and explain what the diagnostic does—and does not—establish.
- A fixed group has \(n=1{,}200\) independent trials with common success probability \(p=0.004\). Find \(\lambda\) and \(np^2\), then assess the screen. (Answer: \(\lambda=4.8\), \(np^2=0.0192\); passes the \(n\geq100,\ p\leq0.05\) screen.)
- A count has \(n=80\) and \(p=0.025\). Find \(\lambda\) and state whether it passes the conservative screen. (Answer: \(\lambda=2\); fails because \(n<100\), although \(p\leq0.05\).)
- Two designs have \((n,p)=(40,0.10)\) and \((400,0.01)\). Confirm their means and compare their variance differences. (Answer: both have \(\lambda=4\); differences are \(0.4\) and \(0.04\), respectively, so the second design has the smaller variance discrepancy and passes the conservative screen.)
- A count has \(n=2{,}000\) and \(p=0.06\). Calculate \(\lambda\), the variance difference, and the percentage of the Poisson variance represented by that difference. Is the conservative screen passed? (Answer: \(\lambda=120\), difference \(=7.2\), percentage \(=6\%\); no, because \(p>0.05\).)
- A manager proposes a Poisson approximation because a count is expected to be 6, but the trials have different success probabilities. Is the ordinary binomial-to-Poisson screening rule sufficient? (Answer: No; a common trial probability is part of the binomial model assumed by this screening rule.)