Why Compare an Approximation With the Exact Probability?
A Poisson model can make a binomial probability calculation easier, but an approximation is useful only if its discrepancy is understood. Comparing the two probabilities for the same event shows whether the Poisson value is above or below the exact binomial value and by how much. The comparison also depends on the event: a point probability and a cumulative probability can have noticeably different discrepancies even when they use the same model parameters.
Throughout this tutorial, suppose \(X\) is the number of successes in \(n\) independent trials, each with success probability \(p\). The exact model is \(X\sim\operatorname{Bin}(n,p)\). The Poisson approximation uses \(Y\sim\operatorname{Pois}(\lambda)\), where \(\lambda=np\). Both probabilities must refer to the same set of integer outcomes.
This sign convention will be used consistently. Some sources define a signed error in the reverse order, so always check the convention before interpreting a reported sign. The absolute difference does not depend on that choice. Neither difference is a relative error: it is measured in probability units.
A Reliable Comparison Procedure
First identify the binomial model and calculate \(\lambda=np\). Then calculate the exact binomial probability and the Poisson probability for the specified event. Subtract in the stated order and take the absolute value only after obtaining the signed difference. Keep intermediate values unrounded so that the final subtraction is not distorted.
Write the event as a set of integer counts, and state \(n\), \(p\), and \(\lambda=np\).
Use the binomial probability formula, a sum of binomial probabilities, or a complement as appropriate.
Use the same event and the Poisson parameter \(\lambda=np\).
Calculate Poisson minus binomial for the signed difference, then report its absolute value. Round only the final results.
For a binomial point probability, the exact formula is $$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,\ldots,n. $$ For the Poisson model, $$ P(Y=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ For an event containing several counts, add the point probabilities for those counts or use a complement if that is shorter.
Worked Comparisons
Worked Example: Comparing a Zero Count and a Short Cumulative Event
A hypothetical inspection program independently checks \(n=20\) components. Each component has probability \(p=0.05\) of requiring rework. Let \(X\) be the number requiring rework. Compare the exact and Poisson probabilities for (a) no components requiring rework and (b) at most one requiring rework. Use \(\lambda=np\), and report signed and absolute differences to four decimal places.
Method and model. The exact count is \(X\sim\operatorname{Bin}(20,0.05)\). Its Poisson approximation is \(Y\sim\operatorname{Pois}(\lambda)\), with \[ \lambda=np=20(0.05)=1. \] For each event, the signed difference is Poisson minus exact binomial. This comparison avoids the mistake of changing the event while comparing the two models.
(a) No components require rework. The exact probability is \[ P(X=0)=\binom{20}{0}(0.05)^0(0.95)^{20}=0.95^{20}=0.3584859224. \] The Poisson probability is \[ P(Y=0)=e^{-1}\frac{1^0}{0!}=e^{-1}=0.3678794412. \] Thus \[ \Delta(\{0\})=0.3678794412-0.3584859224=0.0093935188, \qquad |\Delta(\{0\})|=0.0093935188. \] To four decimal places, the signed difference is \(+0.0094\) and the absolute difference is \(0.0094\). The Poisson value is larger.
(b) At most one component requires rework. For the exact model, \[ P(X\leq 1)=P(X=0)+P(X=1). \] The first term is \(0.95^{20}=0.3584859224\). The second is \[ P(X=1)=\binom{20}{1}(0.05)(0.95)^{19} =1(0.3773536025)=0.3773536025. \] Therefore \(P(X\leq1)=0.3584859224+0.3773536025=0.7358395249\). For the Poisson model, \[ P(Y\leq1)=e^{-1}\left(\frac{1^0}{0!}+\frac{1^1}{1!}\right) =2e^{-1}=0.7357588823. \] It follows that \[ \Delta(\{0,1\})=0.7357588823-0.7358395249=-0.0000806426, \qquad |\Delta(\{0,1\})|=0.0000806426. \] To four decimal places, the signed difference is \(-0.0001\) and the absolute difference is \(0.0001\). The Poisson approximation is slightly smaller for this cumulative event, even though it was larger for the zero-count event.
Check. Both comparisons use \(\lambda=1\), and both compare identical events under the two models. The signed differences have opposite signs, so the close cumulative result is not evidence that each point probability is equally close.
Worked Example: Comparing a High-Threshold Tail
In a hypothetical quality-control process, \(n=50\) items are independently tested, each with probability \(p=0.04\) of a defect. Let \(X\) be the number of defective items. Compare the exact and Poisson probabilities of at least four defects, and report the signed and absolute differences to four decimal places.
Method and model. The exact model is \(X\sim\operatorname{Bin}(50,0.04)\), and \(\lambda=np=50(0.04)=2\), so the approximation is \(Y\sim\operatorname{Pois}(2)\). The event is a tail, so calculate each probability as one minus the probability of counts zero through three.
For the exact calculation, write \(b_k=P(X=k)\). Start with \[ b_0=(0.96)^{50}=0.1298857935. \] The ratios of successive binomial probabilities give \[ b_1=b_0\frac{50(0.04)}{0.96}=0.2705954031, \] \[ b_2=b_1\frac{49(0.04)}{2(0.96)}=0.2762328073, \] \[ b_3=b_2\frac{48(0.04)}{3(0.96)}=0.1841552049. \] Consequently, \[ P(X\geq4)=1-(b_0+b_1+b_2+b_3) =1-0.8608692088=0.1391307912. \]
For the Poisson model, \[ P(Y\geq4)=1-P(Y\leq3) =1-e^{-2}\left(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}\right). \] The sum in parentheses is \(1+2+2+1.3333333333=6.3333333333\). Thus \[ P(Y\geq4)=1-(0.1353352832)(6.3333333333) =1-0.8571234603=0.1428765397. \] The signed difference is \[ \Delta(\{4,5,\ldots\})=0.1428765397-0.1391307912=0.0037457485, \] and the absolute difference is \(0.0037457485\). To four decimal places, these are \(+0.0037\) and \(0.0037\). The Poisson approximation overstates the tail probability.
Check. Each complement includes exactly the counts \(0,1,2,3\), and each tail is at least four. The probability values are between zero and one, and their difference has the sign of the larger Poisson tail.
Worked Example: A Point Probability With the Opposite Error Direction
A hypothetical monitoring system records whether each of \(n=10\) independent devices triggers an alert during a test period. Each device triggers with probability \(p=0.10\). Let \(X\) count alerts. Compare the exact and Poisson probabilities of exactly one alert, and report the signed and absolute differences to four decimal places.
Method and model. Here \(X\sim\operatorname{Bin}(10,0.10)\), and \(\lambda=np=10(0.10)=1\), giving \(Y\sim\operatorname{Pois}(1)\). Calculate the point probability from each model before subtracting.
The exact binomial probability is \[ P(X=1)=\binom{10}{1}(0.10)(0.90)^9 =1(0.387420489)=0.387420489. \] The Poisson probability is \[ P(Y=1)=e^{-1}\frac{1^1}{1!}=0.3678794412. \] Therefore \[ \Delta(\{1\})=0.3678794412-0.3874204890=-0.0195410478, \qquad |\Delta(\{1\})|=0.0195410478. \] To four decimal places, the signed difference is \(-0.0195\) and the absolute difference is \(0.0195\). In this example the Poisson approximation understates the exact probability.
Check and common mistake avoided. The Poisson parameter is \(1\), but that does not mean every point probability is close to its exact binomial counterpart. In particular, the sign here is opposite to the zero-count comparison in the first example. The models were compared at the same value \(k=1\), not by comparing probabilities for different events.
Why Event Choice Can Change the Difference
For an event \(A\) containing several counts, its probability is the sum of the point probabilities for those counts. This gives a direct justification for why signed differences can cancel. If \(A=\{0,1\}\), then \[ \Delta(A)=\big(P(Y=0)-P(X=0)\big)+\big(P(Y=1)-P(X=1)\big). \] The equality follows by adding the two model probabilities within each event and then distributing the subtraction. A positive point difference can therefore be offset by a negative point difference.
The first worked example shows this effect. For count zero, the signed difference is approximately \(+0.0093935\). For count one, it is approximately \[ 0.3678794412-0.3773536025=-0.0094741613. \] Adding the two point differences gives approximately \(-0.0000806426\), the signed difference for the event \(X\leq1\). Thus, a small event-level difference can result from cancellation; it does not imply that every component probability has a small discrepancy.
The absolute difference is always nonnegative, but it is still calculated after summing probabilities over the event. It is not generally the sum of the absolute point differences. In the example above, the absolute event-level difference is about \(0.0000806\), whereas the sum of the two absolute point differences is about \(0.0188677\). These quantities answer different questions.
Common Exam Traps
- Reversing the subtraction order. Under the convention here, calculate Poisson minus binomial. If the exact probability is \(0.40\) and the approximation is \(0.37\), the signed difference is \(0.37-0.40=-0.03\), not \(+0.03\). The absolute difference is \(0.03\).
- Comparing different events. A Poisson probability of exactly three must be compared with the binomial probability of exactly three, not with a binomial probability of at most three.
- Using a different Poisson parameter. For the binomial-to-Poisson approximation in this setting, use \(\lambda=np\). Keep this parameter unchanged across the exact-versus-approximate comparison.
- Taking absolute values too early. Adding absolute point differences discards cancellation and does not give the absolute difference for a range or tail event.
- Rounding before subtraction. Subtract unrounded probabilities when available. If two values are close, premature rounding can change the final reported difference.
Check Your Understanding
For each problem, use the signed-difference convention defined above. Round probabilities and differences to four decimal places where numerical values are requested.
- Let \(X\sim\operatorname{Bin}(25,0.04)\) and use \(Y\sim\operatorname{Pois}(np)\). Compare the probabilities of \(X=0\) and \(Y=0\). State the signed and absolute differences. (Answer: \(P(X=0)=0.3604\), \(P(Y=0)=0.3679\), signed difference \(+0.0075\), absolute difference \(0.0075\).)
- Let \(X\sim\operatorname{Bin}(40,0.025)\). Compare the exact and Poisson-approximation probabilities of exactly one success, showing the value of \(\lambda\). (Answer: \(\lambda=1\), exact probability \(0.3725\), Poisson probability \(0.3679\), signed difference \(-0.0047\), absolute difference \(0.0047\).)
- A hypothetical sensor independently activates with probability \(0.10\) on each of 30 trials. Let \(X\) count activations. Compare the exact and Poisson probabilities of no activations. (Answer: \(\lambda=3\), exact probability \(0.9^{30}=0.0424\), Poisson probability \(e^{-3}=0.0498\), signed difference \(+0.0074\), absolute difference \(0.0074\).)
- Let \(X\sim\operatorname{Bin}(60,0.02)\). Compare the exact and Poisson probabilities of at most one success. Use the sum of the zero- and one-success probabilities for each model. (Answer: \(\lambda=1.2\), exact probability approximately \(0.6619\), Poisson probability approximately \(0.6626\), signed difference \(+0.0007\), absolute difference \(0.0007\).)
- A hypothetical batch has \(n=12\) items, each independently having probability \(0.15\) of a specified flaw. Compare the exact and Poisson probabilities of at least two flawed items. (Answer: \(\lambda=1.8\), exact probability \(1-0.85^{12}-1.8(0.85^{11})\approx0.5565\), Poisson probability \(1-e^{-1.8}(1+1.8)\approx0.5372\), signed difference \(-0.0194\), absolute difference \(0.0194\).)