Comparing Exact Binomial and Poisson Probabilities

Why Compare the Two Models?

A Poisson model is often used to approximate a binomial count when the number of trials is large and the probability of success on each trial is small. Both models can then describe the same kind of count, and the Poisson model uses the binomial mean \(np\) as its parameter. But an approximation is not an identity: the two models can assign different probabilities to the same event. Comparing them directly helps show whether the approximation is accurate enough for a particular calculation.

The accuracy need not be the same for every count or event. A Poisson probability for one exact count might differ noticeably from its binomial counterpart, while a probability for a wider tail differs very little. The relevant comparison is therefore the probability for the event of interest, not a general impression that one model is “close.”

Setting Up a Matched-Mean Comparison

Suppose \(X\) counts successes in \(n\) independent trials, each with the same success probability \(p\). Under the binomial model, \(X\sim\operatorname{Bin}(n,p)\), with possible values \(0,1,\ldots,n\). For comparison, let \(Y\) have a Poisson distribution with parameter \(\lambda=np\). The Poisson probability mass function is \(P(Y=k)=e^{-\lambda}\lambda^k/k!\) for \(k=0,1,2,\ldots\).

The models have the same mean, \(np=\lambda\), but not generally the same variance or probability distribution. The binomial model has variance \(np(1-p)\), while the Poisson model has variance \(\lambda=np\). Matching means makes the comparison useful, but it does not guarantee that the probabilities agree.

Definition: In a matched-mean comparison, calculate an event’s probability under \(X\sim\operatorname{Bin}(n,p)\) and under \(Y\sim\operatorname{Poisson}(\lambda)\), using \(\lambda=np\). The absolute difference is \(\left|P(Y\in A)-P(X\in A)\right|\). The signed difference is \(P(Y\in A)-P(X\in A)\), which indicates whether the Poisson model overestimates or underestimates the binomial probability. A signed relative difference, expressed as a percentage of the exact binomial probability, is defined only when \(P(X\in A)>0\); otherwise, report the absolute difference or use another appropriate comparison.

Here “exact” means calculated from the specified binomial model rather than approximated by a Poisson model. The binomial calculation is not necessarily an exact description of reality; it is exact relative to that model and its assumptions.

Worked Example: Comparing Three Point Probabilities

Suppose a count comes from \(n=20\) independent trials, each with success probability \(p=0.05\). Compare the probabilities of exactly 0, 1, and 2 successes under the exact binomial model and a Poisson model with the same mean.

The binomial assumptions are that there is a fixed number of trials, each trial has two outcomes, the trials are independent, and the success probability is \(0.05\) on each trial. The mean is \(np=20(0.05)=1\), so use \(Y\sim\operatorname{Poisson}(1)\).

For \(k=0\), the binomial probability is

P(X=0)=\binom{20}{0}(0.05)^0(0.95)^{20} =(0.95)^{20}=0.358486.

The matched-mean Poisson probability is \(P(Y=0)=e^{-1}=0.367879\). Thus the signed difference, computed from the unrounded probabilities, is about \(0.009394\): the Poisson model overestimates this probability by about 0.009394, or 0.9394 percentage points.

For \(k=1\),

P(X=1)=\binom{20}{1}(0.05)(0.95)^{19} =20(0.05)(0.95)^{19}=0.377354.

The Poisson calculation gives \(P(Y=1)=e^{-1}(1^1/1!)=0.367879\). Its signed difference from the binomial probability, computed from the unrounded probabilities, is about \(-0.009474\), so it underestimates by about 0.009474.

For \(k=2\),

P(X=2)=\binom{20}{2}(0.05)^2(0.95)^{18} =190(0.0025)(0.95)^{18}=0.188677.

The Poisson probability is \(P(Y=2)=e^{-1}(1^2/2!)=0.183940\). The signed difference is \(0.183940-0.188677=-0.004737\). In this example, the absolute differences for the three point probabilities are approximately 0.009394, 0.009474, and 0.004737, respectively. The probabilities are reasonably close, but they are not identical and the direction of the difference varies.

A Ratio That Explains the Differences

For point probabilities within the binomial model’s support, a ratio gives a direct algebraic comparison. When \(n\geq 1\), \(0<p\leq 1\), and \(0\leq k\leq n\), the Poisson point probability in the denominator is positive, so the ratio is well-defined:

\frac{P(X=k)}{P(Y=k)} = \frac{\binom{n}{k}p^k(1-p)^{n-k}} {e^{-np}(np)^k/k!} = \left[\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)\right] (1-p)^{n-k}e^{np}.

To obtain the product, use \(\binom{n}{k}k!=n(n-1)\cdots(n-k+1)\). Dividing this product by \(n^k\), from the Poisson denominator, gives \(\prod_{j=0}^{k-1}(1-j/n)\). The factors \(p^k\) cancel because \(p>0\), leaving the displayed expression. For \(k=0\), the product is empty and by convention equals 1.

A ratio greater than 1 means the exact binomial point probability is larger; a ratio less than 1 means it is smaller. For the first worked example at \(k=0\), the ratio is \((0.95)^{20}e^1\), approximately 0.9745, consistent with the binomial probability being lower. At \(k=1\), it is \((0.95)^{19}e^1\), approximately 1.0258, consistent with the binomial probability being higher. The ratio concerns point probabilities; it does not directly give the ratio for a sum of several counts or a tail event.

The restriction \(p>0\) matters. If \(p=0\), then \(\lambda=0\); both models put probability 1 at zero and probability 0 at every positive count. For \(k>0\), the ratio of the two probabilities is \(0/0\), not a usable ratio. The matched distributions are nevertheless trivially identical in that degenerate case.

Point Probabilities and Event Probabilities Can Behave Differently

A probability for a range of counts is found by adding the relevant point probabilities, or by subtracting a complementary event from 1. Differences across several counts can partly offset one another. Consequently, comparing one point probability does not determine how well the approximation works for a tail or interval event.

Worked Example: A Tail Probability Can Be Closer Than Its Components

Let \(X\sim\operatorname{Bin}(100,0.01)\). Compare \(P(X\geq2)\) with the probability from the matched-mean model \(Y\sim\operatorname{Poisson}(1)\). The binomial model is appropriate if there are 100 fixed, independent trials with the same success probability \(0.01\); its mean is \(100(0.01)=1\).

First calculate the two binomial probabilities needed for the complement:

P(X=0)=(0.99)^{100}=0.366032, \qquad P(X=1)=\binom{100}{1}(0.01)(0.99)^{99}=0.369730.

Therefore,

P(X\geq2)=1-P(X=0)-P(X=1) =1-0.366032-0.369730=0.264238.

Under the Poisson model, the complementary event consists of counts 0 and 1:

P(Y\geq2)=1-P(Y=0)-P(Y=1) =1-e^{-1}-e^{-1}=1-2e^{-1}=0.264241.

The absolute difference is approximately \(\left|0.264241-0.264238\right|=0.0000031\). Yet the individual probabilities at 0 and 1 differ by about 0.00185 and 0.00185, respectively, in opposite directions. Their effects nearly cancel in the tail calculation. The close agreement for \(X\geq2\) is therefore a conclusion about this particular event, not proof that every point probability is equally close.

When the Approximation Is Less Close

The binomial-to-Poisson approximation is usually most useful when individual success probabilities are small and the mean \(np\) is being kept at a relevant scale. The actual accuracy still depends on the parameters and the event. A comparatively large \(p\), a small number of trials, or a count near the edge of the binomial support can make a difference more apparent. The following comparison shows why it is useful to calculate rather than assume a particular degree of accuracy.

Worked Example: A More Noticeable Difference at Zero

Suppose \(X\sim\operatorname{Bin}(10,0.2)\). This represents 10 independent trials with the same success probability \(0.2\), and has mean \(10(0.2)=2\). Compare the probability of no successes with the matched-mean model \(Y\sim\operatorname{Poisson}(2)\).

The exact binomial probability is

P(X=0)=\binom{10}{0}(0.2)^0(0.8)^{10} =(0.8)^{10}=0.107374.

The Poisson probability is

P(Y=0)=e^{-2}\frac{2^0}{0!}=e^{-2}=0.135335.

The Poisson model overestimates the probability of zero by \(0.135335-0.107374=0.027961\), or about 2.796 percentage points. The signed relative difference is defined here because the exact probability is positive; it is \(0.027961/0.107374\times100\%\), approximately 26.0%. The substantial relative difference is a warning that the approximation is not equally accurate for every event, particularly with these parameters. It does not imply that every event probability differs by 26%.

Choosing a Useful Error Measure

Absolute difference is often the clearest first comparison because it directly reports how many probability units separate the two models. Signed difference adds the direction of the discrepancy. Relative difference can help convey the size of an error compared with the exact probability, but it can be misleading when that probability is very small, and it cannot be calculated at all when the exact probability is zero.

For example, if \(X\sim\operatorname{Bin}(4,0.1)\), the event \(X=5\) is impossible under the binomial model, so \(P(X=5)=0\). The matched Poisson parameter is \(\lambda=0.4\), and

P(Y=5)=e^{-0.4}\frac{0.4^5}{5!} \approx 0.0000572.

The absolute difference for this event is about 0.0000572. But the signed relative difference with the exact binomial probability in the denominator is undefined, because it would require division by zero. A statement that this error is a particular percentage of the exact probability would therefore be incorrect.

This example also illustrates the models’ different supports: a binomial count cannot exceed \(n\), whereas a Poisson count can be any nonnegative integer. That difference matters especially for events beyond the largest possible binomial count. For events within the shared support, make the comparison for the actual event of interest and report enough precision to show whether the numerical discrepancy matters in context.

Common Mistakes

Assuming matched means imply matched probabilities. The models share the mean \(np\), not the entire distribution. Calculate both probabilities for the event being studied.
Comparing only one count and generalizing to all events. Errors at different counts can have different sizes and directions. A tail may be closer or less close than its component point probabilities.
Reporting relative error when the exact probability is zero. Division by zero is undefined. Report the absolute difference, as in the event \(X=5\) when \(X\sim\operatorname{Bin}(4,0.1)\).
Using the point-probability ratio when its denominator is zero. The ratio formula above requires \(p>0\). When \(p=0\) and \(k>0\), both probabilities are zero and their ratio is undefined.
Calling a Poisson calculation “exact” for the binomial model. It is an exact calculation under the Poisson model, but an approximation to the binomial probability when the Poisson model is substituted for the binomial one.

Key takeaway: For a binomial count with parameters \(n\) and \(p\), compare the exact event probability with the Poisson probability using \(\lambda=np\). State the event, calculate both probabilities, and report a difference measure whose denominator and interpretation are valid.

Check Your Understanding

Work through these questions using the definitions and comparisons above.

For \(X\sim\operatorname{Bin}(30,0.02)\), what value of \(\lambda\) should be used for a matched-mean Poisson model?
If the Poisson probability of an event is 0.18 and its exact binomial probability is 0.16, find the signed difference and explain its direction.
Why can a tail probability be closer under the two models than some of the point probabilities that contribute to it?
For \(X\sim\operatorname{Bin}(5,0.1)\), is the relative difference for the event \(X=6\) defined? Explain, and identify a suitable comparison instead.
What does a ratio \(P(X=k)/P(Y=k)\) greater than 1 say about the two point probabilities?