Why Compare Adjacent Probabilities?
A Poisson approximation can be assessed not only by comparing individual probabilities, but also by comparing how probability changes from one count to the next. For a count variable, the ratio \(P(X=k+1)/P(X=k)\) describes this local change: a ratio greater than 1 means the next count is more probable, and a ratio less than 1 means it is less probable. Comparing this ratio under a binomial model with the corresponding Poisson ratio shows whether the two distributions have similar local shape near a chosen count.
This is a diagnostic, not a replacement for comparing probabilities or errors directly. Adjacent ratios can agree while the probabilities themselves differ by a common scale factor. The ratio is most informative when interpreted together with the count value, the parameters, and the approximation's intended use.
The Binomial and Poisson Ratios
Suppose \(X\sim\operatorname{Bin}(n,p)\), where \(n\) is a positive integer and \(0<p<1\). Its probability mass function is \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\), for \(k=0,\ldots,n\). For \(k=0,\ldots,n-1\), both adjacent binomial probabilities are positive, so their ratio is defined. The binomial model has no probability at counts above \(n\).
For the Poisson approximation, take \(Y\sim\operatorname{Poisson}(\lambda)\), with \(\lambda=np\). Its mass function is \(P(Y=k)=e^{-\lambda}\lambda^k/k!\) for \(k=0,1,\ldots\). Because the Poisson model has unbounded support, its adjacent ratio is defined for every nonnegative integer \(k\).
The second ratio is the Poisson benchmark for the first. The factor multiplying it in the binomial ratio, \((1-k/n)/(1-p)\), is the local correction. It measures how the finite number of binomial trials and the factor \(1-p\) alter the step from \(k\) to \(k+1\).
Why the Ratio Formula Holds
The formula follows by dividing successive probability masses. This derivation also explains why the factorial and binomial-coefficient terms simplify.
Use \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\) and the same expression with \(k+1\) in place of \(k\).
The powers of \(p\) contribute \(p\), and the powers of \(1-p\) contribute \(1/(1-p)\). The coefficient quotient is \(\binom{n}{k+1}/\binom{n}{k}=(n-k)/(k+1)\).
Since \(\lambda=np\), replace \(p\) by \(\lambda/n\). The result is the Poisson ratio \(\lambda/(k+1)\) multiplied by \((1-k/n)/(1-p)\).
To determine the direction of the correction, compare \(1-k/n\) with \(1-p\). Their difference is \(p-k/n\). Both quantities are positive for the stated range of \(k\), so the correction exceeds 1 exactly when \(p-k/n>0\), or \(k<np\). Equality occurs at \(k=np\), if that value of \(k\) is an integer. Above \(np\), the correction is below 1. Equivalently, its absolute difference from 1 is \[ \left|\frac{1-k/n}{1-p}-1\right| = \frac{|p-k/n|}{1-p}. \]
This direction describes ratios, not necessarily which model assigns the larger probability to every count. A ratio is relative to the probability at \(k\); to compare actual masses, one must also know a starting probability or calculate the masses directly.
Worked Examples
Worked Example: Checking Ratios at Several Counts
A hypothetical screening process records whether each of 80 independent items has a specified defect. Each item has defect probability \(p=0.025\). Let \(X\) be the defect count and approximate it by a Poisson variable \(Y\) with \(\lambda=np\). Compare the adjacent ratios at \(k=0,1,2,\) and \(5\).
Method. First find \(\lambda=80(0.025)=2\). Use the Poisson ratio \(2/(k+1)\), then multiply by \((1-k/80)/(1-0.025)=(1-k/80)/0.975\) for the binomial ratio.
At \(k=0\), the Poisson ratio is \(2/1=2\). The binomial correction is \(1/0.975=40/39\), so the binomial ratio is \(2(40/39)=80/39=2.0513\).
At \(k=1\), the Poisson ratio is \(2/2=1\). The correction is \(0.9875/0.975=79/78\), so the binomial ratio is \(79/78=1.0128\).
At \(k=2\), the Poisson ratio is \(2/3=0.6667\). Since \(k=np=2\), the correction is \((1-2/80)/0.975=0.975/0.975=1\); the binomial ratio is also \(2/3=0.6667\).
At \(k=5\), the Poisson ratio is \(2/6=1/3=0.3333\). The correction is \(0.9375/0.975=25/26\), so the binomial ratio is \((1/3)(25/26)=25/78=0.3205\).
The ratios agree exactly at \(k=2\), are slightly higher for the binomial model below 2, and are lower for it above 2. The common mistake avoided here is to use \(\lambda/(k+1)\) as the binomial ratio too; that is only the Poisson ratio.
Worked Example: Same Mean, Different Local Agreement
Compare two hypothetical binomial models for a count of flagged transactions. Model A has \(n=80\), \(p=0.025\); Model B has \(n=20\), \(p=0.10\). Both have mean \(\lambda=np=2\). At \(k=5\), calculate each binomial ratio, the matched Poisson ratio, and decide which binomial ratio is closer to it.
Method. The Poisson ratio is \(2/(5+1)=1/3=0.3333\). For Model A, the correction factor is \((1-5/80)/(1-0.025)=0.9375/0.975=25/26\). Thus its ratio is \((1/3)(25/26)=25/78=0.3205\). Its absolute difference from the Poisson ratio is \(1/3-25/78=1/78=0.0128\).
For Model B, the correction factor is \((1-5/20)/(1-0.10)=0.75/0.90=5/6\). Its ratio is \((1/3)(5/6)=5/18=0.2778\). Its absolute difference from the Poisson ratio is \(1/3-5/18=1/18=0.0556\).
Model A has the closer adjacent ratio at \(k=5\), even though both models have the same mean. The difference comes from their distinct \(n\) and \(p\), not from \(\lambda\) alone. The calculation avoids the trap of treating a shared value of \(np\) as proof that the local ratios must match.
Worked Example: Using Ratios to Compare Neighboring Masses
A hypothetical batch has \(n=40\) items, each independently flagged with probability \(p=0.05\). Let \(X\) be the number flagged, and compare the local ordering of the binomial probabilities around counts 1, 2, and 3 with the ordering implied by a Poisson approximation.
Method. Here \(\lambda=np=40(0.05)=2\). For the binomial model, the ratio at \(k=1\) is the Poisson ratio \(2/2=1\) multiplied by \((1-1/40)/0.95=0.975/0.95=39/38\). Thus \(P(X=2)/P(X=1)=39/38=1.0263\), so \(P(X=2)>P(X=1)\).
At \(k=2\), the Poisson ratio is \(2/3=0.6667\). The binomial correction is \((1-2/40)/0.95=0.95/0.95=1\), so \(P(X=3)/P(X=2)=2/3\). Thus \(P(X=3)<P(X=2)\). For comparison, the Poisson ratio at \(k=1\) is exactly 1, so the Poisson masses at 1 and 2 are equal; its ratio at \(k=2\) is also \(2/3\).
The binomial ratios therefore show that its mass increases from 1 to 2 and decreases from 2 to 3, while the Poisson approximation is tied between 1 and 2 before decreasing. The example compares neighboring masses without claiming that ratio agreement establishes equality of the masses themselves.
Interpreting the Diagnostic
The correction factor gives a quick way to organize a local comparison. For counts below \(np\), the binomial ratio is larger than the Poisson ratio. At an integer count equal to \(np\), they match exactly. For counts above \(np\), the binomial ratio is smaller. The size of the correction also depends on \(n\), \(p\), and how far \(k\) is from \(np\).
At \(k=np\), the exact match applies only if \(np\) is an integer and lies in the binomial ratio's valid range. If \(np\) is not an integer, there is no count at which the correction equals 1 exactly. At the upper binomial boundary \(k=n\), the next binomial probability is zero, so the binomial adjacent ratio is not given by the formula above. The Poisson ratio remains defined there, but comparing it with a binomial ratio is no longer possible.
A ratio check is particularly useful for seeing how the approximation changes across counts. It can reveal a direction of local distortion, or show that a model matching well around its mean matches less well farther away. It does not provide a bound on cumulative error, and it does not tell whether an event probability is acceptably approximated. For those questions, compare the relevant probabilities or use an appropriate error measure.
Common Exam Trap
A frequent error is to substitute \(\lambda=np\) into the Poisson adjacent ratio and report \(\lambda/(k+1)\) as the binomial ratio. This drops the finite-binomial correction. For example, with \(n=20\), \(p=0.10\), \(\lambda=2\), and \(k=5\), the incorrect calculation is
That is the Poisson ratio, not the binomial ratio. The correction gives
Another interpretive error is to infer that close adjacent ratios prove close probabilities. The ratios compare changes between consecutive counts; the starting probabilities may still differ. State clearly which ratio is being compared and what conclusion it supports.
Check Your Understanding
For each problem, compare the requested binomial ratio with the Poisson ratio using \(\lambda=np\).
- A hypothetical count follows \(X\sim\operatorname{Bin}(60,0.05)\). Find the Poisson ratio and binomial ratio at \(k=0\). (Answer: \(\lambda=3\); Poisson ratio \(=3.0000\); binomial ratio \(=60/19=3.1579\).)
- A hypothetical count follows \(X\sim\operatorname{Bin}(40,0.025)\). Find both ratios at \(k=4\), and state which is larger. (Answer: \(\lambda=1\); Poisson ratio \(=0.2000\); binomial ratio \(=12/65=0.1846\), so the Poisson ratio is larger.)
- A hypothetical count follows \(X\sim\operatorname{Bin}(25,0.08)\). At \(k=4\), calculate the binomial-to-Poisson ratio quotient and its percentage difference from 1. (Answer: quotient \(=21/23=0.9130\), which is \(8.6957\%\) below 1.)
- Two hypothetical binomial counts have parameters \((n,p)=(100,0.02)\) and \((25,0.08)\). Both have mean 2. At \(k=4\), find each binomial ratio and identify which is closer to the Poisson ratio. (Answer: Poisson ratio \(=0.4000\); the ratios are \(0.3918\) and \(0.3652\), respectively; the \((100,0.02)\) model is closer.)
- A hypothetical count follows \(X\sim\operatorname{Bin}(30,0.10)\). At \(k=5\), calculate the Poisson ratio and the binomial ratio, then state whether the binomial ratio is above or below the Poisson ratio. (Answer: \(\lambda=3\); Poisson ratio \(=0.5000\); binomial ratio \(=25/54=0.4630\), below the Poisson ratio.)