Why Assess Approximation Accuracy?
A Poisson model can simplify calculations for a count of rare events, but “rare” is not by itself a numerical guarantee of accuracy. For a binomial count, the Poisson approximation uses the same expected count, while the approximation error depends on both the number of trials and the success probability. A bound on that error helps determine whether an approximation is adequate for a particular calculation.
This tutorial focuses on assessing binomial-to-Poisson approximation quality and on the corresponding bound for a sum of independent Bernoulli indicators with possibly unequal probabilities. The bound applies to a point probability, a tail, or any other event defined from the count. It is a conservative guarantee, not a prediction of the actual error.
The Poisson Approximation and Its Error Bound
For \(N\sim\operatorname{Binomial}(n,p)\), the exact point probability is \(P(N=k)=\binom{n}{k}p^k(1-p)^{n-k}\), for integers \(k=0,\ldots,n\). The Poisson approximation uses \(P(Z=k)=e^{-np}(np)^k/k!\), for \(k=0,1,2,\ldots\). The parameter is \(np\), not \(p\): it preserves the binomial mean.
The standard Le Cam bound is most naturally stated for independent Bernoulli trials. Let \(I_i\) equal 1 if trial \(i\) produces an event and 0 otherwise, with \(P(I_i=1)=p_i\), and assume the indicators are independent. Set \(W=\sum_{i=1}^n I_i\) and \(\lambda=\sum_{i=1}^n p_i\). For any set \(B\) of nonnegative integers, the bound is
In the equal-probability binomial case, \(p_i=p\), so \(\lambda=np\) and the bound becomes \(np^2\). This is an absolute error bound: if it equals 0.01, the two event probabilities differ by no more than 0.01, or one percentage point. It is not a relative error bound. The same bound works for every event \(B\), including \(B=\{k\}\), \(B=\{0,1,\ldots,r\}\), and \(B=\{r,r+1,\ldots\}\).
The independence assumption matters. If trial outcomes are dependent, this version of the bound does not automatically apply. For unequal probabilities, use \(\lambda=\sum p_i\) and \(\sum p_i^2\), rather than replacing all the probabilities by an unexamined average.
Why the Mean Is the Poisson Parameter
The binomial-to-Poisson limit explains the choice \(\lambda=np\). Consider a sequence of binomial models with \(n\) trials and \(p=\lambda/n\), where \(\lambda\) stays fixed as \(n\) increases. For fixed \(k\), the binomial point probability can be written as
For fixed \(k\), the product of the first two factors tends to \(\lambda^k/k!\): after factoring \(n\) from each of the \(k\) terms, the remaining product tends to 1. Also, \(\left(1-\lambda/n\right)^n\) tends to \(e^{-\lambda}\), while \(\left(1-\lambda/n\right)^{-k}\) tends to 1. Thus
This limiting argument explains the Poisson form, but a limit alone does not say how close a finite model is. Le Cam’s bound supplies a finite-model guarantee: in the binomial case, the maximum absolute error over count events is at most \(np^2\).
Worked Example: Checking a Zero-Claim Probability
A hypothetical portfolio has 100 independent policies. Each policy has probability \(0.01\) of producing one claim during the period. Let \(N\) be the number of claims. Approximate \(P(N=0)\), assess the approximation using Le Cam’s bound, and compare it with the exact probability.
Method. The count is binomial with \(n=100\) and \(p=0.01\). First calculate the mean and bound:
Under the Poisson approximation, \(Z\sim\operatorname{Poisson}(1)\). Therefore,
The exact binomial probability is
The observed absolute difference is approximately \(0.3678794-0.3660323=0.0018471\), which is less than the bound \(0.01\). The Poisson approximation is \(0.3679\) to four decimal places; the exact probability is \(0.3660\) to four decimal places. The bound guarantees that the exact probability lies within \(0.01\) of the unrounded Poisson probability.
Check and trap avoided. The parameter is \(np=1\), not \(p=0.01\). Using \(e^{-0.01}\) for the zero-claim approximation would ignore the 100 opportunities for a claim.
Turning the Bound into a Guaranteed Range
If the Poisson probability for an event is \(q\) and the bound is \(\varepsilon\), then the exact probability \(r\) satisfies \(|r-q|\leq\varepsilon\). Equivalently, \(q-\varepsilon\leq r\leq q+\varepsilon\). Since probabilities must also lie between 0 and 1, the range can be intersected with \([0,1]\) if needed.
When a problem requests a guaranteed range rounded to a specified number of decimals, round the lower endpoint downward and the upper endpoint upward. Ordinary rounding to nearest can make a displayed endpoint slightly narrower than the mathematically guaranteed interval.
Worked Example: Bounding a Zero-Claim Probability Without Replacing the Exact Model
A hypothetical book has 300 independent policies, each with claim probability \(0.004\). Use a Poisson approximation and Le Cam’s bound to give a guaranteed range for the exact probability of zero claims. Report the interval endpoints to four decimal places while preserving the guarantee.
Method. Here \(n=300\), \(p=0.004\), and
For a Poisson random variable with parameter \(1.2\), the probability of zero claims is
The exact binomial zero-claim probability differs from \(q\) by at most \(0.0048\). Thus the unrounded guaranteed range is
Rounding outward to four decimal places gives the guaranteed interval \([0.2963,\ 0.3060]\). The Poisson approximation itself is \(0.3012\) to four decimal places.
Check and trap avoided. The approximation is \(e^{-1.2}=0.3011942\ldots\), which rounds to \(0.3012\), not \(0.3010\). The endpoints must be formed from the unrounded value and the bound, then rounded outward; using a prematurely rounded approximation can produce an interval that is not guaranteed.
Unequal Trial Probabilities
A portfolio need not have the same claim probability for every exposure. If the independent probabilities are \(p_1,\ldots,p_n\), the expected count is \(\lambda=\sum_i p_i\). Matching this mean gives the Poisson approximation, and the error bound is the sum of the squared probabilities. In the binomial case, that sum reduces to \(np^2\).
This formulation also clarifies what makes the approximation more accurate: the aggregate contribution \(\sum_i p_i^2\) should be small. A fixed mean does not by itself ensure a small bound. For example, concentrating that mean in a few relatively large event probabilities can increase the sum of squares.
Worked Example: Unequal Claim Probabilities and a Point Probability
A hypothetical group has three independent exposure types, with one exposure of each type. Their claim probabilities are \(0.10\), \(0.05\), and \(0.02\). Let \(W\) be the total claim count. Approximate \(P(W=1)\), calculate the Le Cam bound, and check the exact probability.
Method. The Poisson parameter is the sum of the event probabilities:
The bound is the sum of their squares:
For \(Z\sim\operatorname{Poisson}(0.17)\), the approximating probability of exactly one claim is
For exactly one claim in the original model, sum the three mutually exclusive possibilities:
The absolute difference is \(|0.1543-0.1434230\ldots|=0.0108770\ldots\), which is less than \(0.0129\). The Poisson approximation is \(0.1434\) to four decimal places.
Check and trap avoided. The probabilities differ across exposure types, so the binomial shortcut \(np^2\) is not the appropriate calculation. The sum-of-squares bound uses each probability separately.
Interpreting the Bound and Avoiding Exam Traps
A small bound gives a strong absolute-error guarantee for every count event. A larger bound does not prove the approximation is poor; it only means this particular guarantee is not tight enough to establish high accuracy. As the first worked example shows, the actual error can be much smaller than the bound.
- Use the expected count as the Poisson parameter. In a binomial model, calculate \(\lambda=np\), not \(p\).
- Square each probability in the error bound. For a binomial count the result is \(np^2\), not \(np\), \(p^2\), or \(np(1-p)\).
- Keep the error absolute. A bound of \(0.01\) means at most one percentage point of absolute difference; it does not mean a one-percent relative difference.
- Check the assumptions. The stated Le Cam bound applies to a sum of independent Bernoulli indicators. It does not automatically cover dependent trials.
- Preserve the guarantee when rounding. Calculate endpoints from unrounded values and round the lower endpoint down and the upper endpoint up.
Assessing the bound is useful whenever the requested answer depends on a particular event probability rather than simply on the count’s mean. The calculation gives a disciplined decision: use the simpler Poisson probability with a quantified error, or retain the exact binomial calculation when the required accuracy is stricter than the bound supports. This material is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, identify the Poisson parameter and apply the stated bound before interpreting the approximation.
- A hypothetical set of 80 independent policies has claim probability \(0.005\) per policy. Approximate the probability of no claims and give a guaranteed range using Le Cam’s bound, with endpoints rounded outward to four decimal places. (Answer: \(\lambda=0.4\), bound \(0.002\), Poisson approximation \(e^{-0.4}=0.6703\); guaranteed range \([0.6683,\,0.6724]\).)
- A portfolio has 250 independent policies, each with claim probability \(0.002\). Approximate the probability of at least one claim and give a guaranteed range rounded outward to four decimal places. (Answer: \(\lambda=0.5\), bound \(0.001\), approximation \(1-e^{-0.5}=0.3935\); guaranteed range \([0.3924,\,0.3945]\).)
- Four independent exposures have claim probabilities \(0.03\), \(0.04\), \(0.01\), and \(0.02\). Calculate the exact and Poisson-approximated probabilities of no claims, and check the bound. (Answer: \(\lambda=0.10\), bound \(0.0030\), exact probability \(0.97(0.96)(0.99)(0.98)=0.90345024\), approximation \(e^{-0.10}=0.9048\); the absolute difference is approximately \(0.0014\), below the bound.)
- A hypothetical binomial claim count has \(n=500\) and \(p=0.001\). Approximate the probability of exactly two claims and give the Le Cam guaranteed range rounded outward to four decimal places. (Answer: \(\lambda=0.5\), bound \(0.0005\), approximation \(e^{-0.5}(0.5)^2/2=0.0758\); guaranteed range \([0.0753,\,0.0764]\).)
- For \(N\sim\operatorname{Binomial}(40,0.02)\), approximate \(P(N\leq1)\) and provide a guaranteed range rounded outward to four decimal places. (Answer: \(\lambda=0.8\), bound \(40(0.02)^2=0.016\), approximation \(e^{-0.8}(1+0.8)=0.8088\); guaranteed range \([0.7927,\,0.8248]\).)