Why a Bound Adds Information
A Poisson distribution can provide a useful approximation to a binomial count when each of many independent trials has a small success probability. Comparing a few individual probabilities can suggest whether the approximation is close, but a formal bound gives a guarantee that applies to every event involving the count. Le Cam’s bound provides that guarantee using only the binomial parameters.
The bound is especially informative when comparing models with the same mean. Even if two binomial counts have the same expected number of successes, their approximation bounds can differ because the numbers of trials and the individual success probabilities differ.
Model, Distance, and Le Cam’s Bound
For two discrete distributions \(P\) and \(Q\), define their total variation distance by
where the supremum is over all sets of possible outcomes. In this tutorial, the outcomes are nonnegative integers. Thus, a bound on total variation controls the absolute difference in probabilities for any event, such as \(X=0\), \(X\geq 4\), or \(X\in\{2,3\}\).
Here \(\mathcal{L}(X)\) denotes the distribution of \(X\). The cap of 1 is valid because the absolute difference between two probabilities cannot exceed 1. It is useful when \(2np^2\) is large, but a bound above 1 provides no practical precision guarantee.
Why the Bound Has an \(np^2\) Term
The dependence on \(n\) and \(p^2\) reflects the accumulation of small discrepancies across trials. For independent Bernoulli trials with probabilities \(p_1,\ldots,p_n\), the corresponding Poisson approximation has mean \(\lambda=\sum_{j=1}^n p_j\). A standard Le Cam inequality bounds the sum of the absolute differences between the two probability mass functions by \(2\sum_{j=1}^n p_j^2\). For equal probabilities \(p_j=p\), this becomes \(2np^2\).
The reason a squared probability appears can be seen by comparing one Bernoulli variable \(B\sim\operatorname{Bernoulli}(p)\) with \(W\sim\operatorname{Pois}(p)\). Their total variation distance is \(p(1-e^{-p})\): at outcome 1, the Bernoulli probability exceeds the Poisson probability by that amount, balancing the total deficit at the other outcomes. Since \(1-e^{-p}\leq p\), this distance is at most \(p^2\). Across \(n\) independent coordinates, these small discrepancies accumulate on the scale \(np^2\); mapping the coordinate vector to its sum cannot increase total variation. This explains why a bound proportional to \(np^2\) is natural. The theorem’s stated factor of 2 is a standard conservative form.
Another convention defines total variation for discrete distributions as half the sum of the absolute differences of their probability masses. That convention equals the supremum-over-events definition used here. Some statements of Le Cam’s inequality instead bound the full sum of absolute differences; checking the convention prevents a factor-of-two mix-up when reporting the result.
Applying the Bound
Confirm that the count is based on \(n\) independent trials with the same success probability \(p\). If the probabilities differ, it is not a binomial model; the generalized bound uses the sum of their squares.
Use \(\lambda=np\), so the approximation is \(Z\sim\operatorname{Pois}(np)\).
Compute \(2np^2\), and cap the result at 1 if needed.
For any event \(A\), the absolute difference between the binomial and Poisson probabilities is no larger than the bound.
The bound concerns absolute error, measured in probability points. It does not say that the relative error is small. For an event whose probability is extremely small, even a small absolute bound can be large relative to the event probability.
Worked Example: Bounding the Error for an At-Least-One Event
A hypothetical quality check inspects \(n=400\) independent components. Each component independently fails the check with probability \(p=0.002\). Let \(X\) be the number of failures. Use Le Cam’s bound to control the error in approximating \(P(X\geq 1)\) with a Poisson probability, and compare the two probabilities. No interest or mortality model is involved; the count model is binomial with parameters \(n=400\) and \(p=0.002\).
Method. Approximate \(X\sim\operatorname{Bin}(400,0.002)\) by \(Z\sim\operatorname{Pois}(\lambda)\), where
The Le Cam bound is
For \(A=\{1,2,3,\ldots\}\), the absolute probability error is therefore at most \(0.0032\). The Poisson approximation gives
The exact binomial probability is
The observed absolute difference is approximately \(|0.551031-0.550671|=0.000360\), which is below the bound \(0.0032\). To four decimal places, the probabilities are \(0.5510\) and \(0.5507\), and their difference is \(0.0004\). Common mistake avoided: the bound is an upper limit, not a prediction that the actual error will equal \(0.0032\).
What Changes When \(n\) and \(p\) Change?
For a binomial model, the bound can also be written in terms of the mean:
At a fixed mean \(\lambda\), reducing the trial probability \(p\) reduces this bound. Maintaining the same mean while reducing \(p\) requires increasing \(n\), so having more trials does not automatically make the approximation worse. The squared individual probability is important.
Worked Example: Same Mean, Different Trial Designs
Two hypothetical monitoring systems count independent alerts. System A has \(n=50\) opportunities for an alert, each with probability \(p=0.04\). System B has \(n=100\) opportunities, each with probability \(p=0.02\). Compare their Poisson means and Le Cam bounds, then compare their probabilities of observing no alerts. Both counts are binomial; the respective parameters are \((50,0.04)\) and \((100,0.02)\).
System A. Its Poisson mean is \(\lambda_A=50(0.04)=2\). Its bound is
The exact no-alert probability is \(P(X_A=0)=(1-0.04)^{50}=0.96^{50}\approx0.1299\). The Poisson value is \(P(Z_A=0)=e^{-2}\approx0.1353\). Their absolute difference is approximately \(0.0054\), within \(0.16\).
System B. Its Poisson mean is \(\lambda_B=100(0.02)=2\), the same as System A. Its bound is
The exact no-alert probability is \(P(X_B=0)=(1-0.02)^{100}=0.98^{100}\approx0.1326\). The Poisson value is again \(e^{-2}\approx0.1353\). Their absolute difference is approximately \(0.0027\), within \(0.08\).
Both systems have mean 2, but System B has the smaller bound because its individual opportunity probability is smaller. The observed differences also satisfy their respective guarantees. Values are rounded to four decimal places. Common mistake avoided: comparing only the means would miss the change in the bound.
Worked Example: Using the Bound for a Count Event
A hypothetical screening program has \(n=120\) independent tests, each with probability \(p=0.01\) of a flagged result. Let \(X\) count flagged results. Find the Poisson approximation to the probability of zero flags, calculate the maximum error guaranteed by Le Cam’s bound, and compare with the exact binomial probability. The model is \(X\sim\operatorname{Bin}(120,0.01)\), approximated by \(Z\sim\operatorname{Pois}(1.2)\).
First, \(\lambda=np=120(0.01)=1.2\). For the event \(A=\{0\}\), the Poisson probability is
The Le Cam bound is
Thus, the absolute error for the zero-flag probability cannot exceed \(0.024\). The exact binomial probability is
The actual difference is approximately \(|0.2994-0.3012|=0.0018\), below \(0.024\). Probabilities and the difference are rounded to four decimal places. Common mistake avoided: the event \(X=0\) is covered by the same bound as every other event; the theorem is not limited to tail probabilities.
Interpreting the Guarantee Carefully
Le Cam’s bound is useful as a parameter-based guarantee, but its size must be interpreted in context. A small bound assures that every event probability is close in absolute terms. A large bound does not prove that the approximation is poor; it only means this particular guarantee is not informative.
- Use \(p^2\), not \(p\), in the error calculation. The bound is \(2np^2\), not \(2np\).
- Keep the Poisson mean at \(np\). The approximation is not generally \(\operatorname{Pois}(n)\) or \(\operatorname{Pois}(p)\).
- Distinguish a bound from an error estimate. The actual difference can be far smaller than the bound.
- Do not infer relative accuracy from absolute accuracy. If the probability being approximated is tiny, the relative error may still be large.
- Check independence and the common success probability. The binomial version assumes both; a count with unequal trial probabilities requires a different model statement.
Check Your Understanding
For each problem, use the stated binomial model and the Le Cam bound \( \min(1,2np^2) \). Report probabilities or bounds to four decimal places where appropriate.
- A device is tested independently \(200\) times, with probability \(0.005\) of a fault on each test. State the Poisson mean and the Le Cam bound. (Answer: \(\lambda=1\); bound \(=0.0100\).)
- A count has distribution \(\operatorname{Bin}(80,0.015)\). What Poisson parameter should be used, and what is the bound on the absolute error for the probability of at least two successes? (Answer: \(\lambda=1.2\); bound \(=0.0360\).)
- A batch contains \(500\) independently assessed items, each flagged with probability \(0.0015\). Find the mean of the Poisson approximation and the Le Cam bound. (Answer: \(\lambda=0.75\); bound \(=0.0023\).)
- Compare \(\operatorname{Bin}(60,0.05)\) and \(\operatorname{Bin}(150,0.02)\). They have the same mean. Find each bound and state which has the smaller bound. (Answer: both means are \(3\); bounds are \(0.3000\) and \(0.1200\), respectively; the second is smaller.)
- For \(n=250\), what is the largest \(p\) that makes the uncapped Le Cam bound no greater than \(0.02\)? (Answer: \(p\leq\sqrt{0.02/(2\cdot250)}=\sqrt{0.00004}\approx0.0063\).)