Maximum Likelihood Estimator of the Poisson Distribution

The maximum likelihood estimator of the Poisson distribution’s rate parameter λ is the sample mean: λ̂ = x̄ = (x₁ + x₂ + … + xₙ) / n. In other words, add up the observed counts and divide by the number of observations. This page derives that result, works it through with real numbers, shows the standard error and a 95% confidence interval, and gives code for Excel, R and Python.

Definition in one paragraph

A maximum likelihood estimator (MLE) is the parameter value that makes the observed data most probable under the assumed model. For n independent counts drawn from a Poisson distribution, the likelihood is largest when λ equals the average of those counts. The NIST/SEMATECH e-Handbook of Statistical Methods states this directly: the maximum likelihood estimator of λ is the sample mean. The same quantity is the natural estimate of the average number of events per interval, which is what λ means in the model. For background on the distribution itself, see our guide to the Poisson distribution.

Formula with every symbol explained

The Poisson probability mass function is p(x; λ) = e^−λ λ^x / x!, for x = 0, 1, 2, …, where λ > 0 is the average number of events in the interval. The derivation takes four steps.

1. Likelihood. For independent observations x₁, …, xₙ: L(λ) = ∏ e^−λ λ^xᵢ / xᵢ! = e^−nλ λ^Σxᵢ / ∏ xᵢ!.
2. Log-likelihood. ℓ(λ) = −nλ + (Σxᵢ) ln λ − Σ ln(xᵢ!). Taking the logarithm turns the product into a sum and does not move the maximum.
3. Score equation. ℓ′(λ) = −n + (Σxᵢ) / λ. Setting it to zero gives λ̂ = (Σxᵢ) / n = x̄.
4. Check it is a maximum. ℓ″(λ) = −(Σxᵢ) / λ². This is negative whenever at least one count is positive, so the stationary point is a maximum.

Symbols: n is the number of observations, xᵢ is the i-th observed count, Σ is the sum over i = 1 to n, λ̂ (“lambda-hat”) is the estimate, and x̄ is the sample mean.

Worked example with real numbers

Suppose a lab records the number of contamination alerts per week for 10 weeks: 3, 5, 2, 4, 6, 1, 3, 4, 2, 5.

• Sum: 3 + 5 + 2 + 4 + 6 + 1 + 3 + 4 + 2 + 5 = 35
• n = 10
• λ̂ = 35 / 10 = 3.5 alerts per week
• Log-likelihood at λ̂ = 3.5: 35 × ln(3.5) − 10 × 3.5 − Σ ln(xᵢ!) = 43.847 − 35 − 27.480 = −18.633
• For comparison, at λ = 3 the log-likelihood is −19.029, lower, as expected for a value that is not the maximum.

Standard error and 95% confidence interval

The estimator is asymptotically normal with mean λ and variance λ/n (StatLect, Poisson distribution: maximum likelihood estimation). Replacing λ with λ̂ gives the standard error se = √(λ̂ / n) = √(3.5 / 10) = √0.35 = 0.592. A Wald 95% interval is λ̂ ± 1.96 × se = 3.5 ± 1.96 × 0.592 = 3.5 ± 1.160, or [2.34, 4.66]. With only 10 observations this large-sample interval is approximate.

Interpretation

λ̂ is the estimated mean count per interval, in the same time or space unit as your data. Because the Poisson mean and variance are both λ (NIST gives the standard deviation as √λ), the estimate also tells you the variance the model expects.

Quantity	Formula	Value in the example
Estimated rate	λ̂ = Σxᵢ / n	3.5 per week
Standard error	√(λ̂ / n)	0.592
Model-implied variance	λ̂	3.5
Model-implied standard deviation	√λ̂	1.871
Wald 95% CI	λ̂ ± 1.96 × se	[2.34, 4.66]

Is the estimator unbiased? Yes: E[x̄] = (1/n) Σ E[xᵢ] = λ, so the sample mean is unbiased for λ. Its variance λ/n shrinks as n grows.

Assumptions and common mistakes

• Independence. The derivation multiplies probabilities, so observations must be independent draws from the same Poisson distribution.
• Equidispersion. The Poisson model forces mean and variance to be equal. If your sample variance is far larger than x̄, the data are overdispersed and a Poisson model will understate uncertainty; see negative binomial regression for overdispersion and quasi-Poisson vs negative binomial.
• Frequency tables. If counts are summarised as “value k occurred f times”, compute Σ(k × f) / Σf, not the plain average of the k values.
• Unequal exposure. If intervals differ in length, the plain sample mean is not the right estimate of a per-unit rate; model the rate with an exposure term, as in a generalized linear model.
• All zeros. If every count is 0, λ̂ = 0, which lies on the boundary of the parameter space; the second-derivative check and the Wald interval do not apply.

How to report it (APA example)

Report the estimate, its precision and the method: “The rate of contamination alerts was estimated by maximum likelihood under a Poisson model, λ̂ = 3.50 alerts per week (SE = 0.59, 95% CI [2.34, 4.66], n = 10 weeks).” State the unit, the sample size and the interval method (Wald, large-sample). Where you compare nested Poisson models, see the likelihood ratio test guide.

How to calculate it in Excel, R or Python

Excel

B1: =AVERAGE(A2:A11)                      (lambda-hat)
B2: =SQRT(AVERAGE(A2:A11)/COUNT(A2:A11))  (standard error)

R

x <- c(3,5,2,4,6,1,3,4,2,5)
lambda_hat <- mean(x)                 # 3.5
se <- sqrt(lambda_hat / length(x))    # 0.592
lambda_hat + c(-1, 1) * 1.96 * se     # 2.34 4.66

Python

import numpy as np
x = np.array([3,5,2,4,6,1,3,4,2,5])
lam = x.mean()                        # 3.5
se = np.sqrt(lam / len(x))            # 0.592
print(lam, lam - 1.96*se, lam + 1.96*se)

FAQ

Is the MLE of Poisson λ the sample mean?

Yes. Setting the derivative of the log-likelihood, −n + Σxᵢ/λ, to zero gives λ̂ = Σxᵢ/n, the sample mean. NIST’s e-Handbook states the same result.

Is it unbiased?

Yes. E[x̄] = λ for any sample size n.

What if all counts are zero?

Then λ̂ = 0. The likelihood e^−nλ is largest at λ = 0, but a zero rate is outside the model’s λ > 0 range, so treat the result as a boundary case and avoid the Wald interval.

How do I get the variance of the estimate?

Var(λ̂) = λ/n, estimated by λ̂/n. Its square root is the standard error.

Sources

• NIST/SEMATECH e-Handbook of Statistical Methods, Poisson distribution (pmf, mean, standard deviation and the maximum likelihood estimator of λ). Date checked: 2026-10-29.
• StatLect, Poisson distribution: maximum likelihood estimation (assumptions, asymptotic variance λ/n). Date checked: 2026-10-29.