Is the Sample Mean of a Poisson Distribution Normal?

If $X_1,\ldots,X_n$ are independent observations from a Poisson distribution with rate $\lambda$, their sample mean is not exactly normally distributed for a finite sample size. Its exact distribution is a scaled Poisson distribution. For a sufficiently large sample, however, the central limit theorem gives a useful normal approximation.

This answer assumes the observations are independent and share the same rate. The case of observations with different Poisson rates is noted below.

Exact distribution of the sample mean

Write the sample mean as

$$\bar X=\frac{X_1+\cdots+X_n}{n}.$$

The sum of independent Poisson random variables is Poisson, with rate equal to the sum of their rates. Thus, if each $X_i\sim\operatorname{Poisson}(\lambda)$, then

$$S=X_1+\cdots+X_n\sim\operatorname{Poisson}(n\lambda), \qquad \bar X=\frac{S}{n}.$$

In particular, the sample mean can take only the values $0,1/n,2/n,\ldots$. Its probability mass function is

$$P\left(\bar X=\frac{k}{n}\right) =e^{-n\lambda}\frac{(n\lambda)^k}{k!}, \qquad k=0,1,2,\ldots.$$

That discrete, equally spaced support is different from the continuous support of a nondegenerate normal distribution. The sample mean is therefore not exactly normal, even though its distribution may look close to normal in suitable cases.

Its expectation and variance are

$$\mathbb{E}[\bar X]=\lambda, \qquad \operatorname{Var}(\bar X)=\frac{\lambda}{n}.$$

The variance shrinks as the sample size increases, so sample means cluster more tightly around $\lambda$.

Why a normal approximation works

A Poisson random variable with rate $\lambda$ has mean and variance both equal to $\lambda$. Applying the central limit theorem to the independent observations gives

$$\frac{\bar X-\lambda}{\sqrt{\lambda/n}} \ \xrightarrow{d}\ N(0,1) \quad\text{as }n\to\infty,$$

for fixed $\lambda>0$. Equivalently, for large $n$,

$$\bar X\approx N\left(\lambda,\frac{\lambda}{n}\right).$$

Here the second parameter is the variance. The corresponding standard deviation is $\sqrt{\lambda/n}$. This is an approximation, not an identity: increasing $n$ makes the standardized sample mean closer in distribution to a standard normal, while the unstandardized sample mean itself becomes more concentrated around $\lambda$.

The moment-generating function also shows the distinction between the exact law and the normal limit. A Poisson variable with rate $\lambda$ has moment-generating function $M_X(t)=\exp(\lambda(e^t-1))$. Independence gives

$$M_{\bar X}(t) =\left[M_X\left(\frac{t}{n}\right)\right]^n =\exp\left(n\lambda\left(e^{t/n}-1\right)\right).$$

This is the moment-generating function of a Poisson variable with rate $n\lambda$, scaled by $1/n$. It is not the moment-generating function of a normal variable for finite $n$. As $n$ grows, the moment-generating function of the standardized mean approaches $e^{t^2/2}$, the standard normal moment-generating function.

Worked example

Suppose you record $n=25$ independent counts, each distributed as $\operatorname{Poisson}(4)$. The total count has the exact distribution

$$S=\sum_{i=1}^{25}X_i\sim\operatorname{Poisson}(100),$$

and $\bar X=S/25$. Therefore the sample mean is not a normal random variable; it takes values in steps of $1/25=0.04$.

Its mean is $4$ and its variance is $4/25=0.16$, so its standard deviation is $0.4$. A normal approximation is

$$\bar X\approx N(4,0.16).$$

For example, to approximate $P(\bar X\leq 4.5)$, translate the event to the sum: $S\leq112.5$. Because $S$ is an integer, this is the same as $S\leq112$. Using the continuity correction, approximate that probability with a normal variable of mean $100$ and standard deviation $10$:

$$P(S\leq112)\approx P\bigl(N(100,100)\leq112.5\bigr) =\Phi\left(\frac{112.5-100}{10}\right) =\Phi(1.25)\approx0.8944.$$

The continuity correction matters because the exact total is discrete while the normal approximation is continuous. The result is approximate; the exact probability is obtained by summing the Poisson probabilities for $S=0,\ldots,112$.

When is the approximation reasonable?

A useful diagnostic is the expected total count $n\lambda$. A Poisson variable’s standardized skewness is $1/\sqrt{n\lambda}$, so larger $n\lambda$ generally means less skew and a better normal approximation. There is no universal cutoff that guarantees a particular accuracy. Check the intended probability or tail if precision matters, especially for small expected counts or probabilities far out in a tail.

The normal approximation can assign some probability to negative values, even though a Poisson sample mean cannot be negative. This is usually unimportant when the mean is many standard deviations above zero, but it can be a warning sign when counts are small.

If the Poisson rates differ

If the observations are independent but $X_i\sim\operatorname{Poisson}(\lambda_i)$ with different rates, then

$$\sum_{i=1}^n X_i\sim\operatorname{Poisson}\left(\sum_{i=1}^n\lambda_i\right).$$

Consequently, the sample mean is still a scaled Poisson variable, but its expectation is $\frac{1}{n}\sum_i\lambda_i$ and its variance is $\frac{1}{n^2}\sum_i\lambda_i$. A normal approximation can again be used when the total expected count is sufficiently large.

Common mistakes

• Calling the sample mean Poisson: The sum is Poisson; dividing it by $n$ scales its possible values. Except when $n=1$, the sample mean is not itself a Poisson variable under the usual definition.
• Calling it exactly normal because of the CLT: The central limit theorem describes a limiting distribution and motivates an approximation for large samples; it does not make a finite-sample Poisson mean exactly normal.
• Using the wrong variance: The sample mean has variance $\lambda/n$, not $\lambda$. The sum has variance $n\lambda$.
• Ignoring discreteness in a probability calculation: When approximating a Poisson sum with a normal distribution, consider a continuity correction.