Deriving the Mean and Variance of a Poisson Variable

Why Derive the Mean and Variance?

For a Poisson count, the parameter \(\lambda\) is not just a setting that determines the probabilities of individual counts. It also equals the distribution’s mean and variance. Deriving these facts from the probability mass function (rather than simply memorizing them) shows how expectations work with an infinite series and why shifting an index can simplify a sum.

The result is useful when describing the center and spread of a Poisson model. It also gives a direct check on calculations: for a Poisson variable with parameter \(\lambda\), the mean must be \(\lambda\), and the variance must be \(\lambda\), though the standard deviation is \(\sqrt{\lambda}\), not \(\lambda\).

Model and Definitions

Let \(X\) be a Poisson random variable with parameter \(\lambda\), where \(\lambda>0\). Its possible values are the nonnegative integers \(0,1,2,\ldots\), and its probability mass function is:

P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}, \qquad k=0,1,2,\ldots

Here \(e\) is the base of the natural logarithm, and \(0!=1\). The parameter \(\lambda\) must be positive for the derivation below; the special case \(\lambda=0\) is addressed afterward.

Definition: For a discrete random variable \(X\), its expected value is \(E[X]=\sum_k kP(X=k)\), when the sum exists. Its variance is \(\operatorname{Var}(X)=E[(X-E[X])^2]\). Equivalently, when the relevant moments are finite, \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\).

The variance measures squared deviations from the mean, so it is in squared units. The standard deviation is the square root of the variance and is in the same units as the count. We will first find \(E[X]\), then find \(E[X^2]\) using an intermediate quantity, and finally combine them to get the variance.

Deriving the Mean

By the definition of expected value and the Poisson probability mass function,

E[X]=\sum_{k=0}^{\infty}kP(X=k) =\sum_{k=0}^{\infty}k e^{-\lambda}\frac{\lambda^k}{k!}.

The \(k=0\) term is zero, so the sum can start at \(k=1\). For every \(k\geq1\), \(k/k!=1/(k-1)!\). Factor out one \(\lambda\), then set \(j=k-1\). As \(k\) runs from 1 upward, \(j\) runs from 0 upward:

\begin{aligned} E[X] &=e^{-\lambda}\sum_{k=1}^{\infty}\frac{k\lambda^k}{k!}\\ &=e^{-\lambda}\sum_{k=1}^{\infty}\frac{\lambda^k}{(k-1)!}\\ &=\lambda e^{-\lambda}\sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{(k-1)!}\\ &=\lambda e^{-\lambda}\sum_{j=0}^{\infty}\frac{\lambda^j}{j!}. \end{aligned}

The exponential series gives \(\sum_{j=0}^{\infty}\lambda^j/j!=e^\lambda\). Therefore,

E[X]=\lambda e^{-\lambda}e^\lambda=\lambda.

The index shift is the key step: it turns the weighted Poisson sum into the exponential series that normalizes the probability mass function. The series converge for every finite \(\lambda\), so these rearrangements and index changes are valid.

Worked Example: Mean of a Poisson Count

Suppose, hypothetically, that the number \(X\) of items arriving at a service desk in a specified observation window follows a Poisson model with \(\lambda=2.5\). This example assumes the stated Poisson model is appropriate; it does not establish that assumption from observations.

Apply the mean derivation. In the general sum, replace \(\lambda\) with 2.5 and shift the index from \(k\) to \(j=k-1\):

\begin{aligned} E[X] &=e^{-2.5}\sum_{k=1}^{\infty}\frac{k(2.5)^k}{k!}\\ &=2.5e^{-2.5}\sum_{j=0}^{\infty}\frac{(2.5)^j}{j!}\\ &=2.5e^{-2.5}e^{2.5}\\ &=2.5. \end{aligned}

The expected number of arriving items in the specified window is 2.5 items. This is a long-run average under the model; an individual window’s count must be a whole number.

Deriving the Variance

To calculate \(E[X^2]\), it is convenient to first find \(E[X(X-1)]\), called the second factorial moment. The identity \(X^2=X(X-1)+X\) then gives the ordinary second moment. For a Poisson variable:

\begin{aligned} E[X(X-1)] &=\sum_{k=0}^{\infty}k(k-1)P(X=k)\\ &=e^{-\lambda}\sum_{k=2}^{\infty} \frac{k(k-1)\lambda^k}{k!}. \end{aligned}

The terms for \(k=0\) and \(k=1\) are zero, which is why the sum begins at 2. For \(k\geq2\), \(k(k-1)/k!=1/(k-2)!\). Factor out \(\lambda^2\) and shift the index by setting \(j=k-2\):

\begin{aligned} E[X(X-1)] &=e^{-\lambda}\sum_{k=2}^{\infty} \frac{\lambda^k}{(k-2)!}\\ &=\lambda^2e^{-\lambda}\sum_{k=2}^{\infty} \frac{\lambda^{k-2}}{(k-2)!}\\ &=\lambda^2e^{-\lambda}\sum_{j=0}^{\infty} \frac{\lambda^j}{j!}\\ &=\lambda^2e^{-\lambda}e^\lambda\\ &=\lambda^2. \end{aligned}

Now use \(X^2=X(X-1)+X\) and the mean already derived:

E[X^2]=E[X(X-1)]+E[X]=\lambda^2+\lambda.

Finally, apply the variance identity. Each quantity is finite because the exponential series converges, as in the sums above:

\begin{aligned} \operatorname{Var}(X) &=E[X^2]-(E[X])^2\\ &=(\lambda^2+\lambda)-\lambda^2\\ &=\lambda. \end{aligned}

Thus both the mean and variance equal \(\lambda\). The equality concerns the numerical values of these parameters, not their units: if \(X\) counts items, the mean is in items and the variance is in items squared.

Theorem: If \(X\sim\operatorname{Poisson}(\lambda)\) with \(\lambda>0\), then \(E[X]=\lambda\), \(E[X(X-1)]=\lambda^2\), \(E[X^2]=\lambda^2+\lambda\), and \(\operatorname{Var}(X)=\lambda\).

Worked Example: Variance and Standard Deviation

Suppose, hypothetically, that the number \(X\) of messages received by a monitoring station in a specified window follows a Poisson model with \(\lambda=4.2\). Take the Poisson model as given for this calculation.

Find the first two moments. The derivation gives \(E[X]=\lambda\) and \(E[X^2]=\lambda^2+\lambda\), so substitution yields:

E[X]=4.2,\qquad E[X^2]=(4.2)^2+4.2=17.64+4.2=21.84.

Calculate the variance.

\operatorname{Var}(X)=E[X^2]-(E[X])^2 =21.84-(4.2)^2 =21.84-17.64 =4.2.

The variance is 4.2 messages squared. The standard deviation is \(\sqrt{4.2}\approx2.049\) messages. The standard deviation is not 4.2 messages: that number is the variance, and it is expressed in squared units.

What the Equality Tells Us

A variance equal to the mean is a distinctive feature of the Poisson model. Once \(\lambda\) is specified, the same parameter determines both the model’s expected count and its variance. It does not mean that every count equals \(\lambda\), or that the count’s standard deviation equals its mean. Individual outcomes vary; \(\lambda\) describes the distribution across repeated observations under the model.

For instance, at \(\lambda=0.8\), the expected count is less than one, which is possible because an expectation need not be a possible individual count. The variance is also 0.8, while the standard deviation is approximately 0.894 counts. These summarize different aspects of the distribution: the expectation is its center, and the standard deviation describes spread on the count scale.

Worked Example: Separating the Mean, Second Moment, and Variance

Consider a hypothetical count \(X\) with a Poisson model and \(\lambda=0.8\). Assume the model applies to this count. Compute the mean, second moment, variance, and standard deviation, keeping the formulas distinct.

Mean: By the shifted-sum derivation, \(E[X]=\lambda=0.8\) counts.

Second moment: The factorial-moment calculation gives \(E[X(X-1)]=\lambda^2=(0.8)^2=0.64\). Since \(X^2=X(X-1)+X\),

E[X^2]=0.64+0.8=1.44\ \text{counts}^2.

Variance: Subtract the square of the mean from the second moment:

\operatorname{Var}(X)=1.44-(0.8)^2=1.44-0.64=0.8\ \text{counts}^2.

Standard deviation: Take the square root of the variance:

\sqrt{\operatorname{Var}(X)}=\sqrt{0.8}\approx0.8944\ \text{counts}.

The second moment, 1.44, is not the variance. The variance is 0.8 after subtracting the squared mean, and the standard deviation is about 0.8944 counts.

Boundary Case and Common Pitfalls

The derivation assumed \(\lambda>0\), so that factoring out \(\lambda\) is straightforward. At \(\lambda=0\), the Poisson distribution assigns probability 1 to \(X=0\): \(P(X=0)=1\), and \(P(X=k)=0\) for every \(k\geq1\). Therefore \(E[X]=0\) and \(\operatorname{Var}(X)=0\), so both formulas still hold at the boundary.

A frequent mistake is to calculate the variance as \(E[X^2]\). The correct formula is \(E[X^2]-(E[X])^2\). For a Poisson variable, \(E[X^2]\) is \(\lambda^2+\lambda\), not \(\lambda\); subtracting \(\lambda^2\) is what leaves the variance \(\lambda\). Another mistake is to report the standard deviation as \(\lambda\). The standard deviation is \(\sqrt{\lambda}\).

Key takeaway: Index shifts turn the Poisson expectation sums into the exponential series. They give \(E[X]=\lambda\) and \(E[X(X-1)]=\lambda^2\); together these imply \(E[X^2]=\lambda^2+\lambda\) and \(\operatorname{Var}(X)=\lambda\).

Check Your Understanding

Use the Poisson probability mass function and the results derived above to answer these questions.

In the mean derivation, why does the sum begin at \(k=1\), and what is the new lower limit after setting \(j=k-1\)?
For a Poisson variable with \(\lambda=3\), give \(E[X]\), \(E[X(X-1)]\), \(E[X^2]\), and \(\operatorname{Var}(X)\).
Explain why \(E[X^2]\) is not the same as \(\operatorname{Var}(X)\) for a Poisson variable with a positive parameter.
If \(\lambda=9\), what are the variance and standard deviation? Include the appropriate units if \(X\) counts events.
What are the mean and variance when \(\lambda=0\), and why does the positive-\(\lambda\) derivation need a separate boundary check?