The Poisson Case at Power One

Why Power One Leads to a Count Model

A count model must describe more than an average. It must assign probabilities to possible outcomes such as zero, one, or several events, while also relating the mean to the variability. In a generalized linear model, the power variance function at \(p=1\) gives the relationship \(V(\mu)=\mu\). The ordinary Poisson distribution is the central count model with this variance function and unit dispersion.

The distinction between the variance function and the full variance is important. In an exponential-dispersion model, the variance is \(\phi V(\mu)\), where \(\phi>0\) is the dispersion parameter. Thus power one gives \(\operatorname{Var}(Y)=\phi\mu\). The ordinary Poisson distribution has \(\phi=1\), so its variance equals its mean. The power alone does not specify the probability support or every detail of a distribution; for counts, the usual Poisson realization has support on the nonnegative integers.

Definition: The power variance function at \(p=1\) is \(V(\mu)=\mu\), for \(\mu>0\). In an exponential-dispersion model with this variance function, \(E[Y]=\mu\) and \(\operatorname{Var}(Y)=\phi\mu\). An ordinary Poisson random variable has \(Y\sim\operatorname{Poisson}(\mu)\), takes values \(0,1,2,\ldots\), and has unit dispersion: \(\phi=1\).

From the Power-One Variance Function to the Log Link

In the canonical parameterization of an exponential-dispersion model, the mean is \(\mu=b'(\theta)\), and the variance function is \(V(\mu)=b''(\theta)\). These identities apply when the cumulant function \(b\) is twice differentiable on the natural-parameter space. Since \(d\mu/d\theta=b''(\theta)\), the power-one choice gives a simple differential equation.

Result: For \(V(\mu)=\mu\), the canonical parameter is \(\theta=\log\mu\), up to the conventional choice of an additive constant, and the cumulant function can be written \(b(\theta)=e^\theta\).

Use the variance identity.
Because \(\mu=b'(\theta)\), differentiation gives \(d\mu/d\theta=b''(\theta)=V(\mu)=\mu\).

Solve the differential equation.
For \(\mu>0\), divide by \(\mu\) and integrate: \(d\mu/\mu=d\theta\), so \(\log\mu=\theta+C\). Choose the standard canonical origin \(C=0\), giving \(\theta=\log\mu\).

Recover the cumulant function.
Since \(b'(\theta)=\mu=e^\theta\), integration gives \(b(\theta)=e^\theta+C_1\). The irrelevant additive constant in \(b\) is conventionally set to zero.

This derivation identifies the canonical link for the mean: \(\theta=\log\mu\). It does not, by itself, prove that any response with variance equal to its mean must be Poisson. A mean and variance do not determine a distribution. The Poisson count model follows when the usual nonnegative-integer support and Poisson probability mass function are used.

The Poisson Probability Mass Function and Its Moments

For \(\mu>0\), the Poisson model assigns probability \(e^{-\mu}\mu^k/k!\) to each nonnegative integer \(k\). The following derivation verifies both that these probabilities sum to one and that their mean and variance match the power-one model.

Theorem: If \(Y\) has probability mass function \(P(Y=k)=e^{-\mu}\mu^k/k!\) for \(k=0,1,2,\ldots\), then the probabilities sum to one, \(E[Y]=\mu\), and \(\operatorname{Var}(Y)=\mu\).

Proof. The exponential series gives

\sum_{k=0}^{\infty} e^{-\mu}\frac{\mu^k}{k!} =e^{-\mu}e^\mu=1.

All terms are nonnegative, so the series defines a probability mass function. For the mean, the \(k=0\) term is zero, and reindexing the remaining absolutely convergent series gives

E[Y] =e^{-\mu}\sum_{k=1}^{\infty} k\frac{\mu^k}{k!} =e^{-\mu}\mu\sum_{j=0}^{\infty}\frac{\mu^j}{j!} =\mu.

Likewise, using \(Y(Y-1)\) and reindexing from \(k=2\),

E[Y(Y-1)] =e^{-\mu}\sum_{k=2}^{\infty} k(k-1)\frac{\mu^k}{k!} =e^{-\mu}\mu^2\sum_{j=0}^{\infty}\frac{\mu^j}{j!} =\mu^2.

Since \(Y^2=Y(Y-1)+Y\), it follows that \(E[Y^2]=\mu^2+\mu\). Therefore

\operatorname{Var}(Y)=E[Y^2]-(E[Y])^2 =(\mu^2+\mu)-\mu^2=\mu.

Thus the ordinary Poisson distribution has the power-one variance function and unit dispersion. \(\square\)

Worked Example: Probability of Exactly Two Events

Suppose the number of events in a fixed interval has an ordinary Poisson distribution with mean \(\mu=2.5\). Find the probability of exactly two events.

Insert \(k=2\) and \(\mu=2.5\) into the Poisson mass function:

P(Y=2)=e^{-2.5}\frac{(2.5)^2}{2!} =e^{-2.5}\frac{6.25}{2} =3.125e^{-2.5}.

Using \(e^{-2.5}\approx0.082085\), the probability is

P(Y=2)\approx3.125(0.082085)=0.256516.

So the chance of exactly two events is about \(0.257\). The factorial in the denominator matters: dividing \(6.25\) by \(2\) gives \(3.125\), not \(1.5625\).

Dispersion and the Meaning of “Poisson”

For power one, the exponential-dispersion variance is \(\phi\mu\), not automatically \(\mu\). One concrete realization for any \(\phi>0\) is a scaled Poisson variable: let \(Z\sim\operatorname{Poisson}(\mu/\phi)\) and set \(Y=\phi Z\). Then \(Y\) has mean \(\mu\) and variance \(\phi\mu\), but its possible values are \(0,\phi,2\phi,\ldots\). Except when \(\phi=1\), this is not an ordinary Poisson count variable on all nonnegative integers.

This distinction makes the unit-dispersion restriction precise. If a response is modeled as an ordinary Poisson count, its variance is fixed to equal its mean. A freely varying \(\phi\) changes that variance relationship and, in the scaled-Poisson realization, changes the support as well. The statement “\(p=1\)” identifies the form \(V(\mu)=\mu\); the statement “ordinary Poisson” additionally specifies count support and unit dispersion.

Worked Example: A Nonunit-Dispersion Power-One Model

Let \(\phi=2\), \(\mu=6\), \(Z\sim\operatorname{Poisson}(\mu/\phi)\), and \(Y=\phi Z\). Find the mean, variance, support, and probability \(P(Y=4)\).

First, \(\mu/\phi=6/2=3\), so \(Z\sim\operatorname{Poisson}(3)\) and \(Y=2Z\). Using the Poisson mean and variance,

E[Y]=2E[Z]=2(3)=6,\qquad \operatorname{Var}(Y)=2^2\operatorname{Var}(Z)=4(3)=12.

The support is \(0,2,4,6,\ldots\), rather than every nonnegative integer. The event \(Y=4\) is exactly the event \(Z=2\), so

P(Y=4)=P(Z=2) =e^{-3}\frac{3^2}{2!} =4.5e^{-3} \approx4.5(0.0497871) \approx0.224042.

The variance \(12\) equals \(\phi\mu=2(6)\), confirming the power-one variance relationship, but it does not equal the mean. This is not an ordinary Poisson count model.

Calculating Count Probabilities

For an ordinary Poisson count with mean \(\mu\), the probability of an exact count \(k\) is obtained directly from the mass function. Probabilities of ranges are sums of the relevant exact probabilities. For an upper tail, it is often shorter to subtract the probability of the complementary lower range from one.

Worked Example: Probability of at Least Five Events

Suppose \(Y\sim\operatorname{Poisson}(3)\). Calculate \(P(Y\geq5)\).

The complement is \(Y\leq4\), so

P(Y\geq5)=1-P(Y\leq4) =1-\sum_{k=0}^{4}e^{-3}\frac{3^k}{k!}.

Evaluate the five terms inside the sum:

\sum_{k=0}^{4}e^{-3}\frac{3^k}{k!} =e^{-3}\left(1+3+\frac{9}{2}+\frac{27}{6}+\frac{81}{24}\right) =e^{-3}(16.375).

Since \(e^{-3}\approx0.0497871\),

P(Y\geq5)=1-(16.375)(0.0497871) \approx1-0.815263 =0.184737.

Thus the probability of five or more events is about \(0.185\). This calculation uses the complement so that only five probabilities need to be summed.

Common Mistake and Why the Distinction Matters

A frequent mistake is to see \(V(\mu)=\mu\) and conclude that the response must be an ordinary Poisson variable, even when the model includes an arbitrary dispersion parameter. That reasoning omits the factor \(\phi\): the model variance is \(\phi V(\mu)=\phi\mu\). The correction is to state both parts of the model. Power one specifies the mean–variance form up to dispersion; an ordinary Poisson count specifies \(\phi=1\) and nonnegative-integer support.

The distinction matters whenever the model is used to calculate probabilities. Under an ordinary Poisson model, probabilities are assigned to integer counts by the Poisson mass function. A nonunit-dispersion power-one model need not have that same support or those same probabilities. It is therefore not valid to insert a nonunit \(\phi\) into the ordinary Poisson formula while still treating the result as an ordinary count distribution.

Key takeaway: At \(p=1\), \(V(\mu)=\mu\) and the canonical link is the log link. The ordinary Poisson count model is the unit-dispersion case, with \(E[Y]=\operatorname{Var}(Y)=\mu\) and \(P(Y=k)=e^{-\mu}\mu^k/k!\).

Check Your Understanding

Use the definitions and calculations above to answer these questions.

For power one and dispersion \(\phi=1.5\), express the variance in terms of the mean.
Why does the equation \(V(\mu)=\mu\) alone not establish that a response has a Poisson distribution?
If \(Y\sim\operatorname{Poisson}(4)\), write the expression for \(P(Y=3)\) before evaluating it.
For \(Z\sim\operatorname{Poisson}(2)\) and \(Y=3Z\), give the support, mean, and variance of \(Y\).
Explain why an ordinary Poisson model uses unit dispersion.

What You'll Learn