The Atom at Zero and Positive Outcomes

Why Zero and Positive Responses Have Different Origins

In a compound Poisson–gamma model, a response is the total amount accumulated across a random number of events. If there are no events, the total is exactly zero. If there is at least one event, each event contributes a positive gamma amount, so the total is positive. This mechanism gives the response two distinct features: a point mass at zero and a continuous distribution over positive values.

The distinction matters when interpreting the model. The zero probability is determined by the chance of observing no events. A positive outcome, by contrast, combines uncertainty about the number of events with uncertainty about the amount contributed by each event. It is therefore generally a mixture of gamma distributions, not just one gamma distribution.

Model and Parameterization

Let \(N\) be a Poisson event count with mean \(\lambda>0\). Given the count, let \(Z_1,Z_2,\ldots\) be independent and identically distributed gamma amounts, independent of \(N\). Use the shape-scale convention: a gamma variable with shape \(a>0\) and scale \(\beta>0\) has density \(z^{a-1}e^{-z/\beta}/(\Gamma(a)\beta^a)\) for \(z>0\).

These parameter formulas match the mean and variance of the Tweedie model. Indeed, for a gamma amount, \(E[Z_j]=\alpha\beta\) and \(E[Z_j^2]=\alpha(\alpha+1)\beta^2\). A compound Poisson sum has \(E[Y]=\lambda E[Z_j]\) and \(\operatorname{Var}(Y)=\lambda E[Z_j^2]\). Substitution gives \(E[Y]=\mu\) and \(\operatorname{Var}(Y)=\phi\mu^p\). We use the representation here to focus on the zero mass and positive outcomes.

The Atom at Zero

The event \(Y=0\) is exactly the event of no arrivals. If \(N=0\), the sum is empty and equals zero. If \(N=k\geq 1\), then \(Y\) is a sum of \(k\) strictly positive gamma amounts, and is therefore strictly positive with probability one.

Proof. A Poisson variable with mean \(\lambda\) satisfies \(P(N=0)=e^{-\lambda}\). When \(N=0\), the definition of the empty sum gives \(Y=0\). For any finite \(k\geq1\), every \(Z_j\) is positive with probability one, so \(\sum_{j=1}^k Z_j>0\) with probability one. Since a Poisson variable is finite with probability one, these facts show that \(Y=0\) if and only if \(N=0\), up to events of probability zero. Hence \(P(Y=0)=P(N=0)=e^{-\lambda}\).

To establish the conditional distribution, first note that the sum of two independent gamma variables with a common scale \(\beta\) has a gamma distribution with that scale and shape equal to the sum of their shapes. For shapes \(a,b>0\), the convolution density at \(y>0\) is \(\frac{e^{-y/\beta}}{\Gamma(a)\Gamma(b)\beta^{a+b}}\int_0^y x^{a-1}(y-x)^{b-1}\,dx\). Set \(u=x/y\); the integral becomes \(y^{a+b-1}\int_0^1 u^{a-1}(1-u)^{b-1}\,du=y^{a+b-1}\Gamma(a)\Gamma(b)/\Gamma(a+b)\), by the beta integral. The convolution is therefore the gamma density with shape \(a+b\) and scale \(\beta\). Applying this result repeatedly to \(k\) independent amounts gives shape \(k\alpha\) and scale \(\beta\). Each such conditional distribution has a density, so it assigns probability zero to any individual positive value. Mixing over the count preserves that property. \(\square\)

The Distribution Given a Positive Response

Because \(Y>0\) and \(N\geq1\) are equivalent up to probability-zero events, conditioning on a positive response amounts to conditioning the Poisson count to be nonzero. For \(k=1,2,\ldots\),

These are the probabilities of a zero-truncated Poisson distribution. They are positive and sum to one, since \(\sum_{k=1}^{\infty}\lambda^k/k!=e^\lambda-1\). Given \(N=k\), the positive response has the gamma distribution established above. Thus the conditional distribution given \(Y>0\) is a mixture: choose \(k\) using the zero-truncated Poisson probabilities, then draw a gamma variable with shape \(k\alpha\) and scale \(\beta\).

To see that this formula is a density, each gamma component integrates to one, and the mixture weights are nonnegative and sum to one. Equivalently, the unconditional density on \(y>0\) is the sum of the component densities weighted by \(P(N=k)\). Dividing by \(P(Y>0)=1-e^{-\lambda}\) gives the displayed conditional density. The sum and integral may be interchanged because all terms are nonnegative, by Tonelli’s theorem.

Conditional Means and Interpretation

Conditioning on a positive outcome changes the mean. Since \(Y=0\) whenever \(Y\) is not positive, \(E[Y]=P(Y>0)E[Y\mid Y>0]\). Using \(E[Y]=\mu\) and \(P(Y>0)=1-e^{-\lambda}\) gives \(E[Y\mid Y>0]=\mu/(1-e^{-\lambda})\). This is larger than the unconditional mean because the zero outcomes contribute nothing to \(E[Y]\).

Common Pitfall: Treating Positive Outcomes as One Gamma Distribution

A tempting shortcut is to say that, once the response is positive, it must have the gamma distribution of a single event amount. That is true only after additionally conditioning on \(N=1\). Given merely \(Y>0\), the event count can be \(1,2,3,\ldots\), and each count gives a different gamma shape. The correct conditional law is the zero-truncated Poisson mixture.

A related error is to use \(1-\lambda\) as the zero probability or as the probability of a positive response. The Poisson probability of zero is exactly \(e^{-\lambda}\); \(1-\lambda\) is not the Poisson probability of at least one event. For example, at \(\lambda=2\), \(1-\lambda=-1\) cannot even be a probability, whereas \(P(Y>0)=1-e^{-2}\approx0.8647\).