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When discrete r.v. $X$ is not Poisson distributed and ${\rm{Var}}X,EX < \infty $, I want to know whether r.v. $X$ with p.g.f. $\exp [\sum\limits_{i = 1}^\infty {{q_i}({z^i}} - 1)],({q_i} \in {\rm{R,}}\sum\limits_{i = 1}^\infty {\left| {{q_i}} \right|} < \infty )$ is overdispersion?

The overdispersion means that the variance is larger than the expectation.

Note that $X$ can be decomposed to $$X = {Z_1} + 2{Z_2} + \cdots + i{Z_i} + \cdots ,$$ where ${Z_i}(i = 1,2, \cdots )$ are independently signed Poisson distributed with parameter ${q_i}$.

Then we have $EX = \sum\limits_{i = 1}^\infty {i{q_i}} ,{\rm{Var}}X = \sum\limits_{i = 1}^\infty {{i^2}{q_i}} $.

It is easily seen that ${\rm{Var}}X \ge EX$ as all ${q_i}$ are non-negative , the equality hold in the case $EX = {q_1} = {\rm{Var}}X$.

How about the case that some ${q_i}$ are negative?

We can also obtain $VarX$ by directly using p.g.f., see my comments.

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  • $\begingroup$ Shouldn't the variance formula have absolute value, $|q_i|$? in that case you trivially get $Var X \geq E[X]$ since the variance remains as in the unsigned case and the expectation can only decrease $\endgroup$
    – Or Zuk
    Commented Apr 12, 2014 at 19:17
  • $\begingroup$ We should't. Let G(z) be the p.g.f. of $X$, then $$G'(z) = (\sum\limits_{i = 1}^\infty {i{q_i}} {z^{i - 1}})G(z),G''(z) = {(\sum\limits_{i = 1}^\infty {i{q_i}} {z^{i - 1}})^2}G(z) + (\sum\limits_{i = 2}^\infty {i(i - 1){q_i}} {z^{i - 2}})G(z)$$. There is the other method to obtain ${\rm{Var}}X$, $$\begin{array}{l} {\rm{Var}}X = EX + EX(X - 1) - {(EX)^2}\\ = [G'(z) + G''(z) - {(G'(z))^2}{\left. ] \right|_{z = 1}}\\ = \sum\limits_{i = 1}^\infty {{i^2}{q_i}} \end{array}$$ $\endgroup$
    – user48365
    Commented Apr 12, 2014 at 20:07

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