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If $X$ is discrete random variable taking values in non-negative integers $\{0,1,\ldots\}$, its probability generating function is defined as follows:

$$G(z)=\mathbb{E}(z^X)=\sum_{x=0}^\infty p(x)z^x$$

Suppose that $X$ has a finite first moment $\mathbb{E}[X]=\mu_X<\infty$.

Consider $z$ that satisfies $0<z<1$.

While one can use Bernoulli's inequality to obtain a lower bound $G(z)\geq 1-(1-z)\mu_X$ in this case, I wonder: is there a non-trivial upper bound for $G(z)$ that can be written in terms of $z$ and $\mu_X$?

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1 Answer 1

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If $X$ is square integrable, the pointwise inequality $z^X\leqslant1-X(1-z)+\frac12X(X-1)(1-z)^2$ yields $G_X(z)\leqslant1-\mu_X(1-z)+\frac12(\sigma_X^2+\mu_X^2-\mu_X)(1-z)^2$ with $\mu_X=E[X]$ and $\sigma_X^2=\mathrm{var}(X)$.

In the other direction, if $X$ is integrable and if $G_X(z)\leqslant1-E[X](1-z)+a(1-z)^2$ for every $z$ in $(0,1)$, for some finite $a$, then $X$ is square integrable and $E[X^2]\leqslant2a+E[X]$.

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