Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mu_X$?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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 $E[X^2]\leqslant2a+E[X]$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.