# Maximum variance of a discrete probability distribution over the non-negative integers.

Suppose $P_{n}$ is a probability distribution over the non-negative integers (i.e. $n=0,1,2,...$). Also, assume that the average \begin{equation} \bar{n} :=\langle n\rangle= \sum_{n=0}^{\infty} n \, P_{n} \end{equation} is fixed. What is the maximum value (in terms for $\bar{n}$) that the variance \begin{equation} \text{Var}[n]:= \langle n^2\rangle -\langle n\rangle^2= \left(\sum_{n=0}^{\infty} n^2 \, P_{n}\right) -\bar{n}^2 \end{equation} can take? In other words, is there an upper-bound on $\text{Var}[n]$ in terms of $\bar{n}$ [i.e. $\text{Var}[n] \le f(\bar{n})$] ?

My guess:$\text{Var}[n] \le 2\bar{n}\left(\bar{n}+1\right)$.

• Not necessarily true. Take $P_n$ proportional to $n^{-(2+\epsilon)}$ for some small $\epsilon>0$. The variance of such distribution is $\infty$. – d.k.o. Jun 18 '15 at 3:00
• Let If (for $n\ge 1$) $P_n$ is a suitable constant times $\frac{1}{n^3}$, then the mean is nice but the variance does not exist. – André Nicolas Jun 18 '15 at 3:03

## 1 Answer

For any integer $n\ge 1$, let $\Pr(X=n)=P_n$, where $$P_n =\frac{1}{\zeta(3)}\cdot \frac{1}{n^3},$$ and $\zeta(3)=1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots$. Note that $\zeta(3)\approx 1.202$. Then $$E(X)=\frac{1}{\zeta(3)}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right)=\frac{\pi^2}{6\zeta(3)}.$$ But $E(X^2)$ does not exist (it is infinite).

Thus there is no bound on $\text{Var}(X)$ in terms of $E(X)$.

If you want examples where the variance exists, let $P_n$ be a suitable constant times $\frac{1}{n^{3+\epsilon}}$, where $\epsilon$ is small positive. The mean of $X$ will be $\lt \frac{\pi^2}{6}$, but the variance of $X$ can be made arbitrarily large positive by choosing $\epsilon$ suitably small.