In one of my papers, I needed an upper bound on the fractional moments of the geometric distribution. The integer moments have a nice-looking upper bound involving factorials. In addition, Mathematica showed that the same upper bound holds for fractional moments once we use Gamma function instead of factorials. The problem was, I could not prove it. Worse, I could not find a proof after much looking on the Internet.

This post contains a proof that for any real , the th moment of a geometric random variable with success probability is at most .

## Geometric distribution and its moments

Definition 1 (Geometric distributions and )Let . Define as the distribution on positive integers satisfying . Define as the distribution on non-negative integers satisfying .

Let and . Semantically, given an ordered list of independent and identical Bernoulli trials, counts *the number of trials to get the first success* while counts *the number of failures before the first success*.

For any real , we can express the moments of in terms of polylogarithm functions, as follows:

where is the polylogarithm function of order . It is defined as

Polylogarithm functions of non-positive integer orders can be found using the recursive rule

Recall from~(1) that for any real , . It follows that

## Main Proposition

*Proof.*

**Case: is an integer.** (This special case has a short proof which we record here for completeness. Note, however, that this case is subsumed by the following cases.)

For any real and such that , the moment generating function of is

We can differentiate the moment generating function from (7) to the order and utilize the fact to get

where are the Eulerian numbers satisfying for any . It follows that

**Case: is a real.** The proof for a real builds on the analytic properties of and . Let us focus on the functions

We claim that for a fixed . Specifically, observe that is a product of , and ; since each of these functions is convex in , so is . Moreover, since , we conclude that . Thus the claim is equivalent to showing for the specified regime of . Suppose for an integer . Let be the first-order approximation of around . Since the derivatives are positive for every , is a lower bound for . Recall that where is the Digamma function defined as . Since (cf. (3) in \cite{diamond2016bounds}), it follows that

On the other hand, the straight-line segment determined by the points and serves as an upper bound on the convex function in the interval . This line is

Observe that the line stays below the line for since we can use (1) and (9) to show that

This proves the claim for every real .

**Case: is a real.** Here, as before, we seek to show that for . The straight-line bounds from the previous case does not work in this case because for . Our argument, instead, relies on the following expression of the negative-order Polylogarithms in term of the Gamma function (cf. (13.1) in \cite{wood1992polylog}):

Using in (12) and further simplification, we can write (11) as

Our plan is to show that for since it would imply that

using .

It remains to show that for . First, observe that we can explicitly compute for using (13). In particular, we can use (3) to compute , and , and further using (13), we get

and, in particular, . Thus to show that for , it suffices to show that is strictly positive for every . The rest of the proof is devoted to this task.

Let and . Then , and . Thus

Since and , the signs of the trigonometric terms decide the sign of the right-hand side above. We claim that

which, in turn, implies that the right-hand side of (14) is strictly positive, as desired. Recall that

if , and

if . Define . Since for all , we have

and consequently, since ,

If we have then both and are strictly negative, satisfying (15). Otherwise, if , exactly one of these terms is strictly negative while the other one is zero; (15) is satisfied in this case as well.

The only remaining case is where ; in this case, the cosine term is negative but the sine term is positive. Let and . If we can show that , (15) will be satisfied. With this in mind, observe that implies , and that . Using this, we calculate

by a direct evaluation, and~(15) is satisfied, as desired.