Expected value of geometric mean of Poisson random variables

Question

I am interested in trying find the expected value of the geometric mean of a set of i.i.d. Poisson random variables.

Say we have $Y_1,\dots,Y_n$, where

$$Y_i \sim Poisson(\lambda) $$

Then, the geometric mean can be expressed as:

$$ GM(Y_1,\dots,Y_n)=(\prod_1^n{Y_i})^{\frac{1}{n}}$$

Then, the expected value of the geometric mean can be expressed as:

$$ E\left[GM\right] = E\left[(\prod_1^n{Y_i})^{\frac{1}{n}}\right] = \left(E\left[Y_i^{\frac{1}{n}}\right]\right)^{n} $$

The last step being a consequence of the i.d.d. assumption.

The crux, then, seems to be figuring out how to calculate $E\left[Y_i^{\frac{1}{n}}\right]$. I am stuck on trying to figure out how to evaluate this. I made several attempts, but made no progress with any.

1. This expected value can be written as $$ E\left[Y_i^{\frac{1}{n}}\right] = \sum_{y=0}^{\infty}y^{\frac{1}{n}} \frac{e^{-\lambda}\lambda^y}{y!} $$ But I cannot find a way to proceed from there.

2. My next thought was to see if I could use moment generating functions, but I don't know how this is possible for a non-integer moment. Traditionally, we find the $i^{th}$ moment by taking the $i^{th}$ derivative of the MGF evaluated at $t=0$, so there is not to my knowledge any straightforward way to evaluate non-integer moments.

3. In lieu of an exact solution, I wondered at the possibility of approximating it using a Taylor expansion, which (I believe) would give us

$$E\left[Y_i^{\frac{1}{n}}\right]=\lambda^{\frac{1}{n}}- \frac{{(\frac{1}{n}-1})\lambda^{\frac{1}{n}-2}}{n2!} + \frac{{(\frac{1}{n}-2)(\frac{1}{n}-1})\lambda^{\frac{1}{n}-3}}{n3!} +\cdots $$

But I'd prefer to find an analytic solution.

Does anybody have any tips on how I should proceed?

Answer 1

It seems that one possible answer to my question is rather straightforward. For a Poisson random variable, the expected value of the geometric mean will be 0. Indeed, for any random variable $X$, if $P(X=0)>0$, then $E(GM)=0$, per Feng et al. (2013). Of course, the sample geometric mean for any particular realization of $n$ Poisson r.v.s may be greater than 0, if none of the realized values of the r.v.s are equal to 0.

This brings up the question of how to think about the expected value of the geometric mean as $\lambda$ increases and $P(X=0)\rightarrow 0$. But that seems like a more complicated question...

I am posting this as an answer, but will refrain from accepting it for now because I'm curious to see if I get any other feedback.

Feng, Changyong; Wang, Hongyue; Tu, Xin M., Geometric mean of nonnegative random variable, Commun. Stat., Theory Methods 42, No. 15, 2714-2717 (2013). ZBL1277.62060.

Answer 2

My best suggestion is to use the approximation in this article(1).

On p.41 this 4th order approximation is given: G_n to 4th order

"...the only general constraint is that the range on either side of the mean should not exceed the mean itself."(1, p.40)

The approximation is valid for all $0<Y<2\lambda$, which might be a limitation for some purposes.

In 2nd order this gives:

$G_n=\lambda(1-\frac12(1-\frac1n) \frac{\sigma^2}{\lambda^2} +...)$ with $n$ the number of samples.

With $\sigma^2=\lambda$ for a Poisson distribution this equals:

$G_n=\lambda(1-\frac1{2\lambda}(1-\frac1n)+...)$

Maybe this is helpful. The article uses $X$ and $\mu$ iso $Y$ and $\lambda$...

The truncation of the Poisson distribution at $2\lambda$ is ok for practical purposes if this truncation starts at $\lambda+3\sigma$ or more. Therefore the minimal value of $\lambda$ is given by:

$\lambda_\min \approx 3\sqrt\lambda_\min$

$\lambda_\min \approx 9$

1. David A. L. Wilson, & Martin, B. (2006). The Distribution of the Geometric Mean. The Mathematical Gazette, 90(517), 40-49. Retrieved July 29, 2021, from http://www.jstor.org/stable/3621411