# What is the definition of the geometric mean of a random variable?

I haven't found any definition online although searching for hours:

So here is the thing: The geometric mean(GM) of an (iid) sample drawn from some random variable is given by:

$$\text{GM}(X_1,...,X_n)=\sqrt[n]{X_1\cdot ...\cdot X_n}$$

and the expected geometric mean for a sample size $$n$$ is given by

$$\mathbb E(X_1^{1/n}\cdot ... \cdot X_n^{1/n})=\left[E\left(X_1^{1/n}\right)\right]^n.$$

So we can see that the (expected) geometric mean is dependent on the sample size $$n$$.

However, some sources like wikipedia talk about geometric moments e.g. here:

https://en.wikipedia.org/wiki/Log-normal_distribution#Geometric_moments

However, there is no definition of it; Now I am curious what the official definition of the geometric mean/geometric first moment actually is; Is it:

$$\lim\limits_{n \to \infty}\left[E\left(X_1^{1/n}\right)\right]^n$$

or does it have some other definition?

You can define the geometric mean of a strictly positive random variable readily enough; an easy method would be to take:

$$GM(X)=\exp(E[\log(X)])$$

For a discrete variable you can write the geometric mean as $$\prod_i x_i^{p_i}$$ where $$p_i={p(X=x_i)}$$.

I have no idea what the "official" answer is, if any, and this might just be a repeat of Glen_b's answer with more calculus-y language (I don't know), but the whole idea is to be analogous to the continuous version of the arithmetic mean, right? Well the arithmetic mean of a finite number of elements can be defined as: $$\sum_{i=1}^n \frac{1}{n} x_i$$

For the arithmetic mean of the values of f(x) of a continuous region of real-number x values starting from a and ending in b, we instead write: $$\frac{\int_a^b f(x)dx}{b-a}$$

The logic I'm missing here is explaining exactly what x values we're summing together, which is something to do with Riemann sums, but the point is that we can do it.

Similarly the geometric mean could be written as: $$\prod_{i=1}^n x_i^{\frac{1}{n}} = e^{\ln\left(\prod_{i=1}^n x_i^{\frac{1}{n}}\right)} = e^{\sum_{i=1}^n \frac{1}{n} \ln(x_i)} = \exp\left( \sum_{i=1}^n \frac{1}{n} ln(x_i) \right)$$

This is just the definition of an arithmetic mean stuck in an exponent. Thus, it would be intuitive at least to generalize and say that the "geometric mean" of the values of f(x) over a continuous region of real-number x values starting with a and ending in b is: $$\exp\left(\frac{\int_a^b \ln(f(x))dx}{b-a}\right)$$

As Glen_b rightly pointed out, this only works if f(x) is always non-negative, because ln(x) is undefined for negative x values, but how often to people use geometric means on negative numbers anyway? It's only even possible if there are an odd number of values you're finding the mean of or if their product is positive, despite some being negative, and even then, I'm not sure what significance the number you would get would have.

To generalize to higher dimensions, i.e. functions of more than one variable, we can either observe that a-b is the length of the line we integrated across, or observe that: $$\frac{\int_a^b f(x)dx}{b-a} = \frac{\int_a^b f(x)dx}{\int_a^b dx}$$ Thus for a double, triple, quadruple, etc. integrals of the values of a function of multiple variables over a continuous region, we divide by the area of that integral by area, volume, or whatever sort of hypervolume of the region we're integrating across: $$\frac{\iint_A f(x,y) dx dy}{\iint_A dx dy}$$

$$\frac{\iiint_V f(x,y,z) dx dy dz}{\iiint_V dx dy dz}$$

$$\frac{\iiiint_{HV} f(x_1,x_2,x_3,x_4) dx dy dz}{\iiiint_{HV} dx_1 dx_2 dx_3 dx_4}$$

$$etc.$$

Thus, we can just do the same with continuous "geometric means": $$\exp\left(\frac{\iint_A \ln(f(x,y)) dx dy}{\iint_A dx dy}\right)$$

$$\exp\left(\frac{\iiint_V \ln{f(x,y,z)} dx dy dz}{\iiint_V dx dy dz}\right)$$

$$\exp\left(\frac{\iiiint_{HV} \ln(f(x_1,x_2,x_3,x_4)) dx dy dz}{\iiiint_{HV} dx_1 dx_2 dx_3 dx_4}\right)$$

$$etc.$$

• This is the correct way to write logs and exps: \ln, \exp to yield $\ln,~\exp$ rather than $ln,~exp.$ I edited but I am seeing you undid them. Please format the same. Feb 27 at 6:31
• You are laboring mightily to re-express a very special case of Glen_b's answer: namely, an unweighted (uniform) GM. You might then appreciate learning about the meaning and definition of expectation because of its power to extend the concept of an average.
– whuber
Feb 27 at 13:48