Question

The strong law of large numbers states that with probability 1, the successive arithmetic averages of a sequence of independent and identically distributed random variables converge to their common mean . What do the successive geometric averages converge to? That is, what is

$\underset{n\to \infty }{\lim }{\left(\prod _{i=1}^n{X}_i\right)}^{1/n}$

Short answer

The successive geometric averages converge to$\underset{n\to \infty }{\lim }{\left(\prod _{i=1}^n{X}_i\right)}^{1/n}=e^{E\left[\ln \left(X_i\right)\right]}$

Step-by-step solution

Given information

The successive arithmetic averages of a sequence of independent and identically distributed random variables converge to their common mean $\mu$.

$\underset{n\to \infty }{\lim }{\left(\prod _{i=1}^n{X}_i\right)}^{1/n}$.

Explanation

Let us consider, $X_1,X_2,X_3\dots$be a set of randomly distributed random variables that are all independent and have the same mean

localid="1650029577411" $$\begin{gathered} \mu =E\left[X_i\right] \\ \ln \left(X_1\right),\ln \left(X_2\right),\ln \left(X_3\right),\dots \end{gathered}$$

is also a set of independently distributed and identically distributed random variables, each with a finite mean $E\left[\ln \left(X_i\right)\right]$

By the strong law of large numbers

$\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^n\ln \left(X_i\right)=E\left[\ln \left(X_i\right)\right]$

On the other hand

localid="1650029585734" $$\begin{gathered} \frac{1}{n}\sum _{i=1}^n\ln \left(X_i\right)=\ln {\left(\prod _{i=1}^n{X}_i\right)}^{1/n} \\ \underset{n\to \infty }{\lim }\ln {\left(\prod _{i=1}^n{X}_i\right)}^{1/n}=E\left[\ln \left(X_i\right)\right] \\ \underset{n\to \infty }{\lim }{\left(\prod _{i=1}^n{X}_i\right)}^{1/n}=e^{E\left[\ln \left(X_i\right)\right]}. \end{gathered}$$