Q. 8.13 - A First Course in Probability

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 

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

Step-by-Step Solution

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 1: Given information

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

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

Step 2: 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

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

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 

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