Q. 8.13 The strong law of large numbers ... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 8: Q. 8.13 (page 394)

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
$\lim_{n 鈫� 鈭�} {({鈭�}_{i = 1}^{n} X_{i})}^{1 / n}$

Short Answer

Expert verified

The successive geometric averages converge to $\lim_{n 鈫� 鈭�} {({鈭�}_{i = 1}^{n} X_{i})}^{1 / n} = e^{E [\ln (X_{i})]}$

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 $渭$ .

$\lim_{n 鈫� 鈭�} {({鈭�}_{i = 1}^{n} X_{i})}^{1 / n}$ .

Explanation

Let us consider, $X_{1}, X_{2}, X_{3} 鈥�$ be a set of randomly distributed random variables that are all independent and have the same mean

localid="1650029577411" $渭 = E [X_{i}] \ln (X_{1}), \ln (X_{2}), \ln (X_{3}), 鈥�$

is also a set of independently distributed and identically distributed random variables, each with a finite mean $E [\ln (X_{i})]$

By the strong law of large numbers

$\lim_{n 鈫� 鈭�} \frac{1}{n} {鈭�}_{i = 1}^{n} \ln (X_{i}) = E [\ln (X_{i})]$

On the other hand

localid="1650029585734" $\frac{1}{n} {鈭�}_{i = 1}^{n} \ln (X_{i}) = \ln {({鈭�}_{i = 1}^{n} X_{i})}^{1 / n} \lim_{n 鈫� 鈭�} \ln {({鈭�}_{i = 1}^{n} X_{i})}^{1 / n} = E [\ln (X_{i})] \lim_{n 鈫� 鈭�} {({鈭�}_{i = 1}^{n} X_{i})}^{1 / n} = e^{E [\ln (X_{i})]} .$

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