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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

limn→∞∏i=1nXi1/n

Short Answer

Expert verified

The successive geometric averages converge tolimn→∞∏i=1nXi1/n=eElnXi

Step by step solution

01

Given information

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

limn→∞∏i=1nXi1/n.

02

Explanation

Let us consider, X1,X2,X3…be a set of randomly distributed random variables that are all independent and have the same mean

localid="1650029577411" μ=EXilnX1,lnX2,lnX3,…

is also a set of independently distributed and identically distributed random variables, each with a finite mean ElnXi

By the strong law of large numbers

limn→∞1n∑i=1nlnXi=ElnXi

On the other hand

localid="1650029585734" 1n∑i=1nlnXi=ln∏i=1nXi1/nlimn→∞ln∏i=1nXi1/n=ElnXilimn→∞∏i=1nXi1/n=eElnXi.

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Most popular questions from this chapter

Many people believe that the daily change in the price of a company’s stock on the stock market is a random variable with a mean of 0and a variance ofσ2. That is if Yn represents the price of the stock on thenth day, then Yn=Yn-1+Xn,n≥1where X1,X2,...are independent and identically distributed random variables with mean 0and varianceσ2. Suppose that the stock’s price today is100. Ifσ2=1, what can you say about the probability that the stock’s price will exceed 105after10days?

Suppose that the number of units produced daily at factory A is a random variable with mean 20and standard deviation 3and the number produced at factory B is a random variable with mean 18and standard deviation of 6. Assuming independence, derive an upper bound for the probability that more units are produced today at factory B than at factory A.

A certain component is critical to the operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and its standard deviation is 30 hours, how many of these components must be in stock so that the probability that the system is in continual operation for the next 2000 hours is at least .95?

The Chernoff bound on a standard normal random variableZgivesP{Z>a}≤e-a2/2,a>0. Show, by considering the densityZ, that the right side of the inequality can be reduced by the factor2. That is, show that

P{Z>a}≤12e-a2/2a>0

From past experience, a professor knows that the test score taking her final examination is a random variable with a mean of75.

(a)Give an upper bound for the probability that a student’s test score will exceed85.

(b)Suppose, in addition, that the professor knows that the variance of a student’s test score is equal25. What can be said about the probability that a student will score between 65and85?

(c)How many students would have to take the examination to ensure a probability of at least .9that the class average would be within 5of75? Do not use the central limit theorem.
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