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Chapter 8: Problem 12

(Chebyshev \(^{8}\) ) Assume that \(X_{1}, X_{2}, \ldots, X_{n}\) are independent random variables with possibly different distributions and let \(S_{n}\) be their sum. Let \(m_{k}=E\left(X_{k}\right)\), \(\sigma_{k}^{2}=V\left(X_{k}\right),\) and \(M_{n}=m_{1}+m_{2}+\cdots+m_{n} .\) Assume that \(\sigma_{k}^{2}0\) $$ P\left(\left|\frac{S_{n}}{n}-\frac{M_{n}}{n}\right|<\epsilon\right) \rightarrow 1 $$ as \(n \rightarrow \infty\)

Short Answer

Expert verified
As \( n \rightarrow \infty \), \( P\left(|\frac{S_n}{n} - \frac{M_n}{n}| < \epsilon \right) \rightarrow 1 \) due to the application of Chebyshev's inequality.

Step by step solution

01

Understanding Chebyshev's Inequality

Chebyshev's inequality for any random variable states that for any \(k > 0\), \( P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \). Here, \(\mu\) and \(\sigma^2\) are the mean and variance of the random variable, respectively. We must use this inequality to show that the probability we are interested in approaches 1 as \(n\rightarrow \infty\).
02

Sum and Average of Random Variables

Consider the sum of n independent random variables, \(S_n = X_1 + X_2 + \cdots + X_n \). The average of these sums is \(\frac{S_n}{n}\), which has mean \(\frac{M_n}{n}\). The focus is on showing that the probability of the difference \( |\frac{S_n}{n} - \frac{M_n}{n}| < \epsilon \) becomes arbitrarily close to 1.
03

Variance of the Sum

The variance of the sum \(S_n = X_1 + X_2 + \cdots + X_n\) is the sum of the individual variances because the variables are independent: \( V(S_n) = \sigma_1^2 + \sigma_2^2 + \cdots + \sigma_n^2 \). We know that each variance \(\sigma_k^2 < R\), thus \( V(S_n) < nR \). Consequently, the variance of the average is \(V\left(\frac{S_n}{n}\right) = \frac{V(S_n)}{n^2} < \frac{nR}{n^2} = \frac{R}{n}\).
04

Applying Chebyshev's Inequality

Using Chebyshev's inequality for \(\frac{S_n}{n}\) with mean \(\frac{M_n}{n}\) and variance \(\frac{R}{n}\), we have: \[P\left( \left| \frac{S_n}{n} - \frac{M_n}{n} \right| \geq \epsilon \right) \leq \frac{V\left(\frac{S_n}{n}\right)}{\epsilon^2} = \frac{R}{n\epsilon^2}.\] As \(n\rightarrow \infty\), \(\frac{R}{n\epsilon^2} \rightarrow 0\).
05

Convergence to 1

Since \(P\left( \left| \frac{S_n}{n} - \frac{M_n}{n} \right| \geq \epsilon \right) \leq \frac{R}{n\epsilon^2}\), it follows that \[P\left( \left| \frac{S_n}{n} - \frac{M_n}{n} \right| < \epsilon \right) = 1 - P\left( \left| \frac{S_n}{n} - \frac{M_n}{n} \right| \geq \epsilon \right) \geq 1 - \frac{R}{n\epsilon^2}.\] As \(n\rightarrow \infty\), \(1 - \frac{R}{n\epsilon^2} \rightarrow 1\), proving the original statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Independent Random Variables
Independent random variables are foundational in probability theory. When dealing with more than one random variable, they can either be dependent or independent. Independent random variables do not influence each other.
For example, if you have two random variables, say the outcome of a die and a coin flip, the result of the die does not affect the coin flip's outcome.
In this context, when you sum these random variables, their independence plays a crucial role. The overall behavior of the sum can be predicted by the characteristics of the individual variables. This allows for simplified calculations, such as computing the variance of the sum.
Therefore, when the problem states that random variables are independent, it means that each one does not impact any other, which helps in using properties like variance sum and applying inequalities such as Chebyshev's.
Analyzing the Variance of the Sum of Random Variables
The variance of the sum of independent random variables is particularly important. When you add up independent random variables, you simply add their variances to find the variance of the sum.
Each variable has its variance denoted as \( \sigma_k^2 \), and collectively for n variables, it becomes \( V(S_n) = \sigma_1^2 + \sigma_2^2 + \cdots + \sigma_n^2 \).
The beauty of this property is in its simplicity鈥攖he independence of the variables ensures no covariance terms in the variance computation. This results in the final variance being the direct sum of individual variances.
In our exercise, knowing that each variance is less than some constant \( R \) leads to a tidy upper bound for \( V(S_n) \). This makes further calculations manageable and helps in evaluating convergence using Chebyshev's inequality.
Understanding Convergence Probability
Convergence probability is a critical concept, especially in statistics involving large samples. It describes how probabilities behave as the number of observations tends towards infinity.
Convergence in probability means that as \( n \) (the number of trials or observations) increases, the probability of a specific difference, like the one in our exercise, being larger than a small positive number \( \epsilon \) becomes negligible.
In this problem, using Chebyshev's inequality helps establish this convergence. The inequality tells us how the probability of deviation from the mean reduces as \( n \) increases. Essentially, for a very large \( n \), this probability can get arbitrarily close to zero, indicating that the sum's average tends to its expected mean, securing convergence to 1 for the probability of the stated interval.
Exploring Mean and Variance Concepts
The concepts of mean and variance are fundamental in statistical analysis. The mean, often denoted as \( m_k \) for random variables \( X_k \), is the expected value, representing a central location or the average outcome.
Variance, denoted \( \sigma_k^2 \), describes how dispersed the values of a variable are around the mean. A smaller variance indicates that the data points are closely clustered around the mean, while a larger variance shows a wider spread.
In the provided exercise, understanding these measures helps in leveraging Chebyshev's inequality effectively, given that the variance controls how probabilities behave around the mean. When computing the distribution of sums and applying probability bounds, the mean and variance allow precise probabilistic predictions, like determining how close \( \frac{S_n}{n} \) is to \( \frac{M_n}{n} \) as \( n \) grows large.

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Most popular questions from this chapter

Let \(Z=X / Y\) where \(X\) and \(Y\) have normal densities with mean 0 and standard deviation \(1 .\) Then it can be shown that \(Z\) has a Cauchy density. (a) Write a program to illustrate this result by plotting a bar graph of 1000 samples obtained by forming the ratio of two standard normal outcomes. Compare your bar graph with the graph of the Cauchy density. Depending upon which computer language you use, you may or may not need to tell the computer how to simulate a normal random variable. A method for doing this was described in Section 5.2 .

A share of common stock in the Pilsdorff beer company has a price \(Y_{n}\) on the \(n\) th business day of the year. Finn observes that the price change \(X_{n}=\) \(Y_{n+1}-Y_{n}\) appears to be a random variable with mean \(\mu=0\) and variance \(\sigma^{2}=1 / 4 .\) If \(Y_{1}=30,\) find a lower bound for the following probabilities, under the assumption that the \(X_{n}\) 's are mutually independent. (a) \(P\left(25 \leq Y_{2} \leq 35\right)\). (b) \(P\left(25 \leq Y_{11} \leq 35\right)\) (c) \(P\left(25 \leq Y_{101} \leq 35\right)\).

Let \(X\) be a continuous random variable with values normally distributed over \((-\infty,+\infty)\) with mean \(\mu=0\) and variance \(\sigma^{2}=1\) (a) Using Chebyshev's Inequality, find upper bounds for the following probabilities: \(P(|X| \geq 1), P(|X| \geq 2),\) and \(P(|X| \geq 3)\) (b) The area under the normal curve between -1 and 1 is .6827 , between -2 and 2 is .9545 , and between -3 and 3 it is .9973 (see the table in Appendix A). Compare your bounds in (a) with these exact values. How good is Chebyshev's Inequality in this case?

The Pilsdorff beer company runs a fleet of trucks along the 100 mile road from Hangtown to Dry Gulch, and maintains a garage halfway in between. Each of the trucks is apt to break down at a point \(X\) miles from Hangtown, where \(X\) is a random variable uniformly distributed over [0,100] (a) Find a lower bound for the probability \(P(|X-50| \leq 10)\). (b) Suppose that in one bad week, 20 trucks break down. Find a lower bound for the probability \(P\left(\left|A_{20}-50\right| \leq 10\right),\) where \(A_{20}\) is the average of the distances from Hangtown at the time of breakdown.

Write a program to toss a coin 10,000 times. Let \(S_{n}\) be the number of heads in the first \(n\) tosses. Have your program print out, after every 1000 tosses, \(S_{n}-n / 2 .\) On the basis of this simulation, is it correct to say that you can expect heads about half of the time when you toss a coin a large number of times?
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