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Chapter 8: Problem 10

Let \(X\) by any random variable which takes on values \(0,1,2, \ldots, n\) and has \(E(X)=V(X)=1\). Show that, for any positive integer \(k\) $$ P(X \geq k+1) \leq \frac{1}{k^{2}} $$

Short Answer

Expert verified
By Chebyshev's inequality, \(P(X \geq k+1) \leq \frac{1}{k^2}\).

Step by step solution

01

Understanding the Given Information

We are given a random variable \(X\) that takes on integer values from \(0\) to \(n\) with both the expected value, \(E(X)\), and the variance, \(V(X)\), equal to 1. We need to show that \(P(X \geq k+1) \leq \frac{1}{k^2}\) for any positive integer \(k\).
02

Recall Chebyshev's Inequality

Chebyshev's inequality provides an upper bound for the probability that a random variable deviates from its mean. The inequality is given by: \(P(|X - \mu| \geq t) \leq \frac{V(X)}{t^2}\), where \(\mu = E(X)\) and \(t\) is any positive value.
03

Apply Chebyshev's Inequality

We set \(\mu = 1\), and since \(V(X) = 1\), for a deviation of \(t = k\), Chebyshev's inequality becomes: \(P(|X - 1| \geq k) \leq \frac{1}{k^2}\). This inequality bounds the probability of \(X\) being outside \((1 - k, 1 + k)\).
04

Relate to the Probability of Interest

We consider the upper bound of the interval, \(1 + k\). Since \(X\) is an integer, \(P(X \geq k+1)\) is naturally a part of \(|X - 1| \geq k\). According to Chebyshev's result, \(P(X \geq k+1) \leq P(|X - 1| \geq k) \leq \frac{1}{k^2}\). Thus, the inequality we wanted to prove is satisfied.

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Key Concepts

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

Chebyshev's Inequality
Chebyshev's Inequality is a powerful tool in probability theory. It provides a way to understand how spread out the values of a random variable can be when they deviate from the mean. This inequality is particularly useful because it works for any random variable, regardless of its distribution.
  • Chebyshev's Inequality states that for a random variable with mean \( \mu \) and variance \( V(X) \), the probability of the variable deviating from its mean by at least \( t \) is bounded by: \[ P(|X - \mu| \geq t) \leq \frac{V(X)}{t^2} \]
  • This formula gives us a way to estimate probabilities even when little is known about the distribution itself.
By using the inequality, we can set bounds on how probable it is that a random variable falls within a certain range around its expected value. This is crucial for understanding behaviors in fields like finance and science, where predicting extreme deviations is key.
Expected Value
The expected value of a random variable is a fundamental concept in statistics and probability, representing the average value or mean that we anticipate over numerous trials.
  • The expected value is denoted as \( E(X) \), and mathematically, for discrete variables, it is calculated as the sum of all possible values of a random variable multiplied by their probabilities: \[ E(X) = \sum_{i} x_i P(x_i) \]
  • For this exercise, the random variable \( X \) has an expected value of 1, which provides a central point around which Chebyshev's inequality is applied.
Understanding expected value helps us predict the long-term results of repeated experiments. It offers a single number summarizing the probability distribution of the random variable, making it a crucial component in both theoretical analysis and practical applications.
Variance
Variance measures how much the values of a random variable fluctuate around the mean, giving us insight into the spread or dispersion of the variable's possible outcomes.
  • Variance is denoted as \( V(X) \), and for discrete variables, it is represented by the expected value of the squared differences from the mean: \[ V(X) = E[(X - \mu)^2] \]
  • In our exercise, variance is given as 1, indicating that the deviations from the mean are relatively small.
  • A smaller variance indicates that values are closely clustered around the mean, while a larger variance signifies a broader spread.
Variance serves as a key component in probabilistic measures like Chebyshev's Inequality, influencing the scale of potential deviations from the expected value.
Random Variables
Random variables are a cornerstone concept in probability theory, playing a crucial role in modeling and analyzing uncertain phenomena.
  • A random variable can be thought of as a function that assigns numerical values to the outcomes of a random process.
  • There are two main types of random variables: discrete and continuous. Discrete random variables have distinct, separate values, while continuous random variables can take any value within a range.
  • In the exercise, \( X \) is a discrete random variable taking values from \( 0 \) to \( n \).
Understanding random variables allows us to apply probabilistic methods to real-world situations, providing a framework to analyze and predict experimental outcomes. They serve as the foundation upon which more complex probabilistic analyses, like those using Chebyshev's inequality, are built.

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Most popular questions from this chapter

Let \(X\) be a random variable with \(E(X)=0\) and \(V(X)=1\). What integer value \(k\) will assure us that \(P(|X| \geq k) \leq .01 ?\)

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?

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?

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) \(P\left(25 \leq Y_{101} \leq 35\right)\).

Show that, if \(X \geq 0,\) then \(P(X \geq a) \leq E(X) / a\)
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