/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Let \(X\) be a random variable w... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 12

Let \(X\) be a random variable with \(\mu=E(X)\) and \(\sigma^{2}=V(X) .\) Define \(X^{*}=\) \((X-\mu) / \sigma\). The random variable \(X^{*}\) is called the standardized random variable associated with \(X\). Show that this standardized random variable has expected value 0 and variance 1 .

Short Answer

Expert verified
The standardized variable \( X^* \) has expected value 0 and variance 1.

Step by step solution

01

Recognize the Standardization Formula

We need to understand the formula for the standardized random variable, which is given as \( X^{*} = \frac{X - \mu}{\sigma} \). This transformation takes a random variable \( X \) with mean \( \mu \) and variance \( \sigma^2 \), and standardizes it to have a new mean and variance. Our task is to find these new mean and variance values.
02

Calculate Expected Value of Standardized Variable

To find the expected value of \( X^* \), compute the expectation: \( E(X^*) = E\left( \frac{X - \mu}{\sigma} \right) \). By the linearity of expectation, this becomes \( \frac{1}{\sigma}E(X - \mu) = \frac{1}{\sigma}(E(X) - \mu) = \frac{1}{\sigma}(\mu - \mu) = 0 \). So, the expected value of \( X^* \) is 0.
03

Calculate Variance of Standardized Variable

The variance of \( X^* \) is found by \( V(X^*) = V\left( \frac{X - \mu}{\sigma} \right) \). Since variance is affected by scaling, we have \( V\left( \frac{X - \mu}{\sigma} \right) = \left( \frac{1}{\sigma} \right)^2 V(X - \mu) = \frac{1}{\sigma^2}V(X) = \frac{\sigma^2}{\sigma^2} = 1 \). So, the variance of \( X^* \) is 1.
04

Conclusion

We have shown that the standardized random variable \( X^* = \frac{X - \mu}{\sigma} \) has an expected value of 0 and a variance of 1, effectively proving that it standardizes the original random variable \( X \). These properties are consistent with what we expect from a standard normal distribution applied in statistics.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Standardized Random Variable
When dealing with random variables, it's often useful to transform them into a standardized form. This is where a "standardized random variable" comes into play. The process of standardization involves changing a random variable so it has a mean of 0 and a variance of 1. This makes it easier to compare different data sets with varied scales. The standardization formula is:
  • Standardized Variable: \( X^{*} = \frac{X - \mu}{\sigma} \)
Here, \( \mu \) is the mean of the original random variable \( X \), and \( \sigma \) is the standard deviation, which is the square root of the variance \( \sigma^2 \). The result of this transformation is a new random variable \( X^* \) that is much easier to handle, especially in statistical analysis.
Expected Value
Expected value, often represented as \( E(X) \), is the long-term average or mean value of a random variable. It's a fundamental concept in statistics representing the central tendency of a probability distribution. If you were to repeat an experiment an infinite number of times, the expected value would be the average result. In the context of the standardized random variable, we calculate the expected value as follows:
  • Expected Value of Standardized Variable: \( E(X^*) = E\left(\frac{X - \mu}{\sigma}\right) \)
Using the linearity of expectation, you can simplify this calculation to find \( E(X^*) = 0 \). This means the standardized variable's expected value is zero, centering it at the origin of its distribution.
Variance
The variance of a random variable measures how much the values differ from the mean on average. It is represented by \( V(X) \) or \( \sigma^2 \). The larger the variance, the more spread out the values of the random variable are around the mean. When we standardize a random variable, we also adjust its variance. For a standardized random variable, the variance is calculated by:
  • Variance of Standardized Variable: \( V(X^*) = V\left(\frac{X - \mu}{\sigma}\right) \)
The computation results in a variance of 1: \( V\left(\frac{X - \mu}{\sigma}\right) = 1 \). This means the spread of the standardized variable is normalized to a unit variance, simplifying further statistical analysis.
Standard Normal Distribution
A standard normal distribution is a normal probability distribution that has been standardized. It has a mean of 0 and a standard deviation of 1. Often represented with the variable \( Z \), it鈥檚 commonly called the "Z-distribution."This concept is particularly useful because:
  • Any normal distribution can be converted to a standard normal distribution using the standardization process.
  • It allows easy comparison and analysis of different data sets, regardless of the original mean and variance.
  • Tables and calculations involving standard normal distribution are widely available, making it easier to perform critical statistical operations.
Understanding standard normal distributions is crucial for hypothesis testing, creating confidence intervals, and general probability assessments in statistics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A coin is tossed three times. Let \(X\) be the number of heads that turn up. Find \(V(X)\) and \(D(X)\)

A die is loaded so that the probability of a face coming up is proportional to the number on that face. The die is rolled with outcome \(X\). Find \(V(X)\) and \(D(X)\)

The Pilsdorff Beer Company runs a fleet of trucks along the 100 mile road from Hangtown to Dry Gulch. The trucks are old, and are apt to break down at any point along the road with equal probability. Where should the company locate a garage so as to minimize the expected distance from a typical breakdown to the garage? In other words, if \(X\) is a random variable giving the location of the breakdown, measured, say, from Hangtown, and \(b\) gives the location of the garage, what choice of \(b\) minimizes \(E(|X-b|) ?\) Now suppose \(X\) is not distributed uniformly over \([0,100],\) but instead has density function \(f_{X}(x)=2 x / 10,000 .\) Then what choice of \(b\) minimizes \(E(|X-b|) ?\)

Recall that in the martingale doubling system (see Exercise 1.1.10), the player doubles his bet each time he loses. Suppose that you are playing roulette in a fair casino where there are no 0 's, and you bet on red each time. You then win with probability \(1 / 2\) each time. Assume that you enter the casino with 100 dollars, start with a 1 -dollar bet and employ the martingale system. You stop as soon as you have won one bet, or in the unlikely event that black turns up six times in a row so that you are down 63 dollars and cannot make the required 64 -dollar bet. Find your expected winnings under this system of play.

A card is drawn at random from a deck of playing cards. If it is red, the player wins 1 dollar; if it is black, the player loses 2 dollars. Find the expected value of the game.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.