/*! 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 21 Let \(X\) be a random variable w... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be a random variable with \(E(X)=\mu\) and \(V(X)=\sigma^{2}\). Show that the function \(f(x)\) defined by $$f(x)=\sum_{\omega}(X(\omega)-x)^{2} p(\omega)$$ has its minimum value when \(x=\mu\).

Short Answer

Expert verified
The function is minimized when \(x=\mu\).

Step by step solution

01

Understanding the Problem

We need to find the value of \( x \) that minimizes the function \( f(x) = \sum_{\omega} (X(\omega) - x)^2 p(\omega) \). Given properties of \( X \) are its expected value \( E(X)=\mu \) and variance \( V(X)=\sigma^2 \).
02

Express Function with Expected Value

Recognize that \( \sum_{\omega} (X(\omega) - x)^2 p(\omega) \) is essentially an expected value: \[ E((X-x)^2) = \sum_{\omega} (X(\omega) - x)^2 p(\omega) \] Therefore, we rewrite the function as an expectation that needs to be minimized.
03

Expanding the Square

Expand the expression inside the expectation:\[(X-x)^2 = (X - \mu + \mu - x)^2 = ((X-\mu) + (\mu-x))^2.\] Using the identity \((a+b)^2 = a^2 + 2ab + b^2\), we have:\[ (X-\mu)^2 + 2(X-\mu)(\mu-x) + (\mu-x)^2. \]
04

Apply Expectation Linearity

Taking expectation of the expanded expression:\[ E((X-x)^2) = E((X-\mu)^2) + E(2(X-\mu)(\mu-x)) + E((\mu-x)^2). \]Given the properties of expected value, we know:\[ E(X-\mu) = 0, \] leading to:\[ E(2(X-\mu)(\mu-x)) = 2(\mu-x)E(X-\mu) = 0. \]
05

Simplifying the Expectation

By substituting, the previous equation becomes:\[ E((X-x)^2) = E((X-\mu)^2) + (\mu-x)^2. \]Therefore, since \( E((X-\mu)^2) = \text{Var}(X) = \sigma^2 \),\[ E((X-x)^2) = \sigma^2 + (\mu-x)^2. \]
06

Minimize the Function

The expression \( \sigma^2 + (\mu-x)^2 \) reaches its minimum at \( (\mu-x)^2 = 0 \) as \( \sigma^2 \) is constant. This occurs when \( x = \mu \). Thus, \( f(x) \) is minimized when \( x = \mu \).

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Key Concepts

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

Random Variable
A random variable, often denoted by capital letters like \(X\), represents a function that assigns a numerical value to each outcome of a random phenomenon. It is used in probability theory to characterize uncertainty in a quantifiable way.
  • Random variables can be discrete, with outcomes ranging over a countable set, or continuous, covering uncountable ranges like intervals within the real numbers.
  • They are governed by a probability distribution that specifies the likelihood of each potential value.
In the context of the lab exercise, \(X\) is a random variable with an expected value of \(\mu\) and a variance \(\sigma^2\). Understanding these aspects of \(X\) is crucial to optimizing functions like \(f(x)\). When analyzing such variables, the focus lies on exploring how their values vary over different trials of the same process.
Variance
Variance, denoted as \(V(X)\) or \(\sigma^2\), measures the spread or dispersion of a set of values. In simpler terms, it tells us how far the values of a random variable \(X\) deviate from its mean \(\mu\).
  • The variance is calculated by taking the expectation of the squared deviation from the mean: \[ V(X) = E((X - \mu)^2). \]
  • A high variance indicates that the values are spread out over a wider range, while a low variance suggests that they are closer to the mean.
In the exercise, the variance is a key component since \(E((X - \mu)^2)\) simplifies to \(\sigma^2\). This is because the deviation \((X - \mu)\) has an expected value of zero, simplifying calculations when minimizing \(f(x)\) since the variance remains constant.
Expectation Formula
The expectation formula, or expected value, signifies the average or mean value of a random variable over numerous trials or experiments. For a random variable \(X\), the expected value is denoted by \(E(X)\) or \(\mu\).
  • The expectation is formally defined as the sum of all potential values, each weighted by its probability of occurrence: \[ E(X) = \sum_{\omega} X(\omega) p(\omega). \]
  • It provides a central value, which serves as an optimal point in many decision-making processes, such as minimizing the function discussed in the lab exercise.
In optimizing the given function \(f(x)\), we leverage the expectation to express \(f(x)\) in a form that easily reveals its minimum value through algebraic manipulation and properties of expectation.
Probability Theory
Probability theory is the mathematical framework for quantifying uncertain phenomena. It provides the foundation for understanding and working with random variables, expectations, and variances. Key principles of probability theory include:
  • Assigning probabilities to events and ensuring they add up to one for certainty across possible outcomes.
  • Using probability distributions to describe how probabilities are distributed among potential outcomes of a random variable.
In the exercise, probability theory underlies the function \(f(x)\). The sum \(\sum_{\omega} (X(\omega) - x)^2 p(\omega)\) involves the likelihood \(p(\omega)\) for each outcome, showing how expected values and variances can indicate the most probable decisions, such as choosing \(x = \mu\) for optimizing \(f(x)\). Understanding these concepts enables deductions about random behavior and makes the analysis of functions like \(f(x)\) effective and straightforward.

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Most popular questions from this chapter

In a second version of roulette in Las Vegas, a player bets on red or black. Half of the numbers from 1 to 36 are red, and half are black. If a player bets a dollar on black, and if the ball stops on a black number, he gets his dollar back and another dollar. If the ball stops on a red number or on 0 or 00 he loses his dollar. Find the expected winnings for this bet.

A long needle of length \(L\) much bigger than 1 is dropped on a grid with horizontal and vertical lines one unit apart. Show that the average number \(a\) of lines crossed is approximately $$a=\frac{4 L}{\pi}$$

Write a program to add random numbers chosen from [0,1] until the first time the sum is greater than one. Have your program repeat this experiment a number of times to estimate the expected number of selections necessary in order that the sum of the chosen numbers first exceeds \(1 .\) On the basis of your experiments, what is your estimate for this number?

Let \(X\) be a random variable taking on values \(a_{1}, a_{2}, \ldots, a_{r}\) with probabilities \(p_{1}, p_{2}, \ldots, p_{r}\) and with \(E(X)=\mu .\) Define the spread of \(X\) as follows: $$\bar{\sigma}=\sum_{i=1}^{r}\left|a_{i}-\mu\right| p_{i}$$ This, like the standard deviation, is a way to quantify the amount that a random variable is spread out around its mean. Recall that the variance of a sum of mutually independent random variables is the sum of the individual variances. The square of the spread corresponds to the variance in a manner similar to the correspondence between the spread and the standard deviation. Show by an example that it is not necessarily true that the square of the spread of the sum of two independent random variables is the sum of the squares of the individual spreads.

Find \(E\left(X^{Y}\right),\) where \(X\) and \(Y\) are independent random variables which are uniform on \([0,1] .\) Then verify your answer by simulation.

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