/*! 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 85 Let \(X\) have a standard beta d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(X\) have a standard beta density with parameters \(\alpha\) and \(\beta\). a. Verify the formula for \(E(X)\) given in the section. b. Compute \(E\left[(1-X)^{m}\right]\). If \(X\) represents the proportion of a substance consisting of a particular ingredient, what is the expected proportion that does not consist of this ingredient?

Short Answer

Expert verified
a. Verified: \(E(X) = \frac{\alpha}{\alpha+\beta}\); b. \(E(1-X) = \frac{\beta}{\alpha+\beta}\).

Step by step solution

01

Recall the Expectation Formula for Beta Distribution

The expectation of a beta-distributed random variable with parameters \(\alpha\) and \(\beta\) is given by the formula \(E(X) = \frac{\alpha}{\alpha+\beta}\). This formula is crucial for solving this problem.
02

Verify the Formula for E(X)

To verify the formula for \(E(X)\), we consider the integral \(E(X) = \int_0^1 x f(x;\alpha,\beta) \, dx\), where \(f(x;\alpha,\beta)\) is the beta density: \(f(x;\alpha,\beta) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha, \beta)}\). Substitute and solve: \(E(X) = \frac{1}{B(\alpha, \beta)}\int_0^1 x^{\alpha}(1-x)^{\beta-1} \, dx\). This is equivalent to the beta function \(B(\alpha+1, \beta)\), so \(E(X) = \frac{B(\alpha+1, \beta)}{B(\alpha, \beta)} = \frac{\alpha}{\alpha+\beta}\). This confirms the given formula.
03

Compute E[(1-X)^{m}]

We want to find \(E[(1-X)^m]\), which requires integration of \((1-x)^m\) over the beta distribution: \(E[(1-X)^m] = \int_0^1 (1-x)^m f(x;\alpha,\beta) \, dx\). This equals \(\frac{1}{B(\alpha,\beta)}\int_0^1 (1-x)^{m+\beta-1} x^{\alpha-1} \, dx\). This integral is a beta function \(B(\alpha, m+\beta)\), leading to the expected value \(E[(1-X)^m] = \frac{B(\alpha, m+\beta)}{B(\alpha, \beta)}\).
04

Interpret the Expected Proportion That Doesn't Consist of the Ingredient

If \(X\) is the portion of a specific ingredient, \((1-X)\) represents the proportion that is not. Therefore, \(E(1-X) = E[(1-X)^1] = \frac{B(\alpha, 1+\beta)}{B(\alpha, \beta)} = \frac{\beta}{\alpha+\beta}\). This shows the expected proportion of what remains not containing the ingredient is \(\frac{\beta}{\alpha+\beta}\).

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 91Ó°ÊÓ!

Key Concepts

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

Expectation of Beta Distribution
In probability theory and statistics, understanding the expectation of a beta distribution is invaluable. The beta distribution is defined with two parameters, \(\alpha\) and \(\beta\), and it is a continuous probability distribution. It is often used to model random variables that are constrained to the interval \([0, 1]\).

The expectation (or mean) of a beta-distributed random variable \(X\) with shape parameters \(\alpha\) and \(\beta\) is given by the straightforward formula:
  • \(E(X) = \frac{\alpha}{\alpha + \beta}\)
This formula signifies the balance between these two shape parameters. A higher \(\alpha\) indicates a greater likelihood of observing values closer to 1, whereas a higher \(\beta\) suggests a tendency towards values near 0. While the formula itself is simple, a deep understanding helps in interpreting data more effectively.

To derive the expectation, we use the integration of the product of \(x\) and the beta density function over its defined limit from 0 to 1. The integral results in the simplification of the beta function, creating a concise and applicable expression for expectation.
Beta Function
The beta function is a fundamental component when dealing with beta distributions. Essentially, it is a special function that is defined by the following integral:
  • \(B(\alpha, \beta) = \int_0^1 x^{\alpha-1} (1-x)^{\beta-1} \, dx\)
This integral expresses how the shape parameters \(\alpha\) and \(\beta\) influence the distribution's behavior across its domain.

The beta function simplifies numerous expressions involving beta-distributed variables. It helps in calculating the expected value and in solving problems related to random variables. Furthermore, the beta function can be connected to the gamma function \((\Gamma)\), where \(B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\). This relationship allows for easier computations and is often used for proofs such as the expectation formula verification.
Random Variable Expectations
In the context of probability distributions, understanding the concept of random variable expectations is crucial. It extends beyond just learning each distribution's mean. Expectation can be thought of as the average outcome if the experiment were repeated an infinite number of times. It provides a central value around which the random variable is distributed.

For the beta distribution, calculating random variable expectations often involves leveraging symmetrical formulas and using mathematical integrations for each specific type of function (such as \((1-X)^m\)). For example:
  • If \(X\) follows the standard beta distribution, \((1-X)\) represents the proportion not occupied by \(X\) in a given problem.
  • The expectation \(E\left[(1-X)^m\right]\) requires integrating the modified beta density function \((1-x)^m\), similar to other expectation calculations.
These calculations provide insights into how the rest of an involved population behaves, which is particularly useful in fields like chemistry or biology, where the concentration of substances plays a critical role. Understanding these principles helps in estimations and hypotheses testing, making the concept of random variable expectations both practical and powerful.

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

Suppose that \(10 \%\) of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let \(X\) denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that \(X\) is a. At most 30 ? b. Less than 30 ? c. Between 15 and 25 (inclusive)?

The paper "Microwave Obsevations of Daily Antarctic SeaIce Edge Expansion and Contribution Rates" (IEEE Geosci. and Remote Sensing Letters, 2006: 54-58) states that "The distribution of the daily sea-ice advance/retreat from each sensor is similar and is approximately double exponential." The proposed double exponential distribution has density function \(f(x)=.5 \lambda e^{-\lambda|x|}\) for \(-\infty

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within \(2 \mathrm{~min}\) after the hour. Let \(X=\) the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of \(X\) is $$ f(x)=\left\\{\begin{array}{cl} k x^{2} & 0 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Find the value of \(k\) and draw the corresponding density curve. [Hint: Total area under the graph of \(f(x)\) is 1.] b. What is the probability that the lecture ends within \(1 \mathrm{~min}\) of the end of the hour? c. What is the probability that the lecture continues beyond the hour for between 60 and \(90 \mathrm{sec}\) ? d. What is the probability that the lecture continues for at least \(90 \mathrm{sec}\) beyond the end of the hour?

The distribution of resistance for resistors of a certain type is known to be normal, with \(10 \%\) of all resistors having a resistance exceeding \(10.256\) ohms and \(5 \%\) having a resistance smaller than \(9.671\) ohms. What are the mean value and standard deviation of the resistance distribution?

Let the ordered sample observations be denoted by \(y_{1}\), \(y_{2}, \ldots, y_{n}\) ( \(y_{1}\) being the smallest and \(y_{n}\) the largest). Our suggested check for normality is to plot the \(\left(\Phi^{-1}((i-.5) / n), y_{i}\right)\) pairs. Suppose we believe that the observations come from a distribution with mean 0 , and let \(w_{1}, \ldots, w_{n}\) be the ordered absolute values of the \(x_{i}\) s. A half-normal plot is a probability plot of the \(w_{i} s\). More specifically, since \(P(|Z| \leq w)=\) \(P(-w \leq Z \leq w)=2 \Phi(w)-1\), a half-normal plot is a plot of the \(\left(\Phi^{-1}\\{[(i-5) / n+1] / 2\\}, w_{i}\right)\) pairs. The virtue of this plot is that small or large outliers in the original sample will now appear only at the upper end of the plot rather than at both ends. Construct a half-normal plot for the following sample of measurement errors, and comment: \(-3.78,-1.27,1.44\), \(-39,12.38,-43.40,1.15,-3.96,-2.34,30.84\).

See all solutions

Recommended explanations on Math Textbooks

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.