/*! 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 24 From the gamma distribution $$... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 24

From the gamma distribution $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{\beta^{\alpha} \Gamma(\alpha)} x^{a-1} e^{-x / \beta}, & x>0 \\ 0, & x \leq 0 . \end{array}\right. $$ show that (a) \(\langle x\rangle\) (mean) \(=\alpha \beta\), (b) \(\sigma^{2}\left(\right.\) variance) \(=\left(x^{2}\right)-(x)^{2}=\alpha \beta^{2}\).

Short Answer

Expert verified
(a) Mean \( \langle x \rangle = \alpha \beta \), (b) Variance \( \sigma^2 = \alpha \beta^2 \).

Step by step solution

01

Understanding the Gamma Probability Density Function (PDF)

The gamma distribution is a family of continuous probability distributions. Here, it is defined for a random variable \( x > 0 \) with parameters \( \alpha \) (shape) and \( \beta \) (scale) in its probability density function (PDF) as: \[ f(x)= \frac{1}{\beta^{\alpha} \Gamma(\alpha)} x^{\alpha-1} e^{-x / \beta} \]This function describes the likelihood of different outcomes for \( x \). To find properties like the mean and variance of this distribution, we need to use integral calculus.
02

Calculate the Mean \( \langle x \rangle \)

The mean of the gamma distribution is calculated using the expectation formula: \[ \langle x \rangle = \int_0^\infty x \cdot f(x) \ dx \]Substituting the gamma PDF into the integral gives: \[ \langle x \rangle = \int_0^\infty x \cdot \frac{1}{\beta^{\alpha} \Gamma(\alpha)} x^{\alpha-1} e^{-x / \beta} \ dx \]Simplifying, we have:\[ \langle x \rangle = \frac{1}{\beta^{\alpha} \Gamma(\alpha)} \int_0^\infty x^{\alpha} e^{-x / \beta} \ dx \]Using the substitution \( u = \frac{x}{\beta} \), we find:\[ \langle x \rangle = \beta \cdot \alpha \]Thus, the mean is \( \alpha \beta \).
03

Calculate \( \langle x^2 \rangle \) for Variance Determination

To find variance, we first find \( \langle x^2 \rangle \):\[ \langle x^2 \rangle = \int_0^\infty x^2 \cdot f(x) \ dx \]Substituting the gamma PDF:\[ \langle x^2 \rangle = \int_0^\infty x^2 \cdot \frac{1}{\beta^\alpha \Gamma(\alpha)} x^{\alpha - 1} e^{-x/\beta} \ dx \]Simplifying this integral gives:\[ \langle x^2 \rangle = \frac{1}{\beta^\alpha \Gamma(\alpha)} \int_0^\infty x^{\alpha + 1} e^{-x/\beta} \ dx \]Substituting \( u = \frac{x}{\beta} \) leads to:\[ \langle x^2 \rangle = \beta^2 (\alpha + 1) \]
04

Determine Variance \( \sigma^2 \)

The variance \( \sigma^2 \) of a random variable is given by:\[ \sigma^2 = \langle x^2 \rangle - \langle x \rangle^2 \]From previous steps, we use \( \langle x \rangle = \alpha \beta \) and \( \langle x^2 \rangle = \beta^2 (\alpha + 1) \).Substitute these into the variance equation:\[ \sigma^2 = \beta^2 (\alpha + 1) - (\alpha \beta)^2 \]Simplifying gives:\[ \sigma^2 = \alpha \beta^2 \]Thus, the variance \( \sigma^2 \) is \( \alpha \beta^2 \).

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.

Probability Density Function
The Probability Density Function (PDF) of a distribution provides a mathematical function that describes the likelihood of a random variable taking on a particular value. In the case of the gamma distribution, we have:
  • When \(x > 0\), the function is defined as \( f(x)= \frac{1}{\beta^{\alpha} \Gamma(\alpha)} x^{\alpha-1} e^{-x / \beta} \). This means the value of the PDF at \(x\) gives the relative likelihood of \(x\), provided it is positive.
  • If \(x \leq 0\), the function equals 0, reflecting that the gamma distribution applies only to positive values.
The parameters \(\alpha\) and \(\beta\), known as the shape and scale parameters respectively, influence the behavior of the distribution. Utilizing the gamma distribution鈥檚 PDF is foundational for further calculations like mean and variance.
Mean Calculation
Mean, often represented as \(\langle x \rangle\), is a measure of central tendency in a probability distribution. For the gamma distribution, it鈥檚 crucial to exploit its Probability Density Function to find the mean:
  • The mean is given by the integral of \( x \cdot f(x) \) over all possible values of \(x\), \[ \langle x \rangle = \int_0^\infty x \cdot f(x) \, dx \]
  • This requires substituting the gamma function into the integral and using calculus to simplify it, ultimately revealing the mean as \(\alpha \beta\).
Understanding the determination of mean from the PDF using integral calculus underlines the relationship between these mathematical elements.
Variance Calculation
Variance, denoted \(\sigma^2\), quantifies the spread or dispersion of a distribution, indicating how much the observations deviate from the mean. For the gamma distribution:First, calculate the expected value of \(x^2\), \[ \langle x^2 \rangle = \int_0^\infty x^2 \cdot f(x) \, dx\]By substituting and simplifying, we arrive at \( \langle x^2 \rangle = \beta^2 (\alpha + 1) \).
  • Variance is derived from the formula: \[ \sigma^2 = \langle x^2 \rangle - \langle x \rangle^2 \] whereby substituting the values for \( \langle x^2 \rangle \) and the previously determined mean \(\alpha \beta\), results in \(\sigma^2 = \alpha \beta^2\).
These steps showcase the necessity of understanding variance calculations through integral calculus and the foundational moments such as \(\langle x \rangle\) and \(\langle x^2 \rangle\).
Integral Calculus
Integral calculus is the mathematical study of continuous accumulation of quantities and the spaces under curves. It is a cornerstone for evaluating expressions like mean and variance of the gamma distribution. Here's how it fits in:
  • For calculating the mean, we perform the integral of \(x \cdot f(x)\) over its range, requiring the manipulation of the gamma PDF with integral methods.
  • Variance computation also relies heavily on integrals, first determining \(\langle x^2 \rangle\) through direct integration of \(x^2 \cdot f(x)\), then applying subtraction in the derived formula for variance.
Understanding integral calculus within these contexts demonstrates its essential role in higher-level probability and statistics, capturing both extent and detailed nature of distributions.

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

Verify (a) \(\int_{0}^{\infty} e^{-r} \ln r d r=-\gamma\). (b) \(\int_{0}^{\infty} r e^{-r} \ln r d r=1-\gamma\). (c) \(\int_{0}^{\infty} r^{n} e^{-r} \ln r d r=(n-1) !+n \int_{0}^{\infty} r^{n-1} e^{-r} \ln r d r, \quad n=1,2,3, \ldots .\) Hint. These may be verified by integration by parts, three parts, or differentiating the integral form of \(n !\) with respect to \(n\).

Rewrite Stirling's series to give \(z !\) instead of \(\ln (z !)\). $$ \text { ANS. } z !=\sqrt{2 \pi} z^{z+1 / 2} e^{-z}\left(1+\frac{1}{12 z}+\frac{1}{288 z^{2}}-\frac{139}{51840 z^{3}}+\cdots\right) \text { . } $$

Without using Stirling's series show that (a) \(\ln (n !)<\int_{1}^{n+1} \ln x d x\), (b) \(\ln (n !)>\int_{1}^{n} \ln x d x ; n\) is an integer \(\geq 2\). Notice that the arithmetic mean of these two integrals gives a good approximation for Stirling's series.

Show that $$ \int_{0}^{\infty} \frac{\sinh ^{\alpha} x}{\cosh ^{\beta} x} d x=\frac{1}{2} B\left(\frac{\alpha+1}{2}, \frac{\beta-\alpha}{2}\right), \quad-1<\alpha<\beta $$ Hint. Let \(\sinh ^{2} x=u\).

The Legendre polynomial may be written as $$ \begin{aligned} P_{n}(\cos \theta)=& 2 \frac{(2 n-1) ! !}{(2 n) ! !}\left\\{\cos n \theta+\frac{1}{1} \cdot \frac{n}{2 n-1} \cos (n-2) \theta\right.\\\ &+\frac{1 \cdot 3}{1 \cdot 2} \frac{n(n-1)}{(2 n-1)(2 n-3)} \cos (n-4) \theta \\\ &\left.+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \frac{n(n-1)(n-2)}{(2 n-1)(2 n-3)(2 n-5)} \cos (n-6) \theta+\cdots\right\\} \end{aligned} $$
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Particle Model of Matter

Read Explanation

Mechanics and Materials

Read Explanation

Electromagnetism

Read Explanation

Electrostatics

Read Explanation

Engineering Physics

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.