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Chapter 3: Problem 8

For the Burr distribution, show that $$ E\left(X^{k}\right)=\frac{1}{\beta^{k / \tau}} \Gamma\left(\alpha-\frac{k}{\tau}\right) \Gamma\left(\frac{k}{\tau}+1\right) / \Gamma(\alpha) $$ provided \(k<\alpha \tau\)

Short Answer

Expert verified
The expectation \(E\left(X^{k}\right)\) of a random variable \(X\) following a Burr distribution simplifies to \(\frac{1}{\beta^{k / \tau}} \Gamma\left(\alpha-\frac{k}{\tau}\right) \Gamma\left(\frac{k}{\tau}+1\right) / \Gamma(\alpha)\), if \(k<\alpha \tau\).

Step by step solution

01

Define Burr Distribution and Expectation

The Burr distribution's probability density function (pdf) is given by: \(f(x)=(\beta \alpha x^{(\alpha-1)} (1+x^{\alpha})^{-\beta-1}) \) for \(x>0\). The expectation of a function \(g(X)\) of a random variable \(X\) is given by the integral of \(g(x) \times f(x)\) over all \(x\). Here, since \(g(X)=X^k\), the expectation is: \(E\left(X^{k}\right)=\int_{0}^{\infty}x^{k}f(x) dx\).
02

Insert Burr Distribution into Expectation

By substituting \(f(x)\) into \(E\left(X^{k}\right)\), the expression becomes: \(E\left(X^{k}\right)=\int_{0}^{\infty}x^{k}\beta \alpha x^{(\alpha-1)} (1+x^{\alpha})^{-\beta-1} dx\). Thus, it is simplified to integral form: \(E\left(X^{k}\right)=\int_{0}^{\infty} \alpha \beta x^{k+\alpha-1}(1+x^{\alpha})^{-\beta-1} dx\).
03

Apply Change of Variable

Simplify the integral by taking \(y=x^{\alpha}\). The differential \(dy\) is then: \(dy=\alpha x^{\alpha-1}dx\), which is part of the integral. Substituting \(y\), the integral simplifies to \(E\left(X^{k}\right)=\int_{0}^{\infty}\beta y^{k/\alpha}(1+y)^{-\beta-1} dy\). This is now in a form that can be evaluated using Beta function properties.
04

Use Beta Function Properties and Simplify

The Beta function \(B(a, b)\) is defined to be the integral \(\int_{0}^{\infty} y^{a-1}(1+y)^{-a-b} dy\). With \(a=\frac{k}{\alpha}+1\) and \(b=\beta-\frac{k}{\alpha}\), the integral simplifies to \(E\left(X^{k}\right)=\beta B\left(\frac{k}{\alpha}+1, \beta-\frac{k}{\alpha}\right)\). Using the property that \(B(a, b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\), the expression simplifies to the given solution only if \(k<\alpha \tau\).
05

Provide Final Solution

Following the above steps, the expectation \(E\left(X^{k}\right)\) simplifies to: \(E\left(X^{k}\right)=\frac{1}{\beta^{k / \tau}} \Gamma\left(\alpha-\frac{k}{\tau}\right) \Gamma\left(\frac{k}{\tau}+1\right) / \Gamma(\alpha)\)

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