If a random variable U has a gamma distribution with

Chapter 6, Problem 112SE

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If a random variable U has a gamma distribution with parameters \(a>0 \text { and } \beta>0\), then \(Y=e^{U}\) [equivalently, \(U=\ln (Y)\)] is said to have a log-gamma distribution. The log-gamma distribution is used by actuaries as part of an important model for the distribution of insurance claims. Let U and Y be as stated.

Show that the density function for Y is

        \(f(y)=\left\{\begin{array}{ll} {\left[\frac{1}{\Gamma(\alpha) \beta^{\alpha}}\right] y^{-(1+\beta) / \beta}(\ln  

                                   y)^{\alpha-1},} & y>1 \\ 0, & \text { elsewhere } \end{array}\right.\)

If \(\beta<1\), show that \(E(Y)=(1-\beta)^{-a}\) [See the hint for part (c).]If \(\beta<.5\), show that \(V(Y)=(1-2 \beta)^{-a}-(1-\beta)^{-2 a}\). [Hint: Recall that \(E(Y)=E\left(e^{U}\right)\) and \(E\left(Y^{2}\right)=E\left(e^{2 U}\right)\), where U is gamma distributed with parameters \(\alpha>0 \text { and } \beta>0\), and that the moment-generating function of a gamma-distributed random variable only exists if \(\mathrm{t}<\beta^{-1}\); see Example 4.13.]

Equation Transcription:

   

 

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Text Transcription:

a>0  and  \beta > 0

Y=e^U

U=ln(Y)

f(y)={1\Gamma(\alpha) \beta^\alpha y^-(1+\beta) / \beta(\ln y)^\alpha-1 & y>1 0, elsewhere

\beta<1

E(Y)=(1-\beta)^-a

\beta<.5

V(Y)=(1-2 \beta)^-a-(1-\beta)^-2 a

E(Y)=E (e^U)

E(Y^2)=E(e^2U)

\alpha>0 and  \beta>0

t<\beta^-1