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I have the moment generating function \begin{equation} M\left(t\right)=\frac{\Gamma\left(\alpha+1\right)\Gamma\left(1-t\right)}{\Gamma\left(\alpha-t+1\right)},\ t<-1. \end{equation} We know that expectation and variance can be found by $E\left(X\right)=\left.\frac{d\ln M\left(t\right)}{dt}\right|_{t=0}$ and $Var\left(X\right)=\left.\frac{d^{2}\ln M\left(t\right)}{dt^{2}}\right|_{t=0}$. How to show that \begin{equation} E\left(X\right)=\psi\left(\alpha+1\right)-\psi\left(1\right)\ \text{and}\ Var\left(X\right)=\psi'\left(1\right)-\psi'\left(\alpha+1\right) \end{equation} where $\psi\left(x\right)=\frac{d}{dx}\ln\Gamma\left(x\right)$ is digamma function.

at first, it seems obvious but; i couldn't get $\psi\left(\alpha+1\right)$ term in expectation.

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\begin{equation} \ln M\left(t\right)=\ln\Gamma\left(\alpha+1\right)+\ln\Gamma\left(1-t\right)-\ln\Gamma\left(\alpha-t+1\right). \end{equation} Hence $$ \frac{d}{dt}\ln M(t)=\frac{d}{dt}\ln\Gamma\left(1-t\right)-\frac{d}{dt}\ln\Gamma\left(\alpha-t+1\right)=-\psi(1-t)+\psi(\alpha-t+1), $$ and at $t=0$ you get $\psi(\alpha+1)-\psi(1)$.

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  • $\begingroup$ thanks. i had a minus mistake. a simple mistake. $\endgroup$
    – mert
    Jan 7, 2016 at 16:15

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