Let $X$ ~ Gamma($\alpha, \lambda$). Let $Y = \ln X$. Find the MGF (moment generating function) of Y.
I know that the MGF of a gamma distribution is $\frac{\lambda^\alpha}{(\lambda + t)^\alpha}$ and the standard MGF function is $E(e^{tX})$. I figured I need to plug in $Y$ into the MGF function and came out with $E(e^{tlnX})$ but I am unsure how to proceed from here. I learned how to do this with another problem where $Y = 3X$ so I could simply just do $E(e^{t3X})$ and then you can make it $M_x(3t)$ and plug the $3t$ into the MGF of the distribution but I didn't think I could do it that way since the natural log of $X$ is different than a simple integer. I was wondering if this problem really is that simple or if there is a way to manipulate the problem to solve using the gamma distribution.
Best Answer
$E(e^{tlnX})=E(X^t)$ so $MGF_Y(t)$ is the $t^{th}$ moment of $X$: $$ MGF_Y(t)=(\lambda^\alpha/\Gamma(\alpha))\int x^tx^{\alpha-1}e^{-\lambda x}dx=(\lambda^\alpha/\Gamma(\alpha))\int x^{t+\alpha-1}e^{-\lambda x}dx\\ =(\lambda^\alpha/\Gamma(\alpha))\Gamma(t+\alpha)/\lambda^{t+\alpha}. $$
We need $t+\alpha>0$ to carry out the integration.