$X\sim \Gamma(p,a)$,find the two dimensional moment generating function of ($X, \ln X$)

What I have done is

$$M_{(X,lnX)}(t_{1},t_{2}) = E(e^{t_{1}x+t_{2}lnx}) = E(e^{t_{1}x}x^{t_{2}}) = \int\int e^{t_{1}x}x^{t_{2}}f_{X,lnX}(x_{1},x_{2})dx_{1}dx_{2}$$

I need help to find the joint distribution of $X$ and $\ln X$.


The gamma distribution is an exponential family and $(x, \log x)$ are the expectation parameters, so just write it in exponential family form and you have the density function of $(x, \log x)$.

$$\begin{align} f_X(x) &= e^{\eta^\intercal T(x) - g(\eta)} \\ \implies \int_{x\in\mathbb R} f_X(x) &= 1 \\ t &= [t_1 t_2]^\intercal \\ \implies \mathbb E(e^{t_1 x + t_2 \log x}) &= \mathbb E(e^{t^\intercal T(x)}) \\ &= \int_{x \in \mathbb R}e^{(\eta + t)^\intercal T(x) - g(\eta)} \\ &= \int_{x \in \mathbb R}e^{(\eta + t)^\intercal T(x) - g(\eta +t) + g(\eta+t) - g(\eta)} \\ &= e^{g(\eta+t) - g(\eta)} \end{align}$$ which gives a general formula for the moment generating function of the sufficient statistics.

For the gamma distribution, the log-normalizer $g(\eta) = \log \Gamma(\eta_2 + 1) - (\eta_2 + 1)\log\left(-\eta_1\right)$

  • 1
    $\begingroup$ Sorry, I don't know what are exponential family and expectation parameters $\endgroup$ – Jakoer Nov 7 '14 at 22:42
  • $\begingroup$ @Jakoer: en.wikipedia.org/wiki/Exponential_family $T(x) = (x, \log x)$ for the Gamma distribution. $\endgroup$ – Neil G Nov 7 '14 at 22:51

The bivariate random variable $(X,\ln X)$ does not have a joint density and so you cannot use a double integral as you have indicated in your question. Instead, the law of the unconscious statistician gives $$M_{(X,\ln X)}(t_{1},t_{2})=E[e^{t_1X+t_2\ln X}] =E[e^{t_1X}X^{t_2}] =\int_0^\infty e^{t_1x}x^{t_2} \cdot \frac{a(ax)^{p-1} e^{-ax}}{\Gamma(p)} \, \mathrm dx$$ which, after suitable change of variables, should give you something of the form $\frac{\Gamma(t_2+p)}{\Gamma(p)}$ times some other stuff involving $a$ and $t_1$ and $t_2$.

  • $\begingroup$ That "suitable change of variables" is the natural parametrization. :) $\endgroup$ – Neil G Nov 8 '14 at 0:59
  • $\begingroup$ @NeilG That stuff is way above my head; I am just an engineer, not a statistician. $\endgroup$ – Dilip Sarwate Nov 8 '14 at 3:14
  • $\begingroup$ Why doesn't it have a joint density? $\endgroup$ – shadowtalker Nov 8 '14 at 14:31
  • 1
    $\begingroup$ @ssdecontrol For two continuous random variables $X$ and $Y$ to have a joint density, it must be that the point $(X,Y)$ can take on all possible values in some region of nonzero area in the plane (with coordinate axes $x$ and $y$): e.g. a rectangle as in $0<x<1, 0 < y < 2$. The point $(X,\ln X)$ perforce lies on the curve $y = \ln x$ in the plane and this is a region of zero area. Thus, the OP's double integral (which computes things over areas) does not work in this case; try setting up limits for the "region" and see. $\endgroup$ – Dilip Sarwate Nov 8 '14 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.