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$X\sim \Gamma(p,a)$,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 $\ln X$.

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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 $g(\eta) = \log \Gamma(\eta_2 + 1) - (\eta_2 + 1)\log\left(-\eta_1\right)$

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    $\begingroup$ 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
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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 $t_2$.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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