How to drive the moment generating function of Gamma distribution using log-partition function?

Suppose $X\sim\Gamma(\alpha,\beta)$, gamma distribution with parameter $(\alpha, \beta)$. Then $X$ has p.d.f $$f_{\alpha,\beta}(X) = \frac{\beta^\alpha}{\gamma(\alpha)}x^{\alpha-1}\exp\{-\beta x\}$$ In the canonical exponential family form, letting $(\eta_1,\eta_2)=(\alpha,-\beta)$ we have $$f_{\eta_1,\eta_2}(x)\propto\eta_1\log(-\eta_2)+(\eta_1-1)\log x-\eta_2x,$$ which means the log-partition function $$A(\eta)=\log\gamma(\eta_1+1)-(\eta_1+1)\log[-\eta_2]$$

My question is how to use the formula $$M_{X}(u)=\frac{\exp{A(\eta+u)}}{\exp{A(\eta)}}$$ to derive the momemtn generating function, which has the result given by $$M_{X}(u)=\left(1-\frac{u}{\beta}\right)^{-\alpha}$$

Particularly, since $A$ is a function of two parameters, i.e. $\eta_1$ and $\eta_2$, I expect the $u$ in $M_{X}(u)=\frac{\exp{A(\eta+u)}}{\exp{A(\eta)}}$ should also have two parameters, i.e. $u_1$ and $u_2$. But moment generating function of $X$ is a function of $u\in\mathbb{R}$, which is one-dimensional.


1 Answer 1


I got it.

The correct formula for Moment generating function should be $$M_{T}(u)=\frac{\exp{A(\eta+u)}}{\exp{A(\eta)}},$$ not $$M_{X}(u)=\frac{\exp{A(\eta+u)}}{\exp{A(\eta)}}$$

Here, $T(X)=(\log X,X)$.

Thus, to get the $M_{X}(u)$, we need to treat $\eta_1$ to be a fixed constant or setting $u_1$ to be 0 in the first equation.

  • $\begingroup$ Could you please explain why it is necessary "to treat $\eta_{1}$ as a fixed constant or set $u_{1}$ to be 0 in the first equation"? It would be greatly appreciated and very helpful. $\endgroup$
    – aliocha
    Commented Mar 21 at 23:48

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