n'th cumulant (of a CGF) for exponential family / exponential dispersion model

Question

The n'th cumulant is defined to be the n'th derivative of the CGF (cumulant generating function).

$$\kappa_n = \frac{d^n K(t)}{dt^n} |_{t=0} $$

But I'm reading in a book (p.215, chapter5, eq. 5.8) now that for the exponential family / exponential dispersion model, this is actually equal to:

$$ K \underset{exp.}{=} \frac{\kappa(\theta+t\phi) - \kappa(\theta)}{\phi}\\ \kappa_n = \phi^{n-1} \frac{d^n\kappa(\theta)}{d\theta^n} $$

Where:

1. $\theta$ is the canonical/natural parameter in exponential family.

2. $K$ is cumulant generating function

3. $\kappa_n$ is the nth cumulant

I'm not really sure how come you get this result. The $\phi^{n-1}$ I understand, but not how the derivative changes from $t$ to $\theta$.

Answer 1

As you did'nt define your terms, I will use the definition of exponential dispersion model used in Definition of exponential family with dispersion parameter, which is $$ f(y|\theta,\phi) = \exp\left(\frac{y\theta - b(\theta)}{a(\phi)} + c(y, \phi)\right) $$ which is an exponential family in $\theta$, for each fixed $\phi$. It seems you have in your notation $a(\phi)=\phi$. Then we can calculate the mgf (moment generating function) as $$ M(t)=\exp\left( \frac{b(t a(\phi)+\theta)-b(\theta)}{a(\phi)} \right) $$ so the cumulant generating function $$ K(t)=\log M(t)= \frac{b(t a(\phi)+\theta)-b(\theta)}{a(\phi)} . $$ Then $$ K'(t)=\frac{b'(t a(\phi)+\theta)\cdot a(\phi)}{a(\phi)}=b'(t a(\phi)+\theta) $$ where $'$ means the derivative of a function with respect to its argument. It does not matter if that argument is named $t$ or $\theta$. Now it is easy to see that we can continue differentiating to get $$ K^{(n)}(t)=b^{(n)}(t a(\phi)+\theta)\cdot a^{n-1}(\phi) $$ by repeated application of the chain rule. Now, if you set $t=0$ and revert to the $\frac{d}{d t}$ notation, you get the result. Now, since on the RHS the argument of $b$ is $\theta$, we differentiate there $b$ with respect to its argument, which explains your confusion.