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


*

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

*$K$ is cumulant generating function

*$\kappa_n$ 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 $\theta$.
 A: 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.
