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kjetil b halvorsen
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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$$\theta$, we differentiate there $b$ with respect to its argument, which explains your confusion.

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.

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.

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kjetil b halvorsen
  • 82.8k
  • 32
  • 201
  • 663

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.