Mean and variance from the cumulant function

Differentiating the displayed density pointwise on its fixed support gives ${\mathrm{\partial }}_{{\theta }_i}{f}_i=f_i(y-b^{\mathrm{\prime }}({\theta }_i))\mathrm{/}a$ and ${\mathrm{\partial }}_{{\theta }_i}^2{f}_i=f_i\{(y-b^{\mathrm{\prime }}({\theta }_i){)}^2\mathrm{/}{a}^2-b^{\mathrm{\prime }\mathrm{\prime }}({\theta }_i)\mathrm{/}a\}$. Off the fixed support the density vanishes outside a fixed ${\nu }_i$-null exception for every ${\theta }_i$, so these derivative identities hold ${\nu }_i$-almost everywhere on the whole response space.

For any sequence of nonzero $h_k\to 0$ with ${\theta }_i+h_k\in J$, the difference quotients $(f_i(\cdot ;{\theta }_i+h_k,\varphi )-f_i(\cdot ;{\theta }_i,\varphi ))\mathrm{/}{h}_k$ converge pointwise almost everywhere to ${\mathrm{\partial }}_{{\theta }_i}{f}_i$. The mean value theorem [F2] bounds their absolute values by the local first-derivative envelope, so [F3] passes the limit through the integral. Each quotient has integral $(1-1)\mathrm{/}{h}_k=0$ by normalization and [F4]; hence $\int {\mathrm{\partial }}_{{\theta }_i}{f}_{i}d{\nu }_i=0$. Since $2\mathrm{\mid }y\mathrm{\mid }\le 1+y^2$, monotonicity and homogeneity [F8], normalization, linearity [F4], and the density identity [F6] give $\mathbb{E}\mathrm{\mid }{Y}_i\mathrm{\mid }\le \frac{1}{2}(1+\mathbb{E}[Y_i^2])<\mathrm{\infty }$. Thus $Y_i$ is integrable by [F7], so its positive and negative parts are integrable. Applying [F6] to those two parts and subtracting by [F4] and [F5] identifies $\mathbb{E}[Y_i]=\int y{f}_{i}d{\nu }_i$. Substituting step 1.1 into $\int {\mathrm{\partial }}_{{\theta }_i}{f}_{i}d{\nu }_i=0$, and using normalization [F1] and linearity [F4], gives $\int y{f}_{i}d{\nu }_i=b^{\mathrm{\prime }}({\theta }_i)$.

Apply the same sequential difference-quotient argument to ${\mathrm{\partial }}_{{\theta }_i}{f}_i$, now using the local second-derivative envelope, to get $\int {\mathrm{\partial }}_{{\theta }_i}^2{f}_{i}d{\nu }_i=0$. Substituting step 1.1 gives $\int (y-b^{\mathrm{\prime }}({\theta }_i){)}^2{f}_{i}d{\nu }_i=a{b}^{\mathrm{\prime }\mathrm{\prime }}({\theta }_i)$. By [F6] this is $\mathbb{E}[(Y_i-b^{\mathrm{\prime }}({\theta }_i){)}^2]$; by step 2.1, $b^{\mathrm{\prime }}({\theta }_i)=\mathbb{E}[Y_i]$, so [F9] identifies the centered second moment with the variance.