Let $X$ be a random variable with density given by an exponential family : $$f(x)=\exp\big({\eta(\theta)T(x)-d(\theta)+S(x)}\big)$$ with $\theta \in U$ an open set and T and S real functions defined on the support S of $X$.
We want to find the expectation of $T(X)$, one way I was given for the identity $\eta(\theta)=\theta$ is the following :
We find the MGT : $$M_T(u)=\int_S\exp\{uT(x)\}\exp\{\theta T(x) -d(\theta)+S(x)\}dx$$
$$ M_T(u)=exp\{d(u+\theta)-d(\theta)\}\int_S \exp{(u+\theta)T(x)-d(u+\theta)+S(x)}dx$$
The integral is integrating the density over the support for a $u$ in an $\epsilon$ interval that is a valid parameter $(u+\theta)$ because $U$ is open so the integral is equal to $1$ and hence, $$M_T(u)=exp\{d(u+\theta)-d(\theta)\} \quad \forall \lvert u \rvert < \epsilon$$
Then we finish by saying that $$\Bbb E(T(X))=\frac{d}{du}M_T(0)=d'(\theta) $$.
Now I need to find the expectation with $\eta$ a $C^1$ function, inversible, with $\eta'(x)\neq 0$ $\forall x$.
By doing the same thing, I got $$M_T(u)=\exp\{d(u+\eta(\theta))-d(\theta)\}$$ $$\Bbb E(T(X))=d'(\eta(\theta))\exp\{d(\eta(\theta))-d(\theta)\}$$
However, the result I want to get is $\Bbb E(T(X))=\frac{d'(\theta)}{\eta'(\theta)}$. Is there something wrong ? Doing the exact same thing was maybe not a good idea, I also did not use the given hypothesis on $\eta$.
Best Answer
Since $f(x)=f(x,\theta)$ is a density function, you have $\int f(x,\theta)\,dx =1$, that is $$ \exp(d(\theta))=\int{\exp\big({\eta(\theta)T(x)+S(x)}\big)}\,dx. $$ Also, you have $\eta \in C^1$ so you can use Leibniz rule and differentiate both sides with respect to $\theta$ to get $$ d'(\theta)\exp(d(\theta))=\int{\eta'(\theta)T(x)\exp\big({\eta(\theta)T(x)+S(x)}\big)}\,dx. $$ This means $$ \frac{d'(\theta)}{\eta'(\theta)}= \int{T(x)\exp\big({\eta(\theta)T(x)-d(\theta)+S(x)}\big)}\,dx = \int{T(x)f(x,\theta)\,dx} = \mathbb{E}(T(X)). $$