I am studying a note on basic probability which introduced the exponential family as
$\left\{p(x;\theta): \theta \in \Theta\right\}$ where
$p(x;\theta) =h(x)c(\theta) \exp\left\{ \sum_{i=1}^{k}w_{i}(\theta)t_i(x)\right\}$
The exponential family is rewritten in terms of a natural logarithm for k=1 as follows:
$p(x;\theta) =h(x)\exp\left\{ \eta t(x) - A(\eta)\right\}$ where $A(\eta) = log\int h(x) exp\left\{ \eta t(x)\right\}dx$
Now, let assume $X$ have an exponential family distribution Then:
$E(t(X)) = A^{\prime}(\eta)$
$Var(t(X)) = A^{\prime\prime}(\eta)$
I am trying to prove the 2 expressions shown above.
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First what is the role of t(X) in this expression? It just assumed X have an exponential family distribution.
and I don't really know how I can derive the above expression: so far I just find the $ A^{\prime}(\eta)= \frac{d}{d\eta}log_b\int h(x)\exp \{\eta t(x)\}dx = log_b{e}\frac{\int h(x)t(x) \exp\left\{ \eta t(x)\right\}dx}{\int h(x)\exp\left\{ \eta t(x)\right\}dx}$
Can anyone help how to approach this problem please?
Thanks in advance.
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Assuming the base of $\log$ function is $e$, I proved $E(t(X))$ in the following way. I am not sure if it is right:
$E(t(X)) = \int (h(x)\exp\{\eta t(x) - A( \eta )\}t(x)dx$
$E(t(X)) = \int h(x)\frac{e^{\eta t(x) }} {e^{ A( \eta )}}t(x)dx$
$E(t(X)) = \frac{1}{e^{ A( \eta )}} \int h(x)e^{\eta t(x) }t(x)dx$
$E(t(X)) = \frac{ \int t(x)h(x)e^{\eta t(x) }dx}{ \exp\{log\int h(x) exp\left\{ \eta t(x)\right\}dx\}}$
$E(t(X)) = \frac{ \int t(x)h(x)e^{\eta t(x) }dx}{\int h(x) e^{\eta t(x) }dx}$