# Identity for exponential family of distributions

Consider the following exponential family of distributions: $$p(x ; \beta)=\frac{1}{Z(\beta)} \pi(x) \exp (-\beta h(x)), \;\beta \in\left[\beta_{n} ; \beta_{0}\right].$$ $${Z(\beta)}$$ -- the normalizing constant. $$x \in \mathbb{R}$$.

I want to prove the following identity: $$\log Z\left(\beta_{n}\right)-\log Z\left(\beta_{0}\right)=\int_{\beta_{n}}^{\beta_{0}}\langle h(x)\rangle_{p(x ; \beta)} d \beta,$$

$$\langle h(x)\rangle_{p(x ; \beta)}=\int_{-\infty}^{\infty}h(x)p(x ; \beta)dx$$ here.

My attempt:

It's clear that $$Z\left(\beta_{}\right)=\int \pi(x) \exp (-\beta h(x))dx$$, it follows from properties of densities.

So, I need to prove $$\log Z\left(\beta_{n}\right)-\log Z\left(\beta_{0}\right)=\int_{\beta_{n}}^{\beta_{0}} \int_{-\infty}^{\infty}h(x)p(x ; \beta)dx d \beta = \int_{\beta_{n}}^{\beta_{0}} \int_{-\infty}^{\infty}h(x) \frac{1}{Z(\beta)} \pi(x) \exp (-\beta h(x)) dx d \beta,$$ Also I noticed that $$\frac{\partial (\pi(x) \exp (-\beta h(x)))}{\partial \beta}=-h(x)\pi(x) \exp (-\beta h(x))$$

But I don't understand what to do next.

For $$p(x;\beta)$$ resp. that clumsy and misleading $$\langle h(x)\rangle_{p(x;\beta)}$$ (this is not a function of $$x$$) I am going to use the slightly different notation $$p_\beta(x)\quad\text{ and }\quad\langle h,p_\beta\rangle\,.$$ Since $$Z(\beta)$$ is the normalizing constant it must satisfy $$Z(\beta)=\int_{-\infty}^{+\infty}\pi(x)\exp(-\beta h(x))\,dx\,.$$ Therefore, $$Z'(\beta)=\int_{-\infty}^{+\infty}-h(x)\,\pi(x)\exp(-\beta h(x))\,dx=-Z(\beta)\int_{-\infty}^{+\infty}h(x)\,p(x;\beta)\,dx=-Z(\beta)\langle h,p_\beta\rangle\,.$$ It follows that $$(\log Z(\beta))'=\frac{Z'(\beta)}{Z(\beta)}=-\langle h,p_\beta\rangle$$ which is equivalent to $$\log Z(\beta)-\log Z(\beta_0)=-\int_{\beta_0}^\beta\langle h,p_\beta\rangle\,d\beta\,$$ which is your formula.