Suppose I have a resource the value of which depends, e.g., on a normal probability density function. That is,
$$Value_\mu(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2}}$$
Now, on the other hand, the value of $\mu$ is given by a random variable governed by another probability distribution, say the standard normal distribution $N(0, 1)$.
How can I calculate the expected curve $Value(x)$? I should point out that I am interested in the general case, for any arbitrary curves.
The model for a typical observation $X$ conditional on unknown parameter $\theta$ is $f(x|\theta)$. As a function of $\theta$, $f(x|\theta)$ is called likelihood. The functional form of $f$ is fully specified up to a parameter $\theta$. The parameter $\theta$ is supported by the parameter space $\Theta$ and considered a random variable. The random variable $\theta$ has a distribution $\pi(\theta)$ that is called the prior. The distribution $h(x,\theta) =f(x|\theta)\pi(\theta)$ is called the joint distribution for $X$ and $\theta$. The joint distribution can also be factorized as $$ h(x,\theta) =\pi(\theta|x)f(x) $$ and the distribution $\pi(\theta|x)$ is called the posterior distribution for $\theta$, given $X=x$. The marginal distribution $m(x)$ can be obtained by integrating out $\theta$ from the joint distribution $h(x,\theta)$ $$ m(x)=\int_{\Theta}h(x,\theta)\mathrm d\theta=\int_{\Theta}f(x|\theta)\pi(\theta)\mathrm d\theta $$ Therefore, the posterior $\pi(\theta|x)=\frac{h(x,\theta)}{m(x)}$ can be expressed as $$ \pi(\theta|x)=\frac{f(x|\theta)\pi(\theta)}{\int_{\Theta}f(x|\theta)\pi(\theta)\mathrm d\theta} $$
Assume that an observation, $X$ is normally distributed with mean $\mu$ and known variance $\sigma^2$. The parameter of interest, $\mu$ also has normal distribution with parameters $\nu$ and $\tau^2$.
So yo have the Bayesian model $X|\mu\sim\mathcal N(\mu,\sigma^2)$ and $\mu\sim\mathcal N(\nu,\tau^2)$. The joint distribution has density $$\begin{align} h(x,\mu)=f(x|\mu)\pi(\mu)&=\frac{1}{2\pi\sigma^2\tau^2}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}-\frac{(\mu-\nu)^2}{2\tau^2}\right)\\ &=\frac{1}{2\pi\sigma^2\tau^2}\exp\left(-\frac{1}{2\rho}\left[\mu-\rho\left(\frac{x}{\sigma^2}+\frac{\nu}{\tau^2}\right)\right]^2-\frac{(x-\nu)^2}{2(\sigma^2+\tau^2)}\right)\\ &=\pi(\mu|x)m(x) \end{align} $$ where $\rho=\frac{\sigma^2\tau^2}{\sigma^2+\tau^2}$. Observing that $\rho\left[\frac{x}{\sigma^2}+\frac{\nu}{\tau^2}\right]=\frac{\tau^2}{\sigma^2+\tau^2}x+\frac{\sigma^2}{\sigma^2+\tau^2}\nu$, we have that $$ \mu|X\sim \mathcal N\left(\frac{\tau^2}{\sigma^2+\tau^2}X+\frac{\sigma^2}{\sigma^2+\tau^2}\nu,\frac{\sigma^2\tau^2}{\sigma^2+\tau^2}\right) $$ and $$ X\sim \mathcal N\left(\nu,\sigma^2+\tau^2\right) $$ You may check that by the elementary properties of conditional expectation, $$ \Bbb E(X)=\Bbb E\left(\Bbb E(X|\mu)\right)=\Bbb E(\mu)=\nu $$ amd $$ \mathrm{Var}(X)=\Bbb E\left(\mathrm{Var}(X|\mu)\right)+\mathrm{Var}\left(\Bbb E(X|\mu)\right)=\Bbb E(\sigma^2)+\mathrm{Var}(\mu)=\sigma^2+\tau^2 $$