I'll detail my context. First of all, if there is something wrong with my procedure, I would like to be corrected. Suppose we have a probability space $(\Omega, \mathcal{F}, \mu)$. Yes, $\mu$ is a probability measure. Suposse $X:\Omega \to \mathbb{R}$ a random variable. We all know that the expectation is defined as
$$E[X]= \int_{\Omega}X d\mu = \int_{\mathbb{R}}x dP^{X} $$
where $P^X$ is the probability distribution of $X$ in $\mathbb{R}$. There is a canonical form to define a random variable to be a absolutly continuous – using $F_X$, the CDF. In other words, $X$ is absolutly continuous if there is a $f$ density such that
$$F_X(x) = \int_{\infty}^x f(t)dt$$
We know that $P^X$ and $F_X$ uniquely determine each other. So in terms of measure, we know can write
\begin{equation}
P^{X}(A) = \int_A f(x)dx
\end{equation}
In other words, $dP^{X} = f dx$. So, we can write
\begin{equation}
E[X] = \int_{\mathbb{R}} xf(x)dx
\end{equation}
However, often in mathematical statistical theory, we have the density depending on a parameter: $X \sim f(x,;\theta)$. So if I want to be more strict, I should write
$$E_\theta [X] = \int_{\mathbb{R}} xf(x;\theta)dx$$
My difficulty lies in understanding whether the measure of probability will also depend on the parameter. In other words, the right would be to write $(\Omega, \mathcal{F}, \mu_\theta)$ and
$$E_\theta[X] = \int_{\Omega}X d\mu_\theta$$
Do you have any idea how to help clarify or help intuition if this is true?
Best Answer
Given a family of probability distributions $\mathcal{P}$ on a measurable space $(\Omega,\mathcal{F})$, the triple $(\Omega,\mathcal{F},\mathcal{P})$ is a statistical model/space. That is, for each $P\in\mathcal{P}$, $(\Omega,\mathcal{F},P)$ is a probability space. Often $\mathcal{P}$ is an indexed family $\mathcal{P}=\{P_{\theta}:\theta\in \Theta\}$, which means that the expectation of $X$ is $\theta$-dependent and the notation $\mathsf{E}_{\theta}X$ helps to avoid confusion.