Is the parameter vector of an indentifiable distribution of a transformed random vector always a subvector...?

Question

I would like, after further considerations about this problem, to reformulate this question of mine again. I kept a record of the past words and remarks as the appendix below. I think the question that really bothers me can be stated as follows:

If $\theta \in \mathbb{R}^{k}$, if $W$ is a random vector in $\mathbb{R}^{r}$ from a distribution having $\theta$ as the value of its parameter vector, if $q \leq k$, if $f: \mathbb{R}^{r} \times \mathbb{R}^{q} \to \mathbb{R}$ is measurable (in the suitable sense), if the distributions $\{ \mathscr{P}_{f(W, b)} \}_{b \in \mathbb{R}^{q}}$ induced by the random variable $f(W, \cdot)$ is identifiable as $\mathbb{R}^{q}$, and if (at no cost of specialization as can be examined) $\beta$ is such that the first moment of $\mathscr{P}_{f(W, \beta)}$ is $= 0$, then is it necessary that $\beta$ is a subvector of $\theta$?

An example of the above more general setting is a orthogonal projection model (identifiable linear model with stochastic regressors).

It took a long while to raise the question in my mind in the current form. I could say it would be what is closest to what I would like to know at the current stage.

Note.

Words and remarks before the above revised version of this question are in the following paragraphs:

Let $p \in \mathbb{N}$; let $\theta \in \mathbb{R}^{p}$; let $(X,Y) \sim F^{\theta}_{X,Y}$ be a random vector in $\mathbb{R}^{2}$ such that $F^{\theta}_{X,Y}$ is the joint CDF of $X$ and $Y$. If there is exactly one $b \in \mathbb{R}$ such that $Y = Xb + U$ and $\mathbb{E}(U \mid X) = 0$, is it necessary that $b$ is a component of $\theta$? How to prove it no matter the answer is affirmative or not? Thanks (for sure at least to Whuber's questions below that pushed to form the current neat form of the question).

It seems that until now did I realize the crux of my question. Thanks again to the feedback providers below, directly or not. Let me use this simple example below to illustrate my confusion. I hope this example would also explain the first paragraph better. Suppose the expectation $\mathbb{E}(Y - Xb)^{2}$ is finite. If $F_{U}^{b}$ is the CDF of $U$, then the expectation can be obtained via two integrals, namely $$ \mathbb{E} (Y - Xb)^{2} = \int_{\mathbb{R}^{2}} (y - xb)^{2} dF_{X,Y}(x,y) = \int_{\mathbb{R}} u^{2} dF^{b}_{U}(u). $$ If the expectation is regarded as the last integral, then it depends on $b$ by assumption. I wonder if $F_{X,Y}$ also depends on $b$ when regarded as the first integral? The notation $\mathbb{E}(Y - Xb)^{2}$ itself provides no information about the distinction, correct? This example does not fit the question title very much; however, I would ask the reader to examine it with a more careless viewpoint unless this mismatch does make the reader completely clueless.

Answer 1

The question puts two functions into evidence, which I will call $\theta$ and $t.$ They are maps from a space $\mathcal F$ of distributions defined on $\mathbb{R}^2$ (or, more generally, any set on which distributions may be defined) into a "parameter space" $\Theta\subset\mathbb{R}^p$ or the real numbers $\mathbb R.$

The parameterization associates the parameter values with the distributions and so may be considered an invertible map

$$\theta:\mathcal{F} \to \Theta\subset \mathbb{R}^p.$$

The regression parameter $b$ of the question is an example of a property of a distribution. That there is exactly one $b$ for each $f\in\mathcal F$ means $b$ can be considered the value $t(f)$ of some function

$$t:\mathcal F \to \mathbb{R}.$$

Note that the components of $\theta$ are automatically properties according to this definition, because the $i^\text{th}$ component is the composition of $\theta$ and the projection $\pi_i:\mathbb{R}^p\to \mathbb{R},$

$$\theta_i:\mathcal F \to \mathbb{R}^p \to \mathbb R\\\theta_i(f)=\pi_i(\theta(f)),$$

and is therefore a real-valued function defined on $\mathcal F.$

The question asks whether, for any possible such $\theta$ and $t,$ it is necessarily the case that there exists a component $i,$ $1\le i \le p,$ such that

$$t(f) = \theta_i(f)$$

for all $f\in \mathcal F.$ That is clearly not true, because distributions have infinitely many properties but there are only a finite number $p$ of components of $\theta.$ For instance, for any number $x$ the map $t+x:\mathcal{F}\to\mathbb R$ given by

$$(t+x)(f) = t(f) + x$$

is not the same property as $t.$

We could generalize the question to ask whether $t$ must depend somehow on the parameter. This is readily shown to be the case because the invertibility of $\theta$ implies that for any parameter $p\in\Theta$ there is a unique $f = \theta^{-1}(p)\in\mathcal F$ associated with $p$ and

$$t(f) = t(\theta^{-1}(p)) = (t\circ \theta^{-1})(p)$$

defines a function

$$t\circ\theta^{-1}:\mathbb{R}^p \to \mathbb{R}.$$

To sum up these two conclusions in words we may say

The parameters are properties of a finitely parameterized family of distributions, but they are not the only properties. All properties are functions of the parameters, however.

---

Although I have been silent about technical issues of continuity (or measurability or differentiability, depending on the application), the same analysis holds quite generally assuming we apply the same criteria to "properties" as we do to "parameters."