# What 's the $(\Omega,\mathcal{F},P_{\theta} )$ those $T_{n}$ defined on?

Definition (Consistency)

Let $$T_1,T_2,\cdots,T_{n},\cdots$$ be a sequence of estimators for the parameter $$g(\theta)$$ where $$T_{n}=T_{n}(X_1,X_2,\cdots,X_{n})$$ is a function of $$X_{1},X_{2},\cdots,X_{n}.$$

The sequence $$T_{n}$$ is a weakly consistent sequence of estimators for $$\theta$$ if for every $$\varepsilon>0,$$ $$\lim_{n\rightarrow\infty}P_{\theta}(|T_{n}-g(\theta)|<\varepsilon)=1.$$ If $$T_{n}$$ converges with probability one or almost surely (a.s.) to $$g(\theta)$$, that is, for every $$\theta\in\Theta$$ $$P_{\theta}\left(\lim_{n\rightarrow\infty}T_{n}=g(\theta)\right)=1,$$ then it is strongly consistent.

Strongly consistency implies weakly consistency.This definition says that as the sample size $$n$$ increases,the probability that $$T_{n}$$ is getting closer to $$\theta$$ is approaching $$1$$.

I am confused about what is the $${\color{Red}{\left(\Omega,\mathcal{F},P_{\theta}\right)}}$$ those $$T_{n},\,n=1,2,\ldots$$, defined on? What's the specific probability measure $${\color{Red} {P_{\theta}}}$$ ?

• The short answer is that the triple $(\Omega, P, \mathcal{F})$ is irrelevant since, for a given problem, you can always provide one if needed. Commented Apr 12, 2023 at 7:05

I deem a generalized framework formalizing the concepts at work is apt here. For more details, refer to $$\rm [I].$$

Let $$(\Omega, \boldsymbol{\mathfrak A}, \Pr)$$ be a probability space. Consider the sequence of probability spaces $$\langle (\mathcal X_i, \boldsymbol{\mathfrak A}_i, \mathbf P_i)\rangle_{i=1}^\infty,$$ where $$(\mathcal X_i, \Vert \cdot \Vert_i)$$ is a normed linear space.

Consider a sequence of rvs $$\langle X_i\rangle_{i=1}^\infty$$ and a sequence of real numbers $$\langle r_i\rangle_{i=1}^\infty.$$ Then $$X_n = o_P(r_n)\iff \Pr[\Vert X_n \Vert_n\leq c|r_n|] = 1, ~\forall c >0.$$

Now consider a sequence of measurable functions $$f_n:\mathcal X_n\to \mathcal R ,~\mathcal R$$ being a normed linear space with Borel $$\sigma$$-field. Define $$T_n := f_n(X_n)$$ and $$T: \Omega \to \mathcal R.$$ Then $$T_n$$ converges in probability to $$T$$ if and only if $$\Vert T_n - T\Vert = o_P(1).$$

Now consider a parametric family of distributions $$\{\mathbf P_\theta\mid \theta\in \Theta \}$$ on a sequence space $$\mathcal X^\infty.$$ Define a measurable function $$g: \Theta\to \mathcal G, ~\mathcal G$$ being a metric space with Borel $$\sigma$$-field. Take $$\mathcal X_n = \mathcal X^n$$ and take measurable functions $$T_n: \mathcal X_n\to \mathcal G.$$ Then $$T_n$$ is consistent for $$g(\theta)$$ if for each $$\theta, ~T_n\overset{\mathbf P}{\to} g(\theta).$$

The simplest and most common instance is taking $$\Omega = \mathbb R^\infty, ~\mathcal X_n = \mathbb R^n.$$

Observe how the underlying probability space is at work here based on the implications of the characterization of the convergence in probability above.

(Also, as a footnote, one can see how in probability can be generalized: Take $$S\subseteq \prod_{i=1}^\infty \mathcal X_i.$$ Then $$S$$ occurs in probability (denoted by $$\mathcal P(S)$$) if, for each $$i,$$ there exists $$S_i(\varepsilon)\in\boldsymbol{\mathfrak A}_i$$ such that $$\prod_{i=1}^\infty S_i(\varepsilon)\subseteq S$$ and for each $$\varepsilon > 0,~\mathbf P_i(S_i(\varepsilon))\geq 1-\varepsilon.$$ To see how powerful it is, consider $$f_n:\mathcal X_n \to \mathbb R$$ and, as above, take $$T_n = f_n(X_n).$$ Now, define $$S:= \left\{\langle x_i\rangle_{i=1}^\infty\mid\lim_{n\to\infty} f_n(x_n) = 0\right\}.$$ Then $$T_n = o_{\mathbf P}(1)\iff \mathcal P(S).$$)

## Reference:

$$\rm [I]$$ Theory of Statistics, Mark J. Schervish, Springer-Verlag, $$1995,$$ sec. $$7.1.2,$$ pp. $$395-398.$$

• super complete answer as usual; +1 to you! Commented Apr 18, 2023 at 7:10

It is customary in probability or mathematical statistics to encounter statements such as

Let $$X$$ be an absolutely continuous random variable with density $$f$$

with no reference to underlying probability space. However, we can always supply an appropriate space as follows.

Take $$\Omega = \mathbb{R}$$, $$\mathcal{F} =$$ Borel sets, $$P(B) = \int_B f(x)\,dx$$ for all $$B\in \mathcal{F}$$. If $$X(w) = \omega$$, $$\omega \in\Omega$$, then $$X$$ is absolutely continuous and has density $$f$$.

In a sense, it does not make any difference how we arrive at $$\Omega$$ and $$P$$; we may equally use a different $$\Omega$$ and different $$P$$ and a different $$X$$, as long as $$X$$ is absolutely continuous with density $$f$$. No matter what construction we use, we get the same essential result, that is

$$P(X\in B) = \int_B f(x)\,dx.$$

Therefore, questions about probabilities of events involving $$X$$ are answered completely by knowledge of the density $$f$$. This implies that probabilities of events of $$T(X_1,\ldots,X_n)$$ are also defined by the density of $$T$$.

• I agree with the basic point; unless the axiom of choice was used there's always a way to set it up, but plenty of common random variables are not continuous, for example binomial ones, also please make it clear that $X = id_\Omega$. Also why use the Borel $\sigma$-algebra instead of the Lebesgue? Commented Apr 12, 2023 at 9:35
• @LukasLohse the substance is the same. For discrete r.v. you can choose a suitable $\Omega$, $\mathcal F$ as the sets of all subsets and the counting measure. Commented Apr 12, 2023 at 10:40
• What exactly do you mean with $X$ "having" a density $f$? I thought RVs were functions from the sample space to, say, the real numbers. As far as I understand, a PDF of random variable $X$ is a function s.t. $\mathbb{P}(a < X < b) = \int_a^b f_X(x) dx$ -- but this would presuppose a probability measure $\mathbb{P}$! I'm confused. Commented Apr 13, 2023 at 15:57

Let's start by setting up each individual $$X_i$$ as a function $$X_i: \Omega_i \to S_i$$, with $$S_i$$ being a set and $$\mathcal{F}_i$$ and $$P_{i, \theta}$$ defined appropriately. Now $$T_n = T_n(X_1, X_2, ..., X_n)$$ is a small abuse of notation as random variables are functions from $$\Omega$$, while the right hand side is a "deterministic" function $$t_n$$ from $$S_1\times S_2\times ...\times S_n$$. So I would rewrite $$T_n := t_n(X_1, X_2, ..., X_n)$$ with $$\omega_i \in \Omega_i$$ and with this we can consider $$t_n(X_1(\omega_1), X_2(\omega_2),..., X_n(\omega_n)) = T_n(\omega_1, \omega_2, ..., \omega_n)$$ which tells us that $$T_n$$ is a function from $$\Omega_1\times \Omega_2\times ...\times \Omega_n$$, which therefore can be used to define $$\Omega$$. Now $$\mathcal{F}$$ and $$P_\theta$$ can be anything as long as $$\mathcal{F}_i$$ and $$P_{i, \theta}$$ are their projections down to the individual $$\Omega_i$$.

To answer the questions in your comments I will be a little bit more explicit: If $$X_i$$ are iid with $$(\Omega_x, P_{x, \theta}, \mathcal{F}_x)$$, then $$\mathcal{F}$$ is actually not $$\mathcal{F}_x^n$$ but instead the $$\sigma$$-Alegabra generated by $$\mathcal{F}_x^n$$ through intersection, compliments and $$\sigma$$-countable unions. $$P_\theta$$ is a probability measure that means a function from $$\mathcal{F}$$ to $$[0, 1]$$ with certain properties. Now $$A \in \mathcal{F}_x^n$$ then $$A = A_1 \times ... \times A_n, A_i \in \mathcal{F}_x$$ and $$P_\theta(A) = P_{x, \theta}(A_1) \cdot ...\cdot P_{x, \theta}(A_n)$$. If $$A\in\mathcal{F}/\mathcal{F}_x^n$$ then $$P_{\theta}(A)$$ is determined by how it was generated from the elements of $$\mathcal{F}_x^n$$

When you write $$P_\theta(|T_n - \theta| < \varepsilon)$$ it really means $$P_\theta(\{\omega \in \Omega: |T_n(\omega) - \theta| < \varepsilon\})$$, with $$\{\omega \in \Omega: |T_n(\omega) - \theta| < \varepsilon\} \in \mathcal{F}$$