what does almost surely mean in this case

Question

Let $X_1,...,X_n$ be some random variables and $\theta$ be a function of these random variables. I was asked to show that $\mathbb{P}(\theta=\hat{\theta}|X_1,...,X_n)=c$ almost surely. I am able to calculate the probability out explicitly, but I don't know what I am supposed to do with almost surely.

Answer 1

Given a particular realization $x_1, \ldots, x_n$ of $X_1\ldots, X_n$, the conditional probability $P(\theta = \hat \theta|x_1\ldots, x_n)$ is a specific number on $[0,1]$. But before we observe the $X_i$, the probability $P(\theta = \hat \theta|X_1\ldots, X_n)$ is a random variable! In other words, without seeing the $X_i$, we can ask about the distribution we expect $\theta$ to have after we see them and condition on them. For $P(\theta = \hat \theta|X_1\ldots, X_n)$ to equal $c$ "almost surely" is probabilist lingo meaning it equals $c$ with probability 1. In notation: $$ P[P(\theta = \hat \theta|X_1\ldots, X_n) = c] = 1. $$ The outer probability is a prior, conditioned on nothing. The inner probability is a random variable representing a probability conditioned on a piece of evidence which is determined randomly.

Answer 2

I guess that you understood wrongly what has been said to you about $\theta $. If $\theta $ is an unknown parameter, that defines a probability distribution, and we want to estimate it using a random sample $X_1,\ldots ,X_n$, then from this sample we construct an estimator for $\theta $, namely a function $\hat \theta :=f(X_1,\ldots ,X_n)$.

Then, by definition of conditional distribution, we have that $$ \mathbb{P}(\hat \theta =\theta |X_1,\ldots ,X_n)=\mathbb{E}(\mathbf{1}_{\{\hat \theta =\theta \}}|X_1,\ldots ,X_n)=\mathbf{1}_{\{\hat \theta =\theta \}} $$ as the function $\mathbf{1}_{\{\hat \theta =\theta \}}$ is $\sigma (X_1,\ldots ,X_n)$-measurable, where $\mathbf{1}_{G}$ is the characteristic function of the event $G$. Thus, as a characteristic function can just take the values zero or one, that $\mathbf{1}_{\{\hat \theta =\theta \}}=c$ almost surely means that $c\in\{0,1\}$, and consequently $\mathbb{P}(\hat \theta =\theta)=\mathbb{E}(\mathbf{1}_{\{\hat \theta =\theta \}})\in\{0,1\}$, that is, that the probability that $\hat \theta $ is equal to $\theta $ is zero or one.