MLE, convergence in probability

Question

The MLE $\tilde{\theta}_n$ of a sample of random variables $X_i$ from parametric model $\{f(x,\theta): x\in\mathbb{R}, \theta \in \Theta\}$ is called consistent if $\tilde{\theta}_n$ converges in probability to $\theta_0$, i.e. $\tilde{\theta}_n$ $\xrightarrow{P}$ $\theta_0$, whenever $X_i$ are generated from $ f(x,\theta_0)$.

My question is regarding the definition of convergence in probability of estimators.

I know that the MLE $\tilde{\theta}_n$ is itself a random variable (i.e. a measurable function). So, does $\tilde{\theta}_n$ $\xrightarrow{P}$ $\theta_0$ mean that $\tilde{\theta}_n$ converges in probability to $\theta_0$ when viewed as a measurable function? That is, does convergence in probability of an estimator mean the following:

$$ \tilde{\theta}_n \xrightarrow{P} \theta_0 \Leftrightarrow \mathbb{P}(\{x \in \mathbb{R}: |\tilde{\theta}_n(x)-\theta_0|\ge \epsilon \})\rightarrow 0 $$

The confusion comes from the fact that everywhere I read about the consistency of the MLE, they seem to treat the sequence ($\tilde{\theta}_n$) like a sequence of real numbers, but then convergence in probability $\tilde{\theta}_n$ $\xrightarrow{P}$ $\theta_0$, only makes sense when talking about measurable functions.

Can anyone please help me clarify the above. Thank you very much.

Answer 1

I think statisticians think of $(\tilde{\theta}_n)$ as a sequence of random variables, for which convergence in probability is well-defined. Random variables themselves can be viewed as measurable functions from a probability space to the real line.

Formally, one should write $P(\{\omega \in \Omega : |\tilde{\theta}_n(\omega) - \theta_0| > \epsilon\}) \to 0$ where $\Omega$ is the underlying probability space, but people often just write $P(|\tilde{\theta}_n - \theta_0| > \epsilon) \to 0$ to mean the same thing.

Answer 2

I am just a student but here an other prove that works if you all ready know by hypothesis that the MLE estimator is unbiased at least when the size sample goes to infinity.

1. $\theta_n $ is the MLE estimator of an i.i.d sample $\left \{ X_1=x_1, X_2=x_2,\ldots,X_n=x_n \right \}$ of size $n$ and $\theta_0$ the parameter to estimate.
   First let note that the sequence $(\theta_n)_n$ is a sequence of r.v. while $\theta_0$ is a parameter (a "constant").
   So it is asking to us to prove: $$\forall \epsilon>0 \Rightarrow \lim_{n \to \infty }\mathbb{P}(|\theta_n - \theta_0| \geq \epsilon)=0$$

2. In order to do that we will use the Markov Inequality with $r=2$ (the condition are verify by question hypothesis):
   $$\forall \epsilon>0 \Rightarrow \mathbb{P}(|\theta_n - \theta_0| \geq \epsilon) \leq \frac{E(|\theta_n-\theta_0|^2)}{\epsilon^2}=\frac{E((\theta_n-\theta_0)^2)}{\epsilon^2}=\frac{E(\theta_n^2+\theta_0^2-2\theta_n \theta_0)}{\epsilon^2}=\frac{E(\theta_n^2)+\theta_0^2-2\theta_0E(\theta_n)}{\epsilon^2}$$

3. As we know that $\theta_n$ is unbiased when $n$ goes to infinity it means that: $$\lim_{n \to \infty }E(\theta_n)-\theta_0=0 \Rightarrow \lim_{n \to \infty }E(\theta_n) = \theta_0 = E(\theta_0)$$

Hence $E(\theta_n)$ converges in law to $E(\theta_0)=\theta_0$. So because $f(x)=x^2$ is a continuous function:$$\lim_{n \to \infty }E(\theta_n^2)=E(\theta_0^2) = \theta_0^2$$

Q.E.D.

I hope this is correct.