MLE, convergence in probability 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.
 A: 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.
A: 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.

*

*$\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: $$\forall \epsilon>0 \Rightarrow \lim_{n \to \infty }\mathbb{P}(|\theta_n - \theta_0| \geq \epsilon)=0$$


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


*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.
