1
$\begingroup$

I want to get into the habit of correctly using rigorous notation where possible. I have tried to do this as best as possible with the maximum likelihood estimator, but I am having some slight confusion as a lot of books will reference $\hat{l}(\theta;x_1,...,x_n)$.

I have been stating the following when talking about the MLE of variable $\theta$:

$\{\hat{\theta}_{MLE} \} \subseteq \{\text{argmax}_{\theta \in\Theta}L(\theta;x_1,...x_n)\}$

Where $L(\theta;x_1,...x_n)$ is the Likelihood function, $x_1,...x_n$ are the sample of iid observations from the distribution, and $\Theta$ refers to the whole parameter space.

Does this look adequate, to me it makes sense, but again I don't know if my notation is off. Any confirmation would be appreciated, thanks!

$\endgroup$
1
$\begingroup$

Since you want to be rigorous, I think your definition might be slightly incorrect. In order to see this, you should think about the definition of an estimator. If $\mathcal{X}$ is the sample space and $\Theta$ is the parameter space, then an estimator, $\hat{\theta}$, is a function from $\mathcal{X}$ to $\Theta$. That is, $\hat{\theta}: \mathcal{X} \rightarrow \Theta$. Your definition doesn't make this clear.

The maximum likelihood estimator is a particular type of estimator. For each $\theta \in \Theta$, let's say that the data has density $f(\mathbb{x}|\theta)$. Now, for each $\mathbb{x} \in \mathcal{X}$, the likelihood function for that point, $L_{\mathbb{x}}$, is a function from $\Theta$ to $\mathbb{R}^{+}$ such that $L_{\mathbb{x}}(\theta)=f(\mathbb{x}|\theta)$. Finally, the maximum likelihood estimator, $\hat{\theta}_{ML}$, is an estimator such that, for every $\mathbb{x} \in \mathcal{X}$ \begin{align*} \hat{\theta}_{ML}(\mathbb{x}) &\in \arg\max_{\theta \in \Theta} L_{\mathbb{x}}(\theta) \end{align*} Of course, my definition also has a flaw from an ultra-rigorous perspective. One might ask how to obtain $f(\mathbb{x}|\theta)$. However, the answer to this question might be too technical for the scope of your question.

$\endgroup$
  • 1
    $\begingroup$ Thanks for your answer. Have a good Christmas! $\endgroup$ – mrhappysmile Dec 23 '17 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.