Now, let’s look at determining the MLE for properties of a distribution. First up, we will take a look at the MLE of the median of a distribution. We will start off with a simple p.d.f with random i.i.d. sampling of \(X_1, X_2, ... ,X_n\):

$$ f(x;\theta) = \frac{2x}{\theta^2}, \quad 0 < x \leq \theta \tag{1} $$

Remembering back to our probability distributions and the definition of the median, we have the following:

$$ P(X \leq x) = \frac{1}2 \quad \Rightarrow \quad \int_{-\infty}^{M}f(x)dx = \frac{1}{2} \tag{2} $$

Similarly here, we need to compute what the median would be in terms of our parameter (in this case \(\theta\)):

$$ \int_{0}^{\psi}\frac{2x}{\theta^2}dx = \left.\frac{x^2}{\theta^2}\right\vert_{0}^{\psi} = \frac{\psi^2}{\theta^2} \quad\Rightarrow \quad \frac{\psi^2}{\theta^2} = \frac{1}2 \tag{3} $$

Now to maximize our inter-variable \(\psi\) we need to use the familiar MLE of the distribution itself. In this case, the MLE of \(\theta\) is

$$ \hat\theta = X_{(n)} $$

I will leave it as an exercise to the reader to validate the above statement. Finally, plugging this result into our above equation, we get:

$$ \frac{\psi^2}{\theta^2} = \frac{1}{2} \quad \Rightarrow \quad \frac{\psi^2}{ {X_{(n)}}^2 } = \frac{1}{2} \quad \Rightarrow \quad \psi = \sqrt{\frac{ {X_{(n)}}^2 }{2}} \tag{4} $$

This finally yields out our MLE for the median of our p.d.f.! I will update this later with working on MLEs of multiple parameters. For now, feel free to comment on anything I may have missed or anything you would like me to go over in a future post.