MLE of Properties/Parameters

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 $X_1, X_2, \ldots, X_n$:

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

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 $\theta$):

\[\int_{0}^{\psi}\frac{2x}{\theta^2}\,dx = \left.\frac{x^2}{\theta^2}\right|_{0}^{\psi} = \frac{\psi^2}{\theta^2} \quad\Rightarrow\quad \frac{\psi^2}{\theta^2} = \frac{1}{2} \tag{3}\]

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 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.

A good video on multiple parameters