Cannot Find Parameter Estimation using Maximum Likelihood Estimation

Question

Given probability density function

$$p(x|\phi)=\begin{cases}\dfrac{\phi}{x^2}&\phi<x<\infty\\0&\text{others}\end{cases}.$$

I want to find an estimation for the $\phi$ parameter with the maximum likelihood estimator.

First, I find the likelihood function as below $$L(\phi)=\dfrac{\phi}{{x_1}^2}\dfrac{\phi}{{x_2}^2}\ldots\dfrac{\phi}{{x_n}^2}=\dfrac{\phi^n}{\prod\limits_{i=1}^n{x_i}^2}.$$

Then I determine the log likelihood $\ln L(\phi)$,

$$\ln L(\phi)=n\ln \phi-2\ln\prod\limits_{i=1}^n{x_i}.$$

Now, I try to find the maximum of the log likelihood,

$$\dfrac{d}{d\phi} \ln L(\phi)=\dfrac{n}{\phi}=0.$$

Then I have $n=0$.

Problem: I can't find the $\hat\phi$. How to find the parameter estimation of $\phi$ using maximum likelihood estimator?

Please help me. Thanks before.

Answer 1

First of all, I would rewrite the density as $$p(x|\phi)=\dfrac{\phi}{x^2}\sigma(x-\phi),$$ in which $\sigma(a)$ is a step function.

If $p$ is a density function then we must check that the integral from $-\infty$ to $\infty$ is equal to $1$. $$\int_{-\infty}^{\infty}\dfrac{\phi}{x^2}\sigma(x-\phi)~dx=\phi\int_{\phi}^{\infty}\dfrac{dx}{x^2}=\phi\lim_{a\to \infty}\left[-\dfrac{1}{a}+\dfrac{1}{\phi} \right]=1.$$

From this condition, we see that $\phi >0$ or we would integrate over a singularity.

Then we note as soon as $\min_{n}\{x_n\}<\phi$ the likelihood will become zero. Hence, we have to assume that $x_n>\phi$ for all $n=1,...,N$.

Your last condition would imply that $\phi \to \infty$ is a solution. Which makes sense when you are looking at the log likelihood. Setting $N=0$ makes no sense as we would not have a data set.

The problem is that we cannot make $\phi \to \infty$ because we would not include our data points. Hence we must use $\hat{\phi} = \min_{n}\{x_n \}$ as a solution. But this only works if $x_n>0$ for all $n=1,...,N$.