Lets say I have the pdf and maximum likelihood function:
$
f_X(x) = \begin{cases}
\frac{\alpha \beta^\alpha}{x^{\alpha+1}}, & x > \beta, \\
0, & x \leq \beta.
\end{cases}
$
$
\begin{aligned}
& l(\alpha, \beta) = \log L(\alpha, \beta) = n \log \alpha + n\alpha \log \beta – (\alpha + 1)\sum_{i=1}^{n} \log x_i
\end{aligned}
$
In the solution they answered:
\begin{aligned}
& \text{To determine the maximum likelihood estimation of } \beta, \text{ we note that the only term containing } \beta \text{ is } n\alpha \log \beta. \text{ It is monotonic and increasing in } \beta, \text{ meaning we maximize } l, \text{ and thus } L, \text{ by choosing } \beta \text{ as large as possible. The constraint we must consider is that } x_i \geq \beta \text{ for each } i = 1, \ldots, n. \text{ Therefore, we set } \hat{\beta} = \min_{1 \leq i \leq n} x_i \text{ to maximize } l \text{ with respect to } \beta.
\end{aligned}
I don't really understand the explanation at all especiall anything after the "The constraint we must consider is that….". I'd really appreciate it if someone could explain it in easier terms.
Best Answer
$\beta$ is the minimum of this Pareto distribution: the question says the density is $0$ below $\beta$ so there is zero probability of observing data below $\beta$.
Suppose for example you knew $\alpha=2$ and you observed the data $10.0, 11.1, 12.4, 14.5$. Then the likelihood would be proportional to this curve as $\beta$ changes:
which illustrates that for $0 \lt \beta \lt 10$ the likelihood is an increasing function of $\beta$. But for $\beta>10$ there is zero likelihood for such $\beta$ given the observations, as some of the data would have to be below the minimum of the distribution. This is the constraint being discussed.
So the likelihood is clearly maximised here at $\hat \beta =10.0$, the minimum of the observations.