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.

  • $\begingroup$ What is the pdf of the random variable? I presume the support of the variables is $(\beta,+\infty)$. $\endgroup$ Commented May 26 at 11:23
  • $\begingroup$ @JulioPuerta sorry I have edited it now. $\endgroup$ Commented May 26 at 12:43

2 Answers 2


$\beta$ 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:

enter image description here

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.

  • 1
    $\begingroup$ This is an example where looking for a zero value of the derivative of the likelihood or of the log-likelihood is not going to find the point of maximum likelihood. $\endgroup$
    – Henry
    Commented May 26 at 20:55

As stated in the solution, the maximum likelihood function, for a fixed $\alpha$, is increasing in $\beta$.

Thus, for some fixed sample $(x_1,\ldots,x_n)$ we want to make $\beta$ as big as possible.

But since the support of the random variables is $(\beta,+\infty)$, we have $x_i\geq \beta$ for $1\leq i\leq n$.

Thus, the maximum value that $\beta$ can have is $\displaystyle\min_{1\leq i\leq n} x_i$, which will be the maximum likelihood estimator of $\beta$.

  • $\begingroup$ what do you mean by "But since the support of the random variables is (β,+∞)"? $\endgroup$ Commented May 26 at 13:24
  • $\begingroup$ @Need_MathHelp The pdf is only nonzero from $\beta$ onwards. Therefore, necessarily $x_i\geq \beta$. $\endgroup$ Commented May 26 at 13:27

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