Maximum likelihood when usual procedure doesn't work

I am trying to get the maximum likelihood estimate for the parameter $$p$$. The distribution is the following:

$$f(x\mid p) = \begin{cases} \frac{p}{x^2} &\text{for} \ p\leq x < \infty \ \\ 0 &\text{if not} \end{cases}$$

The sample has size $$n$$.

The problem is, when I try to estimate it by the procedure I know, I would have to estimate the likelihood function and obtain the derivative of the log-likelihood. We'd have:

$$L(p; x) = \frac{p^n}{\prod_{i=1}^{n} x^2_i}$$ $$\ln L(p;x) = n \ln(p) - \sum_{i=1}^{n} \ln(x_i^2)$$

For the derivative:

$$l'(p;x) = \frac{n}{p} = 0$$

And I am stuck because it has no solution for $$p$$. How do I evaluate this?

Thanks!

EDIT:

So in this case I can use the indicator variable to write:

$$L(p;x) = \frac{p^n}{\prod_{i=1}^{n}x_i^2} I_{(x_i \geq p)}$$

for $$i = 1,2, \ldots n$$ in the indicator variable. So the "closest" non-null value of $$x \in X$$ to $$p$$ is $$min(X_1, \ldots X_n)$$. Is that the point?

• See why differentiation is not valid for other distributions where support depends on parameter: math.stackexchange.com/questions/649678/…, math.stackexchange.com/questions/2944073/…. – StubbornAtom Oct 9 '18 at 20:21
• It would be unwise to say this is the 'usual procedure' since maximum likelihood estimation is not merely about finding critical points of the likelihood function by differentiation. – StubbornAtom Oct 9 '18 at 20:28
• Indeed, the method is about maximizing the likelihood function, not about critical points. Thanks :) – M.Gonzalez Oct 10 '18 at 10:28
• I should explain: I deleted my answer. The reason why is because when $0 < p < 1$, you get a very different answer than when $p > 1$, and I didn't have time to think through the details. Are you provided any restrictions on $p$? – Clarinetist Oct 10 '18 at 12:09
• @M.Gonzalez StubbornAtom is actually correct. If you're setting the derivative equal to $0$, what you are doing is you are finding critical points of the likelihood function by differentiation. – Clarinetist Oct 10 '18 at 12:14

The usual method does not work well when the support of the random variable (in this case, $$[p, \infty)$$) depends on the parameter of interest (which is $$p$$ in this case).

In these situations, you should use indicator functions. Let $$\mathbf{I}$$ denote the indicator function, defined by $$\mathbb{I}(\cdot) = \begin{cases} 1, & \cdot \text{ is true} \\ 0, & \cdot \text{ is false.} \end{cases}$$ Thus, we may write $$f(x \mid p) = \dfrac{p}{x^2}\mathbf{I}(p \leq x)\text{.}$$

(Please read this other answer for details that I will leave unproven here, and for a similar problem to this one.)

Per the link I've put above, you can see that $$L(p \mid \mathbf{x}) = \prod_{i=1}^{n}\dfrac{p}{x_i^2}\mathbf{I}(p \leq x_i)=\dfrac{p^n}{\prod_{i=1}^{n}x_i^2}\mathbf{I}(p \leq x_{(1)})$$ where $$x_{(1)} = \min\limits_{1 \leq i \leq n}x_i$$.

Viewing this as a function of $$p$$, note that if $$p > x_{(1)}$$, then $$\mathbf{I}(p \leq x_{(1)}) = 0 = L(p \mid \mathbf{x})$$, which is obviously not the largest value of $$L$$.

Thus, assume $$p \leq x_{(1)}$$. Disregarding constants of proportionality with respect to $$p$$ (which do not affect the actual maximum likelihood estimator), we obtain $$L(p \mid \mathbf{x}) = \dfrac{p^n}{\prod_{i=1}^{n}x_i^2}\mathbf{I}(p \leq x_{(1)}) \propto p^n\text{.}$$

As long as $$p > 0$$, we know that $$p^n$$ (for $$n$$ fixed) is indeed a monotonically increasing function of $$p$$. Thus, to maximize $$p^n$$, we must seek the largest value of $$p$$. Note that to get to this point, we had to assume $$p \leq x_{(1)}$$. It follows that $$\hat{p}_{\text{MLE}} = X_{(1)}$$ is the maximum likelihood estimator of $$p$$.

• Shouldn't it be $L(p \mid \mathbf{x}) = \prod_{i=1}^{n}\dfrac{p}{x_i^2}\mathbf{I}(p \leq x_i)=\dfrac{p^n}{\prod_{i=1}^{n}x_i^2}\mathbf{I}(p \leq x_{(1)})$ ? – N74 Oct 9 '18 at 20:27
• @N74 Really terrible typo on my part, thank you! – Clarinetist Oct 10 '18 at 1:17
• @Clarinetist: thank you so much! The solution is now very clear to me :) – M.Gonzalez Oct 11 '18 at 20:06
• @M.Gonzalez If this answer was helpful for you, please consider clicking the checkmark. Glad to help! – Clarinetist Oct 12 '18 at 12:15