# 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/…. Oct 9, 2018 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. Oct 9, 2018 at 20:28
• Indeed, the method is about maximizing the likelihood function, not about critical points. Thanks :) Oct 10, 2018 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 $p$? Oct 10, 2018 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. Oct 10, 2018 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, 2018 at 20:27
• @N74 Really terrible typo on my part, thank you! Oct 10, 2018 at 1:17
• @Clarinetist: thank you so much! The solution is now very clear to me :) Oct 11, 2018 at 20:06