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The question arose what to do if the density of a random variable depends on $(\alpha-x)$ (maybe in some degree) on the interval from 0 to $\alpha$, for example. The likelihood function then turns out to depend on the product of $(\alpha-x)^n$. And when the MLE estimator is set to $\hat\alpha=max(X_i)$, it turns out that the likelihood function is reset to zero. What about the MLE estimator in such cases? For example, if r.v. with density $f(x)=\left ( \frac{2\left ( \alpha -x \right )}{\alpha^2}\right ) if x\in \left [ 0;\alpha \right ]$

or the second example in this message The Doctor (https://math.stackexchange.com/users/165662/the-doctor), Existence of Maximum Likelihood Estimator, URL (version: 2017-10-08): Existence of Maximum Likelihood Estimator

When substituting such an estimator $\hat{\mu}=X_{(1)}$, the likelihood function is equal to 0 too - one of the multipliers, where the minimum member of the sample comes across, resets it. What about this estimation then?

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In your example, the lileihood is

$$L(X;a)=\frac{2^n}{a^{2n}} \prod_{i=1}^n (a-x_i) [0\le x_i \le a]$$

In terms of the variable $a$, this gives the domain $a \in [\max(x_i), +\infty)$. Inside this domain, $L(X;a)$ is differentiable. Hence, its global maximum is either a critical point inside the domain, or either lies in the extremes. But the latter is not possible, because $L(X; \max(x_i))=0$. Hence we can seek for critical points.

Using the log likelihood, deriving and equating to zero we get the following equation for the critical point(s):

$$ a =\frac{2n}{\sum \frac{1}{a-x_i}} =2\,{\rm Hm}(a-x_i) $$

where $Hm$ denotes the harmonic mean.

This should be solved numerically (inside the domain).

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  • $\begingroup$ thank you very much! But I don't understand how to solve it if a is both on the left side and on the right under the sum in the denominator $\endgroup$
    – Yuliya
    Oct 23, 2022 at 16:54
  • $\begingroup$ You can't solve that analytically. You must resort for some iterative procedure. For example, here's an iterative calculation, for $n=3$, $x=(10,20,30)$, it gives $a=45.51$ docs.google.com/spreadsheets/d/… $\endgroup$
    – leonbloy
    Oct 23, 2022 at 17:35
  • $\begingroup$ Oh, I see. Thanks again! And with the minimum in the second example, about the same story? $\endgroup$
    – Yuliya
    Oct 23, 2022 at 17:45
  • $\begingroup$ and I wonder why the MLE estimate is so shifted towards the maximum of the sample... Exceeds 1.5 times. Well, maybe because the sample is so small... $\endgroup$
    – Yuliya
    Oct 23, 2022 at 18:01

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