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Let X have an uniform distribution on the segment [a,b]. With the method of maximum likelihood, estimate the parameter d = b-a. Then check if the estimator is consistent.

Hi, i tried solving this problem. I don't know how to continue. Any help will be welcomed :) Thanks in advance

Starting form:

$$f_x(x)= \begin{cases} \frac{1}{\theta} , x \in [a,b] \\ 0, otherwise \end{cases}$$

$$L(\theta) = (\frac{1}{\theta}) ^n$$ $$\frac{\partial{\ln(L(\theta))}}{\partial{\theta}} = \frac{\partial\ln(\frac{1}{\theta}) ^n}{\theta}=0$$ after solving this i have $$\frac{-n}{\theta}=0$$

Angara
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  • Differentiation is just wrong. If you have a sample of $n$ observations, see https://math.stackexchange.com/q/233778/321264. – StubbornAtom May 15 '20 at 21:23
  • Very similar to this post and linked ones as well – EditPiAf May 16 '20 at 17:58

1 Answers1

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This is a case in which you have to find the maximum of the likelihood curve by logic, not by differentiation. (This often happens when the parameter of interest determines the interval of support of the distribution.)

Recall that the likelihood function is the density function viewed as a function of the parameter (here $d)$ for an observed value of the random variable (here $X = 25).$

The graph is made using R statistical software, where dunif is a uniform density function.

x = 25;  a = 20;  d = seq(.01, 40, by=0.01)
like = dunif(x, a, a+d)
plot(d, like, type="l", lwd=2)
abline(h = 0, col="green2")

enter image description here

Thus the MLE of $d$ is $X - a = 25 - 20 = 5.$

Note: The height of the likelihood function is maximized by making the interval of support of the distribution as short as possible.

BruceET
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