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Let the random variable X have a uniform density given by

$f(x;\theta)$~$R(\theta,\theta+1)$

What is the maximum likelihood function according to the samples $X_1,\ldots,X_n$?

The question is much like Likelihood Function for the Uniform Density. But there seems no satisfying answer. So far I get $X_{max}-1\le\theta\le X_{min}$ and my guess is $\hat\theta=\frac{X_{max}+X_{min}-1}{2}$. Am I right? I need a theoretical explaination.

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Hint: When you say that "So far I get $X_{max}-1\le\theta\le X_{min}$" you are actually saying that the likelihood $L(\theta ; x_1 \cdots x_n)$ (regarded as a function of $\theta$) is zero outside that range. Now, what is the likelihood if $\theta$ is inside that range?

Update: you know that the density of each uniform sample is $f(x;\theta) = 1$ if $\theta <x\le \theta+1$, $0$ otherwise.

Then the likelihood $L(\theta;x_1 \cdots x_n) =\prod f(x_i;\theta) $ is $1$ if $\theta <x_i\le \theta+1$ $\forall i$, $0$ otherwise. In terms of $\theta$ as variable, this is equivalent to say that $L(\theta;x_1 \cdots x_n)=1$ in $X_{max}-1\le\theta\le X_{min}$. You need to find the maximum of $L(\theta;x_1 \cdots x_n)$ as a function of $\theta$. But $L(\theta;x_1 \cdots x_n)$ is constant (in that range) (I recommend you to draw it if you don't quite get it). Hence, $\theta_{ML}$ is not well defined, we can pick any value in that range. In particular, $(X_{min}+X_{max})/2$ is as valid as any other.

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  • $\begingroup$ Thanks, "any value" is really beyond my thought... $\endgroup$ – Frazer Mar 22 '14 at 15:20
  • $\begingroup$ @leonbloy: I have a question. What if for $X_1,\dots,X_n$ are such that $X_{min}<X_{max}-1$ ? In this case $L(\theta) = 0$ for all $\theta$. So, what will mle in this case? $\endgroup$ – Kumara Dec 17 '18 at 11:49
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    $\begingroup$ Undefined. Anyway, that cannot happen under the assumed model $\endgroup$ – leonbloy Dec 17 '18 at 14:52
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Here a two thoughts

  • If any $X_i$ lies outside of $[\theta,\theta+1]$, then the likelyhood function is zero
  • If all the $X_i$ lie inside $[\theta_1,\theta_1+1]$ and also inside $[\theta_2,\theta_2+1]$, then the likelyhoods are the same, because $f(x,\theta_1) = f(x,\theta_2)$ for those $x$ which lie in the intersection of the two intervals
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