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Q) Suppose that X1, · · · , Xn form a random sample from a uniform distribution on the interval (θ, θ + 1).where the value of the parameter θ is unknown (−∞ < θ < ∞). Clearly, the density function is f(x|θ) = (1) , for θ ≤ x ≤ θ + 1

My attempt: I never faced a similar question before in all the previous examples that I did I took the product of the pdf/pmf for all Xs then differentiated and equated to 0 but in this question such approach doesn't work.I am thinking maybe I have to manipulate the domain for x.

Thanks in advance for any help

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  • $\begingroup$ +1 for describing the trouble with the differentiability approach. $\endgroup$ – Did Mar 29 '17 at 8:19
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Let's look at an example. I have chosen a value of $\theta$, which I will not reveal to you. Then I have generated $n = 6$ IID observations $\boldsymbol x = (x_1, \ldots, x_6)$ from a $\operatorname{Uniform}(\theta,\theta+1)$ distribution. The values I have observed are: $$\boldsymbol x = \{3.57347, 4.06865, 4.22543, 3.68704, 3.73285, 3.85223 \}.$$ Now, given that data, what can you say about $\theta$? You know that it cannot possibly be larger than the smallest observation in the sample, nor can it be smaller than the largest observation in the sample minus one: i.e., you are constrained to choose $$3.22543 = x_{(n)} - 1 \le \theta \le x_{(1)} = 3.57347.$$ Conversely, any value of $\theta$ in this range is admissible, and moreover, because the likelihood is constant for any $\theta$ in this interval, it follows that any choice of $\hat \theta$ in this interval is a MLE for $\theta$.

Specifically, the likelihood should be properly written as $$\mathcal L(\theta \mid \boldsymbol x) = \mathbb 1(\theta \le x_{(1)} \le x_{(n)} \le \theta+1).$$ Note this likelihood is a function of the sufficient statistic $$\boldsymbol (\boldsymbol x) = (x_{(1)}, x_{(n)}),$$ which comprises the minimum and maximum order statistics of the sample. Then the likelihood is maximized for any $\hat \theta$ for which the indicator function equals $1$, namely whenever $x_{(n)} - 1 \le \hat \theta \le x_{(1)}.$

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  • $\begingroup$ Thank you very much but I have one more question.Assuming I chose theta=Max(X)-1. How can I check if it's unbiased ? $\endgroup$ – Basil Bassam Mar 28 '17 at 23:49
  • $\begingroup$ @BasilBassam You would calculate the expectation of the maximum order statistic $\operatorname{E}[X_{(n)}]$. First, compute the density of $X_{(n)}$, then do the integration. You can observe, for instance, that $$\Pr[X_{(n)} \le x] = \prod_{i=1}^n \Pr[X_i \le x] = \ldots.$$ $\endgroup$ – heropup Mar 28 '17 at 23:55
  • $\begingroup$ I am sorry but I have last one question, How to calculate the density of X(n)? $\endgroup$ – Basil Bassam Mar 28 '17 at 23:59
  • $\begingroup$ @BasilBassam What is the CDF of a single observation? It is $$\Pr[X_i \le x] = \begin{cases} 0, & x < \theta \\ x-\theta, & \theta \le x \le \theta+1 \\ 1, & \theta + 1 < x. \end{cases}$$ Now use my previous comment to compute the CDF of $X_{(n)}$; i.e., $F_{X_{(n)}}(x) = \Pr[X_{(n)} \le x]$. Then differentiate to obtain the density of $X_{(n)}$. $\endgroup$ – heropup Mar 29 '17 at 1:32
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Letting $\mathcal{L}(x;\theta)$ be the likelihood function and $x=(x_i)_{1\leq i \leq n}$ your sample, note that:

$$ \mathcal{L}(x;\theta) = \prod_{i=1}^n{1_{\{\theta \leq x_i \leq \theta +1\}}} $$

This function take only 2 values, $0$ or $1$. For it to be $1$, and hence maximized, one possibility is that all $(x_i)_i$ must be at most within $1$ of their minimal value. Hence, letting $\text{ML}(x) = \{\theta^{\text{ML}}: \theta^{\text{ML}} = \text{argmax}_{\theta} \, \mathcal{L}(x;\theta)\}$ be the set of all maximum likelihood solutions to the problem $-$ see @heropup 's answer:

$$ \hat{\theta}^{\text{ML}} = \min_ix_i \, \in \, \text{ML}(x)$$

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  • $\begingroup$ The MLE for this parametric model is not unique; any choice satisfying $$x_{(n)} - 1 \le \hat \theta \le x_{(1)}$$ yields $\mathcal L (\hat \theta \mid \boldsymbol x) = 1$, therefore it cannot be said that your choice $\hat \theta = x_{(1)}$ is "the" MLE. $\endgroup$ – heropup Mar 28 '17 at 23:30
  • $\begingroup$ Fair enough @heropup, I just thought this was the most "intuitive" solution. I will make some clarifications. $\endgroup$ – Morris Fletcher Mar 29 '17 at 8:11

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