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Let $X_1, X_2, \ldots , X_n$ be a random sample from a distribution with the following pdf

$f(x|\theta) = \begin{cases} 1/(\theta_2−\theta_1), &\quad\text {for}\quad \theta_1 \leq x\leq \theta_2\\ 0 &\quad\text { otherwise}\quad \end{cases}$ Suppose that $\theta_1$ and $\theta_2$ are unknown.

How would I go about writing the likelihood function for this distribution on $\theta_1$ and $\theta_2$.

2 Answers 2

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The easiest way might be to begin by writing the density as it should be written, that is, as $ f(x\mid\theta)=\frac{\mathbf 1_{\theta_1\leqslant x\leqslant\theta_2}}{\theta_2-\theta_1}, $ where $\theta=(\theta_1,\theta_2)$ with $\theta_1\lt\theta_2$. Then the likelihood of an i.i.d. sample $\mathbf x=(x_1,\ldots,x_n)$ is $ f(\mathbf x\mid\theta)=\prod_{k=1}^nf(x_k\mid\theta)=\frac{\mathbf 1_{\theta_1\leqslant m_n(\mathbf x),s_n(\mathbf x)\leqslant\theta_2}}{(\theta_2-\theta_1)^n}, $ where $ m_n(\mathbf x)=\min\{x_k\mid 1\leqslant k\leqslant n\}, \qquad s_n(\mathbf x)=\max\{x_k\mid 1\leqslant k\leqslant n\}. $ For every fixed $\mathbf x$, $f(\mathbf x\mid\theta)$ is maximal when $\theta_2-\theta_1$ is as small as possible hence the MLE for $\theta=(\theta_1,\theta_2)$ based on $\mathbf x$ is $ \widehat\theta(\mathbf x)=(m_n(\mathbf x),s_n(\mathbf x)). $

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    I'm confused about the following: "for every fixed $ \theta_2 $, the same value of $ \theta_1 $ yields the maximum (restricted) likelihood."2012-11-10
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$ L(\theta_1,\theta_2) = \begin{cases} \frac{1}{(\theta_2-\theta_1)^n} & \text{for }\theta_2\ge\max\text{ and }\theta_1 \le\min \\[10pt] 0 & \text{for other values of }(\theta_1,\theta_2) \end{cases} $ where $\max=\max\{X_1,\ldots,X_n\}$ and $\min=\min\{X_1,\ldots,X_n\}$.

Draw the picture in the $(\theta_1,\theta_2)$ plane.

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    If \min<\max, then the MLEs for $\theta_1$ and $\theta_2$ are different.2012-11-10