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Let $X_1,X_2,X_3....X_n$ be a random sample from a population having probability density function
$$f(x)=\dfrac{1}{\theta} e^\frac{-(x-\theta)}{\theta} ;x\geq\theta$$

where $\theta\in(0,\infty)$. Find the MLE of $\theta$.

In this question when i followed normal method of finding MLE(taking log likelihood differentiating with respect to $\theta$)I found $\hat\theta =\bar x$. But like in uniform distribution case we have here parameter dependent on $x$ . Now i am not sure if MLE is $\bar x$ or $x_{(1)}$.

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    $\begingroup$ Any maximum likelihood estimator cannot be greater than $x_{(1)}$ here as that would lead to a likelihood of $0$. It might be clearer if you wrote $f(x)=\dfrac{1}{\theta} e^\frac{-(x-\theta)}{\theta} I_{x \ge \theta}$ for all $x$ where $I_{x \ge \theta}$ is an indicator function $\endgroup$
    – Henry
    Feb 1, 2018 at 16:16
  • $\begingroup$ Likelihood in this $0 \leq \theta \leq x_{(1)}$ part will still give $\bar x $ i don't get it . $\endgroup$
    – Daman
    Feb 1, 2018 at 16:37
  • $\begingroup$ @Damn1o1 No, because $x_{(1)}$ is always less than or equal to $\bar x$ $\endgroup$
    – heropup
    Feb 1, 2018 at 16:40
  • $\begingroup$ @Henry After writing first comment you never comeback. $\endgroup$
    – Daman
    Feb 1, 2018 at 16:56
  • $\begingroup$ Damn1o1 - I am not permanently on the site. $\hat{\theta} = \bar x$ will have a zero likelihood because it is almost surely greater than $x_{(1)}$. @heropup gave the details $\endgroup$
    – Henry
    Feb 1, 2018 at 17:39

1 Answer 1

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Note that $$f(x) = \frac{1}{\theta} \exp\left( - \left(\tfrac{x}{\theta} - 1 \right) \right) \mathbb 1 (x \ge \theta > 0) = \frac{e}{\theta} e^{-x/\theta} \mathbb 1 (x \ge \theta > 0).$$ Hence the likelihood given a sample $\boldsymbol x$ is proportional to $$\mathcal L(\theta \mid \boldsymbol x) = \theta^{-n} e^{-n \bar x/\theta} \mathbb 1(x_{(1)} \ge \theta > 0).$$ The log-likelihood is $$\ell(\theta \mid \boldsymbol x) = \begin{cases} - n \log \theta - \frac{n \bar x}{\theta}, & 0 < \theta \le x_{(1)} \\ -\infty, & \text{otherwise.}\end{cases}$$ But since the critical point satisfying $$\frac{\partial \ell}{\partial \theta} = 0 = -\frac{n}{\theta} \left( 1 - \frac{\bar x}{\theta}\right)$$ is $\theta = \bar x$, and $$\bar x = \frac{1}{n} \sum_{i=1}^n x_i \ge \frac{1}{n} \sum_{i=1}^n x_{(1)} = x_{(1)},$$ with equality if and only if $x_i = x_{(1)}$ for all $i = 1, \ldots, n$, we conclude that the log-likelihood has no critical point in $\theta \in (0, x_{(1)})$; hence it is maximized when $\theta = x_{(1)}$ as $\partial \ell/\partial \theta < 0$ for all such $\theta$ (i.e., it is strictly decreasing).

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  • $\begingroup$ Thank you so much i get it now. $\endgroup$
    – Daman
    Feb 1, 2018 at 16:43
  • $\begingroup$ So $\bar x $ is nothing but $x_{(1)}$ because there nothing else except $x_{(1)}$ according to $\theta \in (0, x_{(1)})$. I am thinking it this way did i follow it right? $\endgroup$
    – Daman
    Feb 1, 2018 at 16:48
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    $\begingroup$ Not really. Unless all of your observations are the same (which for such a distribution happens with zero probability for a sample size greater than 1), the sample mean is always greater than the minimum order statistic, hence it cannot be chosen as the MLE. The likelihood is zero that $\theta$ is greater than the minimum order statistic, because you cannot choose the parameter to exceed a value that you observed. $\endgroup$
    – heropup
    Feb 1, 2018 at 16:50

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