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Given is a random sample $X_1, .., X_n$ drawn from a distribution with the pdf $$ f(x; \theta) = \left\{ \begin{array}{ll} \dfrac{1}{4} e^{-\dfrac{1}{4}(x-\theta)} & \theta < x \\ 0 & \text{otherwise} \\ \end{array} \right. $$

with an unknown paramter $\theta$ $(0 < \theta)$. Find the maximum likelihood estimator (MLE) and determine whether or not it's unbiased.

So for the MLE I found $X_{1:N}$, since the derivative of the likelihood function is $\dfrac{n}{4}$ with $n>0$, which is always positive so the likelihood function is increasing, which means we should choose the maximum value of $\theta$ possible to maximise the function. Our restriction is $\theta < x$, so the maximum we can choose is $X_{1:n}$.

However, for determining its unbiasedness I'm not completely sure. We want to show I guess that $E[X_{1:n}] = \theta$, but I don't really know what to do with the expression $E[X_{1:n}]$ to be able to write it in an appropriate form including $\theta$. How should this be done?

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It should be intuitively obvious that such an estimator is necessarily biased, because it can never be smaller than the true value of $\theta$. If it were, then you would observe $$\hat\theta_{\text{MLE}} = X_{1:n} = \min_i X_i \le \theta,$$ which is absurd. So if there is a nonzero probability that the MLE is greater than $\theta$ (which of course is the case), it must be biased since $\Pr[\hat \theta_{\text{MLE}} < \theta] = 0.$

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    $\begingroup$ Why is $\theta \geq \min X_i$ absurd, exactly? I'm missing something here. $\endgroup$ May 17, 2017 at 20:09
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    $\begingroup$ Your PDF stipulates that $\Pr[X < \theta] = 0$, does it not? That implies no realizations you draw from such a distribution can ever be less than $\theta$; consequently the minimum order statistic also cannot be less than $\theta$, and your MLE is also never less than $\theta$. So how can the MLE be unbiased; i.e., $\operatorname{E}[\hat\theta_{\text{MLE}}] = \theta$ if it is greater than $\theta$ with positive probability but never less than it? $\endgroup$
    – heropup
    May 17, 2017 at 20:16
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    $\begingroup$ This is true, I used this property to obtain the MLE. But now I'm confused because the restriction is $\theta < x$, not $\theta < X_i$ for all $i$, which I read it as... $\endgroup$ May 17, 2017 at 20:20
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    $\begingroup$ $X_i$ is simply the $i^{\rm th}$ observation in your sample. It too follows the same distribution as $X$. The point is, no observations in your sample can be smaller than $\theta$. So your MLE, which relies on picking the smallest observation in your sample, is always bigger than the parameter $\theta$, therefore its expectation cannot be equal to $\theta$. $\endgroup$
    – heropup
    May 17, 2017 at 20:27
  • $\begingroup$ You're right, the notation confused me again. I get it now. $\endgroup$ May 17, 2017 at 20:32

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