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Given random sample $X_1, X_2, ..., X_n$ with the distribution function $$f(x|\theta) = \left \{ \begin{aligned} e^{-(x-\theta)}, \ \ \theta < x < \infty \\ 0, \text{ otherwise.} \end{aligned} \right. $$

where $\theta \in (-\infty, \infty).$ Show that the estimator $\theta_1 = \min\left\{X_1, X_2, ..., X_n \right\} $ is unbiased. I only got as far as defining $\theta_1 = \mathrm{function}(sample) \ \ \min(sample)$, what should I do next?

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    $\begingroup$ Why do you think $e^{-(x-\theta)}, \ \ 0 < x < \theta $ is a density function? $\endgroup$ – Masoud Mar 11 at 21:06
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    $\begingroup$ I don't see how $\theta_1$ could possibly be unbiased, because when $\theta \gt 0$ it is guaranteed to be less than $\theta$! Also, it makes no sense to write "$0\lt x\lt \theta$" when $\theta$ is negative. You must have a typo (or several) somewhere in your question--please fix it. $\endgroup$ – whuber Mar 11 at 21:36
  • $\begingroup$ @whuber I fixed it. $\endgroup$ – use1883 Mar 11 at 21:42
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    $\begingroup$ @Masoud That is not the case whenever $\theta$ is negative. use1883: you can find the distribution of $\theta_1$ explicitly. $\endgroup$ – whuber Mar 11 at 21:45
  • $\begingroup$ stats.stackexchange.com/q/406181/119261 $\endgroup$ – StubbornAtom 21 hours ago
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I assume $X_i$ are independent. So

$Y=\min (X_1,\cdots X_n) $

$$F_Y(y)=1-P(Y>y)=1-P(X_1>y,\cdots, X_n>y)=1-e^{-n(y-\theta)}$$ so

$$f_Y(y)=ne^{-n(y-\theta)} \hspace{1cm} \theta < y$$

so $E(Y)=\theta + \frac{1}{n}$.

That is, $\min (X_1,\cdots X_n)$ is biased.

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    $\begingroup$ Could you explain why you assume $\theta$ is positive? $\endgroup$ – whuber Mar 12 at 2:30
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    $\begingroup$ I think this is right also if $\theta\in (-\infty , \infty)$ $\endgroup$ – Masoud Mar 12 at 6:39
  • $\begingroup$ $X-\theta$ is exponential distribution. so $\theta \in R$ $\endgroup$ – Masoud Mar 12 at 7:34
  • $\begingroup$ @masoud How did you arrived at expected value for Y? Also in this case Y is minimum of each sample, but what is y? $\endgroup$ – use1883 Mar 12 at 18:57
  • $\begingroup$ $Y-\theta$ has exponential distribution,$Exponential(n)$, so$E(Y-\theta)=\frac{1}{n}$. $\endgroup$ – Masoud Mar 12 at 19:45

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