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Consider $X_1, X_2, \dots, X_n$ as a random sample from a distribution with the probability density function (pdf):

$$ f(x) = \begin{cases} e^{-(x - \theta)}, & \text{for } x < \theta, -\infty<\theta<\infty \\ 0, & \text{otherwise}. \end{cases}$$

If the estimator for the parameter $\theta$ is $Y_n = \min\{X_1, X_2, \dots, X_n\}$, then is the estimator $Y_n$ unbiased for $\theta$?

I know that it is unbiased if $E(Y_n)=\theta$ and I have found the $f_{Y_n}(x)=-ne^{n(\theta-x)}$. Now what I'm confuse is to find the $E(Y_n)$ what will be the lower and upper bound for the integral $\int_{-\infty}^{\infty}-nxe^{n(\theta-x)}dx$?

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    $\begingroup$ Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. $\endgroup$ Commented Apr 9 at 14:16
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    $\begingroup$ Is this a valid pdf? It does not seem to integrate to 1 (its integral is infinity?) As $x$ becomes large and negative, the pdf also increases unboundedly. $\endgroup$ Commented Apr 9 at 14:26

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It seems that there is a typo in the task; this should be $$ f(x) = \begin{cases} e^{-(x - \theta)}, & \text{for } x \color{red}{>} \theta, -\infty<\theta<\infty \\ 0, & \text{otherwise}. \end{cases} $$ A way is to find the density of $Y_n=\min\{X_1, X_2, \dots, X_n\}$, then compute the expectation.

Alternatively, notice that $X'_i:=X_i-\theta$ has an exponential distribution with parameter $1$. Letting $Y'_n=\min\{X'_1, X'_2, \dots, X'_n\}$, we have $Y_n=\theta+Y'_n$. Thus the biais is $\mathbb E[Y'_n]$, which is not zero.

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    $\begingroup$ +1. You could say that $Y_n \ge \theta$ with probability $1$ with a positive probability that $Y_n> \theta$ (in fact also $1$) so you must have $\mathbb E[Y] > \theta$, equivalent to saying $Y'_n \ge 0$ with probability $1$ with a positive probability that $Y'_n> 0$ so you must have $\mathbb E[Y'] > 0$. $\endgroup$ Commented Apr 10 at 10:48
  • $\begingroup$ Thank you. Yes, turns out the question contains a typo, that's why I was confused $\endgroup$ Commented Apr 10 at 14:11

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