Say we have a shifted exponential distribution with common density $$f(x|\theta)=\left\{\begin{matrix} e^{-(x-\theta)} & x\geq \theta\\ 0 & x<\theta \end{matrix}\right.$$

We have $\theta$ a real number as the unknown shift parameter and $\textbf{X}=(X_1,...,X_n)$ a random sample. If $X_{(1)}=min \left \{ X_1,...,X_n \right \}$, then the density $f_{(1)}(x)=ne^{-n(x-\theta)}$, where $x\geq \theta$.

I have tried to determine if this estimator $\hat{\theta}=X_{(1)}$ is unbiased or not. I got that $E[\hat{\theta}]=e^{-n\theta}(\theta+\frac{1}{n})$ which is not equal to $\theta$ so there is bias, but I am not sure if I have done this correctly.

If this estimator isn't biased, how does one then determine an unbiased estimator by making an adjustment to the estimator $\hat{\theta}=X_{(1)}$.


1 Answer 1


Defining $Y_{i}=X_{i}-\theta$ random variable $Y_{i}$ has standard (i.e. $\lambda=1$) exponential distribution.

Then $\min\left\{ Y_{1},\dots,Y_{n}\right\} $ has exponential distribution with parameter $\lambda=n$.

Based on: $$X_{\left(1\right)}=\min\left\{ X_{1},\dots,X_{n}\right\} =\min\left\{ \theta+Y_{1},\dots,\theta+Y_{n}\right\} =\theta+\min\left\{ Y_{1},\dots,Y_{n}\right\} $$

we find:

$$\mathbb{E}X_{\left(1\right)}=\theta+\mathbb{E}\min\left\{ Y_{1},\dots,Y_{n}\right\} =\theta+\frac{1}{n}$$

So apparantly $X_{(1)}-\frac1{n}$ will serve as unbiased estimator of $\theta$.

  • $\begingroup$ I follow but I'm trying to understand why when I worked out $E[X_{(1)}]$ from the pdf, I got $e^{-n\theta}(\theta+\frac{1}{n})$, whilst you got $\theta+\frac{1}{n}$, which makes sense. Can you explain this to me if I went wrong in my calculation? $\endgroup$
    – dsakiocxla
    May 26, 2020 at 17:07
  • 1
    $\begingroup$ Your PDF is correct. Finding expectation comes to finding integral $\int_{\theta}xne^{-nx+n\theta}dx$. Apply substitution $x=\frac1{n}u+\theta$. As long as I do not see your calculation I cannot tell what's wrong with it. $\endgroup$
    – drhab
    May 27, 2020 at 9:18

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