# Determining an unbiased estimator

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 $$\hat{\theta}=X_{(1)}$$.

• May 26, 2020 at 16:39

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 $$\theta$$.

• 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? May 26, 2020 at 17:07
• 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. May 27, 2020 at 9:18