# How to check if an estimator is unbiased?

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?

• Why do you think $e^{-(x-\theta)}, \ \ 0 < x < \theta$ is a density function? – Masoud Mar 11 at 21:06
• 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. – whuber Mar 11 at 21:36
• @whuber I fixed it. – use1883 Mar 11 at 21:42
• @Masoud That is not the case whenever $\theta$ is negative. use1883: you can find the distribution of $\theta_1$ explicitly. – whuber Mar 11 at 21:45
• stats.stackexchange.com/q/406181/119261 – StubbornAtom 21 hours ago

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.

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