1

My attempt: We observe that since $X_i\sim$ Uni$[\theta,0]$, then $X_i=-\theta U_i$ where $U_i\sim$ Uni$[-1,0]$, with $U_i$ independent. We estimate $\theta$ considering the minimum of the random sample. Since $U_i$ are i.i.d, we observe that for $-1\le x \le 0$, $P(U_{(1)}\leq x)=1-P(U_{(1)}>x)=1-\prod\limits_{i=1}^{n} (1-P(U_{i}\leq x))=1-\prod\limits_{i=1}^{n} (1-(x+1))=1-(-x)^n$, where $U_{(1)}=\min\{U_{i}: i=1,2,..,N\}$. Thus, the density of $U_{(1)}$ in $[-1,0]$ will be given by: $f_{U_{(1)}}(x)=n(-x)^{n-1}$. We thus obtain, $\mathbb{E}(X_{1})=-\theta\mathbb{E}(U_{(1)})=-\theta \int\limits_{-1}^{0}xf_{U_{(1)}}(x)dx=-n\theta(-1)^{n-1} \int\limits_{-1}^{0} x^n dx=\frac{-n\theta(-1)^{n-1}}{n+1}(-1)^{n+1}=\frac{-n\theta}{n+1}$. Thus, $Z=-\frac{n+1}{n}X_{(1)}$ is an unbiased estimator. But the answer to the problem indicates that $-Z$ is unbiased. Any comments on where I might have a mistake ? Thank you in advance.

GaloisTH
  • 137
  • note that $Z$ is always positive since the smallest of the $X_i$ is always going be negative, so $Z$ is a poor estimator for $\theta$ here since $\theta <0$ – WeakLearner Nov 06 '22 at 15:41
  • I understand that, but any ideas on what is wrong in the above solution ? – GaloisTH Nov 06 '22 at 16:37

1 Answers1

1

For $n >-1$: \begin{align*} \mathbb{E}(X_{(1)}) = -\theta \int_{-1}^0 nx(-x)^{n-1}dx = n \theta \int_{-1}^0 (-x)(-x)^{n-1} dx = n \theta \int_{-1}^0 (-x)^{n} dx = n \theta \frac{1}{n+1} \end{align*}

WeakLearner
  • 5,982