# Uniform distribution unbiased estimator

Let $x_i$ be independent and identically distributed observations in a sample from a uniform distribution over $[0,θ]$. Now I need to estimate $θ$ based on $N$ observations and I want the estimator to be unbiased.

I thought about simple estimator $\hatθ =\min(x_i)$.

Based on simulation it is not biased, yet I couldn't show it analytically.

Could anyone, please, show how can I get it unbiased?

• The minimum is an unbiased estimator of a parameter which is greater than the maximum with full probability? I do not think so.
– Did
Commented Apr 10, 2014 at 8:36
• Note that taking the minium is not very wise, as this discredits all other observations of being possible obervations from the distribution. In other words: $\theta$ must be at least as high as the maximum of your observations.
– Marc
Commented Apr 10, 2014 at 8:40
• According to the article about the German tank problem, the minimum-variance unbiased estimator is given by $\hat\theta=m(1+k^{-1})-1$ if the distribution is discrete and $\hat\theta=m(1+k^{-1})$ if the distribution is continuous, where $m$ is the sample maximum and $k$ is the sample size. The sampling is done without replacement. Commented Apr 10, 2014 at 9:40
• math.stackexchange.com/q/60497/321264 Commented May 10, 2020 at 8:03

I propose the $\hat{\theta}_N=\frac{N+1}{N}\max(X_1,\ldots,X_N)$. Indeed, let $Y=\max(X_1,\ldots,X_N)$ then $$\Bbb{P}(Y\leq t)=\left(\frac{t}{\theta}\right)^N,$$ Hence, the density of $Y$ is given by $f_Y(t)=\frac{N}{\theta^N}t^{N-1}{\bf 1}_{[0,\theta]}(t)$ and $$\Bbb{E}(\hat\theta_N)=\frac{N+1}{N}\Bbb{E}(Y)=\frac{N+1}{N}\frac{N}{\theta^N}\int_0^\theta t^Ndt=\theta.$$ and we are done.