Unbiased Estimator for a Uniform Variable Support Let $ x_i $ be iid observations in a sample from a uniform distribution over $ \left[ 0, \theta \right] $.  Now I need to estimate $ \theta $ based on $N$ observations and I want the estimator to be unbiased.
I thought about simple estimator $ \hat{\theta} = \max \left( x_i \right) $.
Based on simulation it is not biased, yet I couldn't show it analytically.
Could anyone, please, show it is unbiased?
BTW, I could easily find another, easy to prove, unbiased estimator, $ \hat{\theta} = 2 \mathrm{mean} \left( {x}_{i} \right) $
 A: To derive Didier's result, observe that the cumulative distribution function for the maximum $m$ is given by the ratio of the volume of $[0,m]^N$ to the volume of $[0,\theta]^N$, which is $(m/\theta)^N$. So the density is the derivative of that, and we can calculate the expectation value of $m$ as
$$
\begin{eqnarray}
\int_0^\theta m\frac{\mathrm d}{\mathrm dm}\left(\frac m\theta\right)^N\mathrm dm
&=&
\frac N{\theta^N}\int_0^\theta m^N\mathrm dm
\\
&=&
\frac N{N+1}\theta\;,
\end{eqnarray}
$$
so we get an unbiased estimator for $\theta$ by multiplying by $(N+1)/N$.
A: It may be worth noticing a couple of things about the other unbiased estimator mentioned in the question: $2\times\text{the sample mean}$.
(1) If one finds the conditional expected value of $2\times\text{the sample mean}$ given the sample maximum, one gets $(N+1)/N$ times the sample maximum.
(2) In some cases, $2\times\text{the sample mean}$ is actually smaller than the sample maximum.  Thus although it is on average the right amount, there are instances where the data themselves tell you that it's nowhere near the right amount.
A: Of course $\hat\theta=\max\{x_i\}$ is biased, simulations or not, since $\hat\theta<\theta$ with full probability hence one can be sure that, for every $\theta$, $E_\theta(\hat\theta)<\theta$.
One usually rather considers $\hat\theta_N=\frac{N+1}N\cdot\max\{x_i\}$, then $E_\theta(\hat\theta_N)=\theta$ for every $\theta$.
To show that $\hat\theta_N$ is unbiased, one must compute the expectation of $M_N=\max\{x_i\}$, and the simplest way to do that might be to note that for every $x$ in $(0,\theta)$, $P_\theta(M_N\le x)=(x/\theta)^N$, hence
$$
E_\theta(M_N)=\int\limits_0^{\theta}P_\theta(M_N\ge x)\text{d}x=\int\limits_0^{\theta}(1-(x/\theta)^N)\text{d}x=\theta-\theta/(N+1)=\theta N/(N+1).
$$
Edit (Below is a remark due to @cardinal, which completes nicely this post.)
It may be worth noting that the maximum $M_N=\max\limits_{1\le k\le N}X_k$ of an i.i.d. Uniform$(0,θ)$ random sample is a sufficient statistic for $θ$ and that it is one of the few statistics for distributions outside the exponential family which is also complete. 
An immediate consequence is that $\hat\theta_N=(N+1)M_N/N$ is the uniformly minimum variance unbiased estimator (UMVUE) for $θ$, that is, that any other unbiased estimator for $θ$ is a worse estimator in the $L^2$ sense. 
A: Another way to look at the result derived by Joriki:
It is known that if you order $N$ uniform observations in a sample from a given interval, then the resulting partition of the interval gives a point sampled uniformly on the $(N+1)$-simplex.
In particular, the remaining difference $\theta-\hat\theta$ has the same distribution of any component of such a random point. In particular, it is clear that it has expectancy $\theta/(N+1)$ (since by symmetry, all components have the same expectation).
All in all:
$$\mathbb{E}[\theta-\hat\theta] = \frac{\theta}{N+1},$$
and indeed
$$\mathbb{E}[\hat\theta] = \frac{N}{N+1}\theta.$$
Edit:
If the minimum of the interval is unknown as well (say $[a,b]$), then by calling $\hat a$ the minimum of the sample and $\hat b$ the maximum, the same reasoning shows that 
$$\mathbb{E}[b-\hat b]=\frac{b-a}{N+1},$$
$$\mathbb{E}[\hat b]=\frac{N}{N+1}b+\frac{1}{N+1}a.$$
And so to have an unbiased estimator of the maximum $b$, one could use for example $$\frac{N\hat b-\hat a}{N-1}.$$
