Find unbiased estimator 
Let $X_i$ be observations from $U[0, \theta]$ (continuous uniform distribution). Find unbiased estimator for $\frac{1}{\theta}$

What i did is: let $\theta^* = g(X)$ be an estimator. Than, to be unbiased, the following must hold: $$\int\limits_{0}^{\theta} \theta^*(x) \frac{1}{\theta} dx =\frac{1}{\theta}\int\limits_{0}^{\theta} g(x) dx = \frac{1}{\theta}\big[G(\theta) - G(0) \big]$$
where $G' = g$, so the clue is to find some function $g$ with such properties. This is the part where i need help. 
 A: Can you find an estimator for $\theta$?  One natural example is the maximum likelihood estimator, $$\hat \theta = \max_i x_i = x_{(n)}.$$  Another example is the method of moments estimator, $$\tilde \theta = 2 \bar x,$$ that is to say, twice the sample mean.  Are these biased, or unbiased?
Then, you could choose estimators for $1/\theta$ such as $1/\hat \theta$ or $1/\tilde \theta$.  But what is the expectation of each one?  How would you go about calculating $$\operatorname{E}\left[\frac{1}{\max_i X_i}\right]?$$
One way would be to compute the PDF of the order statistic $X_{(n)}$, then define the transformed random variable $$Y = 1/X_{(n)}, \quad f_Y(y) = \frac{f_X(1/y)}{y^2}, \quad \frac{1}{\theta} \le y < \infty.$$  This gives you the PDF of the estimator $1/\hat \theta$.  Then $$\operatorname{E}\left[\frac{1}{\max_i X_i}\right] = \operatorname{E}[Y] = \int_{y = 1/\theta}^\infty y f_Y(y) \, dy.$$  If $1/\hat \theta$ were unbiased for $1/\theta$, what should be the result of this calculation?  If it is biased, how can we "adjust" this estimator to make it unbiased?
Here's a hint:  do not try this method on $1/\tilde \theta$.  The PDF of the sample mean for a uniform distribution is not a nice density to work with.
A: Fill in the steps


*

*Find an unbiased estimator of $\theta$ (consider scaled version of complete, sufficient statistic (c.s.s) for $Unif[0,\theta]$) 

*Find $f(T)$, the pdf of the c.s.s., call it $T=T(X_1,...,X_n)$ (note the support)

*Now consider $h(T)$, which is an unbiased estimator of $g(\theta)=1/\theta$, which is differentiable on $(0,\infty)$

*That is $h(T)$ must satisfy, $E[h(T)]=g(\theta)$.

*$E[h(T)]=\int_{-\infty}^{\infty}h(t)f(t)dt= g(\theta)$. Simplify and differentiate both sides. Be careful of differentiation of the integral. 

*Solve for $h(\theta)$ to get an expression for $h(T)$, the unbiased estimator of $g(\theta)=1/\theta$

