Find unbiased for $\theta$ given a uniform distribution on the interval $[0, \theta]$. Let $Y_{(1)} = \min(Y_1,\ldots, y_n)$ Show that $\hat{\theta_1} = (n+1) ~y_{(1)}$ is unbiased for $\theta$.
So far what I got is 
$f(y) = 1/\theta$ for , $0 \le y \le \theta$
So $F(y) = P[Y \le y] = \displaystyle \int_\theta^y 1/\theta dy = y/\theta |_\theta^y = 1 - y/\theta$.
Then $g_{(1)} (y) = n [ -y/\theta]^{(n-1)} (y/\theta)$ 
Are these right? If so, how to go from here? If so, could you show your approach?
Thanks
 A: Let $W$ be the minimum. The minimum is $\gt w$ iff all the $Y_i$ are $\gt w$.
The probability a particular $Y_i$ is $\gt w$ is, for $0\lt w\lt \theta$, equal to $1-w/\theta$. Thus the probability all the $Y_i$ are $\gt w$ is $(1-w/\theta)^n$, and therefore 
$$F_W(w)=1-(1-w/\theta)^n,$$
(for $0\lt w\lt \theta$).
Thus the minimum has density function $(1/\theta)n(1-w/\theta)^{n-1}$ in our interval. The mean of $W$ is therefore
$$\int_0^{\theta} w(1/\theta)n(1-w/\theta)^{n-1}\,dw.$$
Integrate. I suggest making the substitution $x=1-w/\theta$. Integration by parts will also work well. 
If you know a formula for expectation in terms of an integral that involves the cumulative distribution function, use that, it will be somewhat simpler. 
Added: Let $1-w/\theta=x$. Then $-dw/\theta=dx$, so $dw=-\theta \,dx$. Also, $w=(1-x)\theta$. Substitute. We get
$$\int_{x=1}^0 (1-x)\theta(1/\theta)nx^{n-1}(-\theta)\,dx.$$
Simplify a bit. We end up with
$$\int_0^1 (\theta n)(x^{n-1}-x^n)\,dx.$$
Integrate. The integral of $x^{n-1}-x^n$ from $0$ to $1$ is $\frac{1}{n}-\frac{1}{n+1}$, which is $\frac{1}{n(n+1)}$. Multiply by $n\theta$. we get $\frac{\theta}{n+1}$.
So the expectation of $W$ is $\frac{\theta}{n+1}$. Thus the expectation of our estimator $(n+1)W$ is $\theta$. It follows that $(n+1)W$ is an unbiased estimator of $\theta$. 
A: Actually you are close.
The cdf of $Y_{(1)} =\min(y_1, …, y_n)$ looks like: 
$$F_Y(y)=1-(1-F_{y(1)}(y))^n=1-\left(1-\frac{y}{\theta}\right)^n,$$ then the pdf $$w_Y(y)=\frac{n}{\theta }\left(1-\frac{y}{\theta }\right)^{n-1}.$$
So $$\mathbb{E}\{\hat{\theta}\}=(n+1)\int_0^\theta y \  w_Y(y)\mathrm dy=(n+1)\int_0^\theta y \  \frac{n}{\theta }\left(1-\frac{y}{\theta }\right)^{n-1}\mathrm dy=\theta$$
And since the mean value of your estimator ($\hat{\theta}$) equals to $\theta$ it is unbiased.
