The distribution is uniform with bounds $\theta$ to $\theta + 1$
I am able to get the following as its pdf
$$n(y-\theta)^{n-1}$$
I know if I want to find the expected value I need to integrate it
$$\int ny(y-\theta)^{n-1}$$
but I don't know where to go from here. How do I integrate this integral?
I am hoping that once I can do that I will also be able to handle finding the second moment as well.
Best Answer
For the first question regarding integration, let $u = y$ and $dv = n(y- \theta)^{n-1}\, dy.$ It follows that $du = dy$ and $v = (y - \theta)^n$, and integrating by parts we obtain
$$\int ny(y - \theta)^{n-1} \, dy = y(y- \theta)^n - \int(y - \theta)^n \, dy \\ = y(y- \theta)^n - \frac{(y - \theta)^{n+1}}{n+1} + C.$$
Using this anti-derivative, the expected value of the nth order statistic is
$$\begin{align}E(y_{(n)}) &= \int_\theta^{\theta+1} ny (y- \theta)^{n-1} \, dy \\ &= \left. \left[y(y- \theta)^n - \frac{(y - \theta)^{n+1}}{n+1}\right] \right|_\theta^{\theta + 1} \\ &= \theta + \frac{n}{n+1}\end{align} $$
The second moment is
$$\begin{align} E(y_{(n)}^2) &= \int_\theta^{\theta+1} ny^2 (y- \theta)^{n-1} \, dy \end{align}.$$
A first integration by parts yields
$$\begin{align} E(y_{(n)}^2) &= \left.y^2(y- \theta)^n\right|_\theta^{\theta+1} - \int_\theta^{\theta+1} 2y (y- \theta)^{n} \, dy \\ &= (\theta+1)^2 - \int_\theta^{\theta+1} 2y (y- \theta)^{n} \, dy \end{align}$$
A second integration by parts yields
$$\begin{align} E(y_{(n)}^2) &= (\theta+1)^2 - \left. \frac{2y(y - \theta)^{n+1}}{n+1} \right|_\theta^{\theta + 1} + \frac{2}{n+1}\int_\theta^{\theta+1} (y- \theta)^{n+1} \, dy \\ &= (\theta+1)^2 - \frac{2(\theta+1)}{n+1} + \frac{2}{(n+1)(n+2)}\end{align}$$
The variance of the nth order statistic is then
$$E(y_{(n)}^2) - E(y_{(n)})^2 = \frac{n}{(n+1)^2} \underset{n \to \infty}{\longrightarrow} 0$$