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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.

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    Integrate by parts with $u = y$ and $dv = n(y- \theta)^{n-1} dy$. – 2017-02-24
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    Could you post a solution? – 2017-02-24

2 Answers 2

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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

$$E(y_{(n)}^2) - E(y_{(n)})^2 = \frac{n}{(n+1)^2} \underset{n \to \infty}{\longrightarrow} 0$$

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    I am sorry I am confused by your solution, could you put more steps in please? – 2017-02-24
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    Do you know integration by parts? $\int u \, dv = uv - \int v \, du$ If you are tackling order statistics, then I imagine you learned this in calculus. – 2017-02-24
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    Is there maybe something else wrong with the way I approach the problem? Getting the variance to prove that it goes to zero of this is a consistent estimator of theta seems a bit tedious. – 2017-02-24
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    The expected value tends to $\theta +1$ and the variance tends to $0$ as $n \to \infty.$ – 2017-02-24
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The last line of the previous answer seems incorrect

$E(y_{(n)}^2) - E(y_{(n)})^2 = \frac{n}{(n+1)^2(n+2)}$

There is a nice algebraic trick to simplifying this expression.

$E(y_{(n)})^2 = \left(\theta + \frac{n}{n+1}\right)^2 = \left((\theta + 1) + (\frac{n}{n+1} - 1)\right)^2 = \left((\theta + 1) - \left(\frac{1}{n+1}\right)\right)^2$