Variance of unbiased estimator of a random sample from the uniform distribution?

Question

Let X1,...,Xn be a random sample from the uniform distribution on the interval from 0 to theta for some theta>0. I want to find the variance of the unbiased estimator. I know the unbiased estimator is Y= [(n+1)/n]*X(n) where X(n)=Max(X1,...X) but am lost on how to find the variance of Y. Thanks so much.

Answer 1

The variance of $Y$ is $\left(\frac{n+1}{n}\right)^2$ times the variance of $X$. Now we will find the variance of $X$.

Recall that $X$ has cdf $\left(\frac{x}{\theta}\right)^n$ on the interval $(0,\theta)$. So $X$ has density function $\frac{n}{\theta^n}x^{n-1}$.

Now we can compute $E(X^2)$ by using $$E(X^2)=\int_0^\theta x^2\cdot \frac{n}{\theta^n}x^{n-1}\,dx,$$ The integration is easy.

We can also compute $E(X)$ in a similar way (but you already know it).

Finally, the variance of $X$ is $E(X^2)-(E(X))^2$.