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

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.

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 $E(X^2)-(E(X))^2$.
• For $E(X^2)$, we are integrating $\frac{n}{\theta^n}x^{n+1}$. When I do the calculation I get $\frac{n}{n+2}\theta^2$. And the expectation of $X$ was $\frac{n}{n+1}\theta$, which is why multiplying by $\frac{n+1}{n}$ gives us an unbiased estimator of $\theta$. Now the rest of the calculation of the variance of $Y$ is plain algebra. – André Nicolas Sep 23 '15 at 1:26