Show that a given estimator is unbiased

Question

I am working on the following exercise:

Let $(X_1,\ldots,X_n)$ be a random sample from a uniform distribution $U[0,\theta]$. Show that the estimator $$A_n = \frac{n+1}{n} \max\{X_1,\ldots,X_n\}$$ is unbiased, i.e. $E(A_n) = \theta$.

I first computed the CDF of $\max\{X_1,\ldots,X_n\}$ as

$$\prod_{i=1}^{n} P(X_i \le x) = x^n$$

and so the PDF is $nx^{n-1}$. But when I try to calculate the expected value I get:

$$E(A_n) = \frac{n+1}{n} \cdot E(\max\{X_1,\ldots,X_n\} = \frac{n+1}{n} \cdot \int_{0}^{\theta}nx^{n-1} \cdot x \ dx = \theta^{n+1}.$$

Could you tell me what I am doing wrong?

Answer 1

You have the wrong cdf for the $X_i$. The $X_i$ are not uniformly distributed on $[0,1]$ but on $[0,\theta]$.

Let $X\sim\text{Unif}[0,\theta]$. Then the cdf of $X$ is given by $P(X\leq x)=x/\theta$ for $0\leq x\leq \theta$. In particular if $Y$ is the maximum of $X_1,\dotsc, X_n\stackrel{\text{iid}}{\sim}\text{Unif}[0,\theta]$, we have that the cdf of $Y$ $$ P(Y\leq y)=P(X_1\leq y,\dotsc, X_n\leq y)=\prod_{i=1}^nP(X_i\leq y)=\frac{x^n}{\theta^n}\quad (0\leq x\leq \theta) $$ where we use the independence of the $X_i$ and the identical distributed assumption. So the density of $Y$ is given by $$ f_Y(y)=n\frac{x^{n-1}}{\theta^n} \quad (0\leq x\leq \theta) $$ In particular $$ E\left(\frac{n+1}{n}Y\right)=\frac{n+1}{n}EY=\frac{n+1}{n}\int_0^\theta yn\frac{y^{n-1}}{\theta^n}\, dy =(n+1)\int_0^\theta\frac{y^n}{\theta^n}\, dy =\frac{n+1}{\theta^n}\frac{\theta^{n+1}}{n+1}=\theta. $$