# Statistics - Show that $\hat{\theta}$ hat is a biased estimator of $\theta$

I'm asked to solve this exercise, but I can't manage to find something satisfying. Any help/hint would be much appreciated.

Let $Y_1, Y_2,\dots, Y_n$ denote a random variable sample of size n from a population whose density is given by :

$$f(y)=\begin{cases}\frac{\alpha y^{\alpha - 1}}{\Theta^\alpha} & 0\leq y \leq \Theta \\ 0 & y \not\in[0,\Theta]\end{cases}$$

Where $\alpha > 0$ is a known fixed value but $\theta$ is unknown.

We consider $\hat{\theta} = \max(Y_1,Y_2,...,Y_n)$. How can one show that $\hat{\theta}$ is a biased estimator for $\theta$.

My try was to compute the distribution function $F(\hat{\theta})$ in order to calculate the probability density function and then simply prove that $E(\hat{\theta}) \neq \theta$ but it didn't get me anywhere.

I suppose the sample is iid. You can find the distribution of $F_{\hat\theta}(t)$ by noticing

$$F_{\hat\theta}(t)=P(\hat\theta\leq t)=P(Y\leq t)^n$$

Thus you can find $f_{\hat\theta}$, and then its expected value

$$E(\hat\theta)=\int_{[0,\Theta]} t f_{\hat\theta}(t)\;\mathrm{d}t$$

Here you have

$$P(Y\leq t)=\int_0^t f_Y(u)\;\mathrm{d}u=\int_0^t \frac{\alpha u^{\alpha-1}}{\Theta^\alpha}\;\mathrm{d}u=\frac{t^\alpha}{\Theta^\alpha}$$

Hence

$$F_{\hat\theta}(t)=\frac{t^{n\alpha}}{\Theta^{n\alpha}}$$

$$f_{\hat\theta}(t)=n\alpha\frac{t^{n\alpha-1}}{\Theta^{n\alpha}}$$

And finally

$$E(\hat\theta)=\int_0^\Theta n\alpha\frac{t^{n\alpha}}{\Theta^{n\alpha}}\;\mathrm{d}t=\frac{n\alpha}{n\alpha+1}\Theta\neq\Theta$$

Hence the estimator $\hat\theta$ is biased.