Proving that a MLE is an asymptotically unbiased estimator Let $X_1,\ldots,X_n$ be a sample in a space with PDF $f_X(x; \theta) = \frac{3}{\theta^3}x^2 I(0\le x \le \theta)$ then caclulate the MLE for $\theta$ and prove that it is an asymptotically unbiased estimator.
So far, I managed to calculate $$ \theta_m (\mathrm{MLE}) = \max(X_i),$$ but proving that it is an asymptotically unbiased estimator isn't working out. I've tried integrating $\max(X_i)$, but the integral ends up not working out for me. Should I still use an integral for proving that $\operatorname{E}(\theta_m) = \theta$ or not? And if so, how should I integrate/calculate it? Thanks.
 A: Since the MLE is the maximum observed value, it is reasonably simple in this case to find the sampling distribution of the MLE, and from this the bias.  The distribution function is given by:
$$F_\hat{\theta}(t | \theta) = \mathbb{P}(\hat{\theta} \leqslant t | \theta) = F_X(t | \theta)^n = \Big( \frac{t}{\theta} \Big)^{3n} \quad \text{for } 0 \leqslant t \leqslant \theta.$$
Differentiating gives the density function:$^\dagger$
$$f_\hat{\theta}(t | \theta) = 3n \cdot \frac{t^{3n-1}}{\theta^{3n}} \quad \text{for } 0 \leqslant t \leqslant \theta.$$
The estimator $\hat{\theta}$ therefore has expectation:
$$\mathbb{E}(\hat{\theta} | \theta) = \int \limits_0^\theta t f_\hat{\theta}(t | \theta) dt = \frac{3n}{\theta^{3n}} \int \limits_0^\theta t^{3n} dt = \frac{3n}{3n+1} \cdot \theta .$$
Hence, we see that $\mathbb{E}(\hat{\theta} | \theta) \rightarrow \theta$ as $n \rightarrow \infty$.  This estimator is negatively biased, but asymptotically unbiased.

$^\dagger$ We take the density to be defined at the endpoints of the interval, which is valid (since the density is any Radon-Nikodym derivative of the measure induced by the distribution function).
