Suppose that $X_1, \ldots, X_n$ is a random sample from the probability density function
$$f(x\mid \theta) = \frac{rx^{r−1}}{\theta} e^{-x^{r}/\theta} \quad\,, x>0$$
where $r > 1$ is a known constant and $\theta > 0$ is an unknown parameter.
Based on the MLE of $\theta$, find an unbiased estimator of $2\theta$.
I got the mle to be $\hat{\theta} = \frac{\sum{X_i^{r}}}{n}$
I don't know how to get the expected value of $E[x^{r}]$ other than of $$ E[x^{r}]=\int_0^{\infty}x^{r}\frac{rx^{r−1}}{\theta} e^{-x^{r}/\theta} dx$$
Is there a simpler more straight forward way to get the expected value of $E[x^{r}]$?
Best Answer
Let's consider the likelihood score an let $T=\hat{\theta}$
$$l^*(\theta)=-\frac{n}{\theta}+\frac{\sum_x x^r}{\theta^2}$$
It' is well known (First Bartlett identity) that the expectation of the score is zero, so
$$\mathbb{E}[-\frac{n}{\theta}+\frac{nT}{\theta^2}]=0$$
Thus immediately you get that
$$\mathbb{E}[T]=\theta$$
... the rest is self evident