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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 $E[x^{r}]$?

StubbornAtom
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Daniel
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2 Answers2

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Is there a simpler more straight forward way to get the expected value of E[xr]x^{r}]?

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

tommik
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You can evaluate this integral directly using the substitution $u=\frac{x^r}{\theta}$ to get $E(X^r)=\theta$. Are you assuming that $r>\frac{1}{2}$?