# Is $\sum_{i=1}^{n} y_i^r/n$ the MVUE of $\theta$?

Let $Y_1, Y_2, ..., Y_n$ denote a random sample from the density function given by

$$f(y|\theta)= \begin{cases} \frac{1}{\theta}ry^{r-1}e^{-y^r/\theta}, & \theta>0, y>0, \\ 0, & \text{elsewhere,} \end{cases}$$

where $r$ is a known positive constant. What is the MLE of $\theta$? In this case, is the MLE also the MVUE for $\theta$?

I started with

\begin{split} L(\theta) & = \prod_{i=1}^{n} \frac{1}{\theta}ry_i^{r-1}e^{-y_i^r/\theta} \\ & = \Big[r\prod_{i=1}^{n}y_i^{r-1}\Big]\Big[\frac{1}{\theta}e^{-\sum_{i=1}^{n}y_i^r/\theta}\Big] \\ & = \Big[h\Big(y_1,y_2,...,y_n\Big)\Big]\Big[g\Big(\sum_{i=1}^{n}y_i^r, \theta\Big)\Big]. \end{split}

So I have a sufficient statistic of $\sum_{i=1}^{n}y_i^r$. Then to get the MLE we take the derivative $$\frac{d}{d\theta}\ell(\theta) = \frac{\sum_{i=1}^{n}y_i^r}{\theta^2}-n\theta \text{.}$$ Thus the MLE for $\theta$ is $\sum_{i=1}^{n}y_i^r/n$. Normally I would take the expected value of the statistic to check if it is unbiased but I was unsuccessful. How would I prove or disprove that $\sum_{i=1}^{n}y_i^r/n$ is an unbiased estimator of $\theta$?

I figured out the answer to my question. First we let $u=y^r$ and $du=ry^{r-1}dy$ then we have an exponential distribution with mean $\theta$:
$$f(u)=\begin{cases} \frac{1}{\theta}e^{-u/\theta}, & \quad u>0 \\ 0, & \quad \text{elsewhere}. \end{cases}$$
which means that $\sum_{i=1}^{n} Y_i^r/n$ is an unbiased estimator and function of the sufficient statistic for $\theta$. Therefore $\sum_{i=1}^{n} Y_i^r/n$ is the MVUE for $\theta$. $\blacksquare$
• Correct explanation! There is a mistake in the second part when you switch to $Y_i^2$ instead of $Y_i^r$. – Xi'an Feb 15 '17 at 7:58