Expected value and variance of estimator with a sum in the denominator I have the following estimators:
$\hat{\theta_1} = \frac{n}{\sum_{i=1}^{n} x_i^{^{\delta}}} $ and
$\hat{\theta_2} = -\frac{n}{\sum_{i=1}^{n} \ln (x_i^{^{}})} $
I want to find the expected value and variance of those estimators to determine whether they are biased and/or consistent. Yet, I'm somehow stuck on how to deal with the sum in the denominator.
$\mathbf{E}\left( \frac{n}{\sum_{i=1}^{n} x_i^{^{\delta}}} \right)$ = n $\mathbf{E}\left( \frac{1}{\sum_{i=1}^{n} x_i^{^{\delta}}} \right)$
and
$\mathbf{V}\left( \frac{n}{\sum_{i=1}^{n} x_i^{^{\delta}}} \right) = {n}^2$ $\mathbf{V}\left( \frac{1}{\sum_{i=1}^{n} x_i^{^{\delta}}} \right)$
The same holds for $\hat{\theta_2}$, but at this point I don't know how to proceed.
 A: The first "expression" you wrote in the comment is
$$f_X(x)=\theta \delta x^{\delta-1}e^{-\delta \theta x}$$
please provide the support because i.e. if $X>0$ this is NOT a density.
Example: Set $\theta=2$ and $\delta=3$ and you get
$$\int_0^{\infty}6x^2e^{-6x}dx \ne 1$$

The second expression is a Beta.
$$f_X(x)=\theta x^{\theta-1}$$
$\theta>0$
$x \in [0;1 ]$
The ML estimator for $\theta$ you showed is wrong. The correct one is
$$\hat{\theta}=\frac{n}{\Sigma_i(-\log X_i)}$$
It is very easy to verify that
$$Y=-\log(X)\sim  Exp(\theta)$$
Thus $\Sigma_i Y_i\sim Gamma(n;\theta)$ with expectation $E(\Sigma_i Y_i)=n/\theta$
Thus, using Jensen's inequality you get
$$\mathbb{E}[\hat{\theta}]\ne \frac{n}{n/\theta}=\theta$$
so it is biased for $\theta$
On the other hand, as $\hat{\theta}$ is MLE, it is asymptotically unbiased and consistent for $\theta$ (these are basic properties of MLE)
If you want to calulate mean and variance of your MLE analithically it is not difficult... you have to solve some integrals connected with gamma distribution.

The first "modified " density is now clear: it's a Weibull. Now, assuming $\delta$ known,
$$\hat{\theta}=\frac{n}{\Sigma_x x^\delta}$$
is the MLE for $\theta$ thus it is asimptotically unbiased and consistent. To check unbiasedness let's take the likelihood
$$L(\theta)\propto \theta^n e^{-\theta \Sigma_x x^\delta}$$
$$l(\theta)=n\log\theta-\theta\Sigma_x x^{\delta}$$
The score is
$$l^*(\theta)=\frac{n}{\theta}-\Sigma_x x^{\delta}$$
Now remembering that $E(l^*)=0$ you get
$$E\Big(\frac{n}{\Sigma_x x^\delta}\Big)=\theta$$
Thus $\hat{\theta}$ is unbiased
