I have a distribution:

$$f_\theta(x) =\theta x^{\theta−1},0< x <1, \theta > 0 \qquad (1)$$

and I need to show that there is a function $\tau(\theta)$ for which there is an unbiased estimator whose variance attains the Cramer-Rao lower bound.

I know that the lowerbound is attained if and only if:

$$a(\theta)[W(X) - \tau(\theta)] = \frac{d}{d\theta}ln\left(f_\theta(X|\theta)\right) \qquad (2)$$

where $W(X)$ is an unbiased estimator of $\tau(\theta)$.

Starting with the MLE as an estimator of $\theta$:

$$\hat \theta = \frac{-n}{\sum_{i=1}^n log(X_i)}$$

To show that it's an unbiased estimator:

\begin{align} E\left[\frac{-n}{\sum_{i=1}^n log(X_i)}\right] &= \frac{-n}{\sum_{i=1}^n E[log(X_i)]} \\ &= \frac{-n}{\frac{-n}{\theta}}\\ &= \theta \end{align}

To show that the estimator attains the Cramer-Rao lower bound:

\begin{align} a(\theta)\left[\frac{-n}{\sum_{i=1}^n log(X_i)} - \theta \right] = \frac{n}{\theta} + \sum_{i=1}^nlog(X_i) \end{align}

Solving for $a(\theta)$:

$$a(\theta) = \frac{-\sum_{i=1}^n log(X_i)}{\theta}$$

This indicates the variance of $W(X)$ attains the following Cramer-Rao lowerbound: \begin{align} Var(W(X)) &=\frac{\left(\frac{d}{d\theta}E\left[W(X)\right]\right)^2}{E\left[\left(\frac{d}{d\theta}ln\left(f(X|\theta)\right)\right)^2\right]} \\ &=\frac{1}{E\left[\left(\frac{d}{d\theta}ln\left(f(X|\theta)\right)\right)^2\right]} \\ &=\frac{1}{E\left[\left(\frac{n}{\theta} + \sum_{i=1}^nlog(X_i)\right)^2\right]} \end{align}

Correct me if I am wrong, but isn't this variance different from that described here?

  • $\begingroup$ MLE is NOT unbiased and there is no need to bring it here. Working with $(2)$ a suitable $\tau(\theta)$ is simply $1/\theta$. $\endgroup$ Nov 19 '19 at 2:12
  • $\begingroup$ Why is MLE not unbiased ? Did I make a mistake when computing the expectation? Also, I will try $1/\theta$. $\endgroup$
    – N.B.
    Nov 19 '19 at 2:20

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