I have the next problem:
$f(x; \theta) = \theta x^{\theta - 1}$
I obtained the maximum likelihood estimator, which is: $\hat{\theta} = \frac{-n}{\sum_1^n \ln(x_i)}$.
Knowing that $\ln(1/x) > 1/ \ln(x)$ for all $x ∈ (0, 1),$ determine whether the estimator ˆθMV is unbiased. Note that an estimator is said to be unbiased if E(ˆθ) = θ, otherwise the estimator is said to be biased. Also, explain and analyze your results.
How can I prove that the estimator is unbiased?
First we will find distribution of $Y=-Ln(X)$
$Y=-Ln(X) \Rightarrow x=e^{-y}, dx=-e^{-y}\Rightarrow f_Y(y)=f_X(e^{-y})|-e^{-y}|=\theta (e^{-y})^{\theta-1}e^{-y}=\theta e^{-\theta y}$ So Y has exponential distribution with parameter $\theta$ we know if $Y\sim E(\theta)$ then $U=\sum_{i=1}^{n}Y_i \sim \Gamma(n,\theta)=\frac{\theta^n}{\Gamma(n)}u^{n-1}e^{-\theta u}$ So $E(\hat{\theta})=E(\frac{n}{u})=nE(\frac{1}{u})=n\int_{0}^{\infty}\frac{\theta^n}{\Gamma(n)u}u^{n-1}e^{-\theta u}du=\frac{n\theta}{n}\int_{0}^{\infty}\frac{\theta^{n-1}}{\Gamma(n-1)u}u^{(n-1)-1}e^{-\theta u}du$ result of integral is 1 because integrate on $\Gamma(n-1,\theta)$ So $E(\hat{\theta})=\theta$ Therefore $\hat{\theta}$ is unbiased for $\theta$
