How to show that estimator is unbiased

Question

Let $f(x;\theta) = (1/\theta)x^{(1-\theta)/\theta} $$\hspace{20 mm}$ $ 0 <x <1 ,\hspace{5 mm} 0 <\theta<\infty$

The maximum likelihood estimator of $\theta $ is $\hat{\theta}$ = $-(1/n) \sum ln X_i$

How can I show that this estimator is unbiased?

Answer 1

Let $W--\frac{1}{n}\sum_1^n \ln(X_i)$. We want to show that $E(W)=\theta$. By the linearity of expectation, it is enough to show that $E(-\ln(X_i))=\theta$.

We find $E(-\ln(X_i))$ as usual, by finding $$\int_0^1 (-\ln x)\left(\frac{1}{\theta} x^{(1-\theta)/\theta}\right) \,dx.\tag{1}$$ For $\theta\gt 1$, we get an improper integral. We will treat it as if it were an ordinary integral, though really we should integrate from $\epsilon$ to $1$ and let $\epsilon\to 0^+$.

We do the calculation of (1) by integration by parts. Let $du=-\frac{1}{\theta} x^{(1-\theta)/\theta}\,dx$ and let $v=\ln x$. Then we can take $u=-x^{1/\theta}$ and we have $dv=\frac{1}{x}\,dx$. Thus Integral (1) is equal to $$\left.\left(-x^{1/\theta}\ln x\right)\right|_0^1 +\int_0^1 x^{1/\theta-1}\,dx.$$ The first part above is $0$, and straightforward integration shows that $\int_0^1 x^{1/\theta-1}\,dx=\theta$.