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Let $\hat{\theta}_n= -\frac{n}{\sum_{i=1}^n \log(X_i)}$, where $X_i$ are i.i.d. samples from distribution with pdf $\theta x^{\theta-1}$ for $x \in (0,1)$. How to prove that $\hat{\theta}_n$ is consistent estimator of $\theta$.

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    $\begingroup$ Hint: Show that $\hat\theta_n$ is the MLE of $\theta$. You can do this directly, or by finding the distribution of $Y_i = -\log X_i$ which turns the problem into something a bit more familiar. $\endgroup$
    – knrumsey
    Mar 17 '18 at 16:23
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Probability limits distribute, so

$$\text{plim}\left (\frac {1}{\hat \theta}\right) = \frac {1}{\text{plim}\hat \theta}$$

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

The sample is i.i.d. and so ergodic and so

$$\text{plim}\left (\frac {1}{\hat \theta}\right) = \text{plim} \left(-\frac 1n\sum_{i=1}^n \log(X_i)\right) = E\left[-\log(X_i)\right]$$

So you just have to find the distribution of $Y = -\log(X_i)$ as suggested in a comment, and calculate its expected value.

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We can also use the sufficient condition of consistency showing that $E_{\theta}(\hat\theta_n)\to\theta$ and $\operatorname{Var}_{\theta}(\hat\theta_n)\to0$ as $n\to\infty$ to prove that $\hat\theta_n$ is consistent for $\theta$. But then again, one needs to know the distribution of the sufficient statistic $\sum_{i=1}^n\ln X_i$.

Since the population DF is of the form $F_{\theta}(x)=x^{\theta}$ for $0<x<1$, we have $X^{\theta}\sim\mathcal U(0,1)$.

Thus,

\begin{align} -\ln X^{\theta}&\sim\text{Exp}(1) \\\text{ or },\,-\theta\ln X&\sim\text{Exp}(1) \end{align}

Let $T=\sum_{i=1}^n\ln X_i$ where $-\theta \ln X_i$ are i.i.d $\text{Exp}(1)$ variables.

Then, $-\theta \,T\sim\mathcal{G}(1,n)$ with density $$g(t)=\frac{e^{-t}t^{n-1}}{\Gamma(n)}\mathbf1_{t>0}$$

The MLE of $\theta$ is $$\hat\theta_n=-\frac nT$$

Now,

\begin{align} E(\hat\theta_n)&=n\theta E\left(\frac{1}{-\theta T}\right) \\&=n\theta \int_0^\infty \frac{1}{t}\frac{e^{-t}t^{n-1}}{\Gamma(n)}\,dt \\&=n\theta\,\frac{\Gamma(n-1)}{\Gamma(n)} \\&=\frac{n\theta}{n-1} \\&=\frac{\theta}{1-\frac{1}{n}}\longrightarrow\theta\quad\text{ as }n\to\infty \end{align}

Similarly,

\begin{align}E\left(\hat{\theta_n}^2\right)&=n^2\theta^2E\left(\frac{1}{\theta^2T^2}\right) \\&=\frac{n^2\theta^2}{(n-1)(n-2)} \end{align}

Therefore,

\begin{align} \operatorname{Var}(\hat\theta_n)&=E\left(\hat\theta_n^2\right)-\left(E(\hat\theta_n)\right)^2 \\&=\frac{n^2\theta^2}{(n-1)^2(n-2)} \\&=\frac{\theta^2}{\left(1-\frac{1}{n}\right)^2(n-2)}\longrightarrow0\quad\text{ as }n\to\infty \end{align}

Hence, $\hat\theta_n$ is a consistent estimator of $\theta$.

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