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 $\theta$.

  • 2
    $\begingroup$ 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, 2018 at 16:23

2 Answers 2


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.


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 $X^{\theta}\sim\mathcal U(0,1)$.


\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 $$\hat\theta_n=-\frac nT$$


\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}


\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}


\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 $\theta$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.