Is this MLE estimator unbiased? - Cross Validated most recent 30 from stats.stackexchange.com 2019-08-23T18:29:08Z https://stats.stackexchange.com/feeds/question/272922 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://stats.stackexchange.com/q/272922 5 Is this MLE estimator unbiased? dietervdf https://stats.stackexchange.com/users/66287 2017-04-10T13:53:14Z 2018-11-07T09:33:46Z <p>From 'Modern Mathematical Statistics with Applications' (Devore and Beck) pg 377</p> <blockquote> <p>Let <span class="math-container">$X_1, X_2 \ldots$</span> be a random sample from the disbribution <span class="math-container">$f(x,\theta) = \theta x^{\theta-1}$</span> for <span class="math-container">$x\in {]0,1[}$</span> and <span class="math-container">$\theta &gt;0$</span>.</p> <p>The MLE is given by <span class="math-container">$\hat \theta = \dfrac{-n}{\sum_i\ln X_i}$</span>.</p> </blockquote> <p>It also shows how <span class="math-container">$$\sqrt{n}(\hat \theta - \theta) \stackrel{\text{D}}{\rightarrow}N(0,\theta)$$</span></p> <p>Using the fact that the MLE is consistent and the CLT.</p> <p>I wonder if this estimator is also unbiased, I want to show how <span class="math-container">$E[\hat \theta] = \theta$</span>. Any <strong>ideas</strong> (no full solutions please) on how to (dis)prove this.</p> <p>Here are some things I've tried:</p> <ul> <li><p>Notice how <span class="math-container">$E[\ln X] = \frac{-1}{\theta}$</span> or <span class="math-container">$E[\sum_i \ln X_i]= \dfrac{-n}{\theta}$</span></p></li> <li><p>Calculating the expected value seems cumbersome:</p></li> </ul> <p><span class="math-container">$$E\left[\frac{-n}{\sum_i \ln X_i}\right] = \int_0^1\ldots \int_0^1 \frac{-n}{\sum_i \ln x_i} \cdot \theta^n x_1^{\theta-1}x_2^{\theta-1}\ldots x_n^{\theta-1}\operatorname d x_1\ldots \operatorname dx_n$$</span></p> <h3>Edit after the responses from JohnK and Alecos Papadopoulos</h3> <ul> <li><p>Direct calculation: Okay, this is pretty cool. I worked it out and found how <span class="math-container">$\sum_i -\ln X_i = \sum Y_i \stackrel{\text{d}}{=} \Gamma(n,\theta)$</span> (through the hints supplied byJohnK), then I immedialty used LOTUS and found <span class="math-container">$E[\hat \theta] = \dfrac{n}{n-1}\theta &gt; \theta$</span>.</p></li> <li><p>Jensen's reasoning: I guess I should use Jensen's inequality as follows, since <span class="math-container">$E\left[\sum_i \ln X_i\right] = \dfrac{-n}{\theta}$</span> and because <span class="math-container">$\sum_i \ln X_i \in {]-\infty, 0[}$</span> I should look at the <em>left</em> part of the function <span class="math-container">$g:x\mapsto \dfrac{-n}{x}$</span> which is stictly convex. Jensen's concludes: <span class="math-container">$$E[g(X)] &gt; g(E[X])$$</span> this would imply here: <span class="math-container">$$E[\hat \theta] &gt; \theta$$</span></p></li> </ul> https://stats.stackexchange.com/questions/272922/-/272925#272925 5 Answer by JohnK for Is this MLE estimator unbiased? JohnK https://stats.stackexchange.com/users/31420 2017-04-10T14:01:59Z 2017-04-10T16:39:57Z <p>Here is a sequence of steps that will help:</p> <p>-Find the distribution of $Y=-\log X$</p> <p>-Find the distribution of $\sum_{i=1}^n Y_i = - \sum_{i=1}^n \log X_i$</p> <p>-Lastly, evaluate the expectation of $Z = \frac{1}{\bar{Y}}$. You can do this directly using the distribution of the previous step and <a href="https://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician" rel="nofollow noreferrer">LOTUS</a> or by first finding the distribution of $Z$.</p> <p><strong>Hint</strong>: the gamma family is a truly large family of distributions.</p>