Let $X_1, ..., X_n$ be $i.i.d$ random variables, uniformly distributed over $(\theta,2 \theta)$. Find a sufficient statistic for $\theta$, and compute $\widehat{\theta}_{MLE}$.


The joint density for the $X_1, ..., X_n$ is:

\begin{align} f_\theta(x_1, ..., x_ n ) &= \prod_{i=1}^{n}\frac{1}{\theta}\mathbb{1}(\theta \leq x_i\leq 2\theta )\\ &= \frac{1}{\theta^n}\mathbb{1}(\underset{i}{min} \; x_i\geq\theta, \underset{i}{max} \; x_i\leq 2\theta ) \end{align}

Hence, $T = (T_1, T_2) = (\underset{i}{min} \; X_i,\: \underset{i}{max} \; X_i)$ is a joint sufficient statistics for $\theta$.

My questions are:

1. Can I interpret this as: "We need two statistics, $\underset{i}{min} \; X_i$ and $\underset{i}{max} \; X_i$, to provide full information about the value of the parameter $\theta$ given the data $X_1, ..., X_n$".

We cannot say that $T_1=\underset{i}{min} \; X_i$ is sufficient, because other statistic can be calculated from the same sample, i.e. $T_2=\underset{i}{max} \; X_i$, so that $T = (T_1, T_2)$ provides additional information about the parameter $\theta$.

2. Why $\widehat{\theta}_{MLE}$ is equal to $\frac{max_i X_i}{2}$?


Regarding 1., note that interpretation of a sufficient statistic is:

"no other statistic that can be calculated from the same sample provides any additional information as to the value of the parameter".

So, we need both the statistics to provide the "most information" possible about the value of the parameter. Not necessarily "full information".

Regarding 2.

Note that your likelihood can be written as

$$ f_{\theta} (x_1, \dots, x_n) = \dfrac{1}{\theta^n} I\left( \dfrac{\max x_i}{2} \leq \theta \leq \min x_i \right)\,.$$

The MLE is the statistic that maximizes the likelihood. You can show that the likelihood is a decreasing function of $\theta$. So the likelihood is maximized on the smallest value $\theta$ is allowed to take, which is $(\max x_i)/2$

  • $\begingroup$ That's a brilliant answer, very clear! Many thanks @Greenparker, that's very helpful! $\endgroup$ – AlexMe Oct 24 '18 at 10:59

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