Exercise :

Let $X_1, \dots, X_n$ be a random sample $(n>1)$ from the distribution with pdf $f(x) = \theta x^{-2}, \; \; 0 < \theta \leq x < \infty$, where $\theta$ an unknown parameter. Find the Maximum Likelihood Estimator $\hat{\theta}$ of $\theta$ and determine if it's an unbiased estimator for the parameter $\theta$.

Attempt :

The likelihood function is :

$$L(x;\theta) = \prod_{i=1}^n \theta x^{-2} \mathbb{I}_{[\theta, + \infty)}(x_i) = \theta^n \mathbb{I}_{[\theta, + \infty)}(\min x_i)$$

and thus the MLE is : $\hat{\theta} = \min x_i$.

Now, in order to determine if it's an unbiased estimator for $\theta$, I have to find :

$$\mathbb{E} [\min xi |\theta] $$

and determine whether it's equal to $\theta$ or not.

How does one proceed with calculating $\mathbb{E} [\min xi |\theta] $ ? Or is there another way of determining if $\hat{\theta}$ is an unbiased estimator for $\theta$ ?

  • $\begingroup$ You have to find $E(\min X_i)$. So find the distribution of $\min X_i$. $\endgroup$ – StubbornAtom May 27 '18 at 7:26
  • $\begingroup$ @StubbornAtom I have found it to be $f_T(t,\theta) = n\theta^{n}t^{-n-1}$. Should I integrate from $0$ to $+ \infty$ now to check if it's equal to $\theta$ ? $\endgroup$ – Rebellos May 27 '18 at 7:27
  • $\begingroup$ Just check whether $\int _{\theta}^\infty tf_T(t)\,dt$ is $\theta$ or not. $\endgroup$ – StubbornAtom May 27 '18 at 7:30

Let $Z = \min\{X_1, X_2, \ldots, X_n\}$.

$$P(Z \leq z | \theta) = 1 - P(Z > z|\theta) = 1 - P(X_i > z|\theta)^n = 1 - \left(\frac{\theta}{z}\right)^n$$

Hence, $$f_{Z|\theta} = n \frac{\theta^n}{z^{n+1}}$$

Therefore, $$\mathbb{E}[Z | \theta] = \int_{\theta}^{\infty} z n \frac{\theta^n}{z^{n+1}}dz = n \theta^n \int_{\theta}^{\infty} \frac{1}{z^{n}}dz = \frac{n}{n-1} \theta$$

Thus, your estimator is biased.


Since $\mathbb P(X_i \geq \theta)=1$ for each $i$, the same holds for $\hat{\theta}$: $\min X_i \geq \theta$ with probability $1$. For this case $\mathbb E_{\theta}[\min X_i]\geq \theta$. Moreover, the equality can hold iff $\mathbb P(\min X_i=\theta)=1$. But this is not the case since all samples have non-degenerate distribution.

You can also find CDF of $\min X_i$ and then calculate its expectation $\mathbb E\min X_i = \frac{n}{n-1}\theta>\theta$.


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