# MLE of median is a consistent estimator - Problem 6.1.4 (Hogg, McKean, Craig)

Suppose $$X_1, . . .,X_n$$ are iid with pdf $$f(x; θ) = 2x/θ^2$$, $$0 < x ≤ θ$$, zero elsewhere. Find:

1. The MLE $$\hat\theta$$ for $$\theta$$,
2. The constant $$c$$ so that $$E[c\hat\theta] = \theta$$,
3. The MLE of the median of the distribution. Show that it is a consistent estimator.

I need help with completing part (c) of the problem. I did the rest, but please check my solution:

1. The likelihood is $$L(\theta) = \frac{2^n}{\theta^{2n}}\prod_i X_i$$ To maximize $$L(\theta)$$ we minimize $$\theta$$. We have $$0 \le x_i \le \theta$$ for every $$i$$. So, $$\theta \ge \max_i x_i$$. $$Y = \hat\theta = \max_{1\le i \le n} X_i$$
2. The pdf of $$Y$$ (I'm skipping steps here) turns out to be $$f_Y(y) = \frac{2ny^{2n-1}}{\theta^{2n}}, \quad 0 < y \le \theta$$ So $$E[Y] = \frac{2n\theta}{2n+1}$$, and $$c = \frac{2n+1}{2n}$$.
3. The median is $$\theta/\sqrt 2$$, since $$x^2/\theta^2 = 1/2 \implies x = \theta/\sqrt 2$$. The MLE of $$\theta/\sqrt2$$ is $$Y/\sqrt 2$$. Why is it a consistent estimator though?
• Please add the self-study tag. And check the many answers on this forum about moments or expectation of `order-statistics' Commented Apr 16, 2021 at 13:21
• @Xi'an Can you please point me to some particular ones? Commented Apr 16, 2021 at 13:22
• Sure: a search for median CLT, for instance, turns up stats.stackexchange.com/questions/45124 and stats.stackexchange.com/questions/196658.
– whuber
Commented Apr 16, 2021 at 15:51

Foreword: I'm going to stick within the context of the textbook. And I will be leaving in a few gaps given that this is a self-study thing. When referring to your textbook, I have the international 6th edition o of the textbook to which I think you are referring. Though there is some discrepancy in the problem you give, and the one I see. The consistency isn't requested.

Let's start with the definition of a consistent estimator. From chapter 4 of the version of the textbook that I am looking:

Let $$X$$ be a random variable with cdf $$F(x,\theta), \theta \in \Omega$$. Let $$X_1,...,X_n$$ be a sample from the distribution of $$X$$ and let $$T_n$$ denote a statistic. We say that $$T_n$$ is a consistent estimator of $$\theta$$ if $$T_n \mathop\to\limits^P \theta.$$

That is $$T_n$$ converges in probability to $$\theta$$.

You have a theorem somewhere in the section on convergence in probability that states if $$X_n \mathop \to\limits^P a$$ and $$g$$ is a real and continuous function at $$a$$, then $$g(X_n) \mathop \to\limits^P g(a)$$. I think multiplying by a constant counts as a continuous function, if memory serves.

Therefore if you prove $$Y$$ is consistent for $$\theta$$, then $$Y/\sqrt 2$$ is consistent for the median.

Which would mean, according to the definition of convergence in probability, you would have to show $$\lim_{n\to\infty}P(|Y - \theta| < \epsilon) = 1$$

or equivalently

$$\lim_{n\to\infty}P(|Y - \theta| \geq \epsilon) = 0$$ for any $$\epsilon >0$$. Which doesn't look so bad given the pdf for $$Y$$ looks relatively straight forward to integrate since it is effectively of the form $$\int x^k$$. You would "just" need to specify appropriate bound and show the limit trends to 1 or 0 depending on which you choose. The difficulty does not really differ.

This actually isn't too hard of a process as long as you keep tabs on the fact that $$P(|Y-\theta|<\epsilon) = P(\theta - Y<\epsilon)$$ because of the boundaries on $$y$$ in the pdf. I forgot about that and ended up down some rabbit hole involving trying to get binomial polynomials to cancel out. I figure I will try to save you the pain by pointing it out.

Side Note: It is tempting to use a corollary in the chapter on MLEs that allows you to say that any MLE is a consistent estimator. However there are regulatory conditions and this distribution violates one of them. The support of the pdf depends on $$\theta$$.