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

Question

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?

Answer 1

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 $\theta$.