Asymptotic Distribution and Quantile (Inverse) Function

Question

For a concrete example:

Question. Suppose I have iid random variables $X_1, \dots, X_n$ with pdf $f(x) = \frac{1}{6|x|^{2/3}}$ for $x \in [-1, 0) \cup (0, 1]$. Find the asymptotic distribution of $n^{3/2}\tilde{X}_n$, where $\tilde{X}_n$ is the sample median.

It is well known that for $0 < q < 1$ the $q^{\text{th}}$-quantile of $F_x$, denoted as $\theta$, has asymp. dist.

$$\sqrt{n}(Y_n - \theta) \stackrel{d}{\longrightarrow} N\left(0, \frac{q(1-q)}{f(\theta)^2}\right),$$

where $Y_n = \frac{1}{n}\sum_{i=1}^n \mathbb{1}\{X_i \leq x\}$. However, for the above question there is no value $\theta$ such that $F_x(\theta) = q = 0.5$. Instead, I propose we make the following logical transformation:

$$U_i = F_x(X_i),~~~~X_i = F_x^{-1}(U_i),~~~~U_i \sim \text{Uniform}(0,1)~~~\forall i=1,\dots,n$$

Clearly, if $\tilde{U}_n$ is the sample median of the uniform distribution and $q = 0.5$ then $$\sqrt{n}(\tilde{U}_n - \theta) \stackrel{d}{\longrightarrow} N\left(0, \frac{q(1-q)}{f_u(\theta)^2}\right) = N\left(0, \frac{1}{4f_u(\theta)} \right).$$

But I'm unsure how to relate this back to $n^{3/2}\tilde{X}_n$? Also, I assume this method would work for any random variables like those defined in the question?

Answer 1

There is a nice distribution for $n^{3/2}$ times the median: it has a mean of 0 and a variance of 15, and is highly spiked at 0. The graph below compares this distribution with the normal with the same mean and variance.

To derive this, let $f(x)=|x|^{-2/3}/6$, and $F(x)=(1\pm|x|^{1/3})/2$

If $y>0$, the probability that $y$ is the median among $n$ samples of $X$ is

$$f_{median}(y)=n\binom{n-1}{(n-1)/2}f(y)F(y)^{(n-1)/2}(1-F(y))^{(n-1)/2}$$

i.e. we first choose which of the $n$ samples will be the median, then we choose which of the remaining samples will be less than the median, and we multiply the number of those choices by the probability that the first chosen sample is equal to $y$, the probability that the next chosen samples are less than $y$, and the probability that the remaining chosen samples are greater than $y$.

In the limit of high $n$, we can approximate the factorials in the binomial using Simpson's approximation, and expand $f(y)$ and $F(y)$ to write this as

$$f_{median}(y)\sim\frac{\sqrt{n}(1-y^{2/3})^{(n-1)/2}}{3\sqrt{2\pi}y^{2/3}}$$

The same formula obviously holds for $f_{median}(-y)$.

So by a change of variables with $z=n^{3/2}y$,

$$f_{n^{3/2}median}(z)\sim\frac{(1-|z|^{2/3}/n)^{(n-1)/2}}{3\sqrt{2\pi}|z|^{2/3}}$$

In the limit of high $n$ we can ignore the factor of $(1-|z|^{2/3}/n)^{-1/2}$, and use the exponential limit to write this as

$$f_{n^{3/2}median}(z)\sim\frac{\exp(-|z|^{2/3}/2)}{3\sqrt{2\pi}|z|^{2/3}}$$