# Asymptotic Distribution and Quantile (Inverse) Function

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?

• Re "However:" your distribution is symmetric and supported on $[-1,1],$ whence $0$ is the unique median: $F_x(0) = 1/2.$
– whuber
Apr 25, 2020 at 0:54
• @whuber Yes, you right. But dividing by $f(\theta)^2$ is an issue when we apply the Delta Method if we take $\theta = 0$. Apr 25, 2020 at 1:51
• That's right--and it explains why you should suspect the usual asymptotic result might not hold. Note that the delta method doesn't apply because $f$ is not differentiable at $0.$ This situation deserves a separate analysis. Some of the fundamental ideas are brought out in the thread at stats.stackexchange.com/questions/45124.
– whuber
Apr 25, 2020 at 14:34
• What is the cumulative distribution function of this random variable? May 2, 2020 at 19:37

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

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