Asymptotic Distribution of Sample Quantile

Question

My question is related to the proof below. Why can we say that $F^{-1}(Y_n(x))=X_{[np]}$? I would like to understand better the reason for that.

Any help would be appreciated.

I've also asked here.I wasn't sure which one was better suited.

Answer 1

Quantiles are values of the support of a random variable that partition probability mass in equal lumps. So, say, the 4-quantiles (called the "quartile"), partition probability mass in four areas each holding 25% of probability mass. Note again: they are the cut-off points, not the intervals created thus, but it is common usage to say "1st quartile" and refer to the interval defined by the quantile.

Which are those values can be calculated from the inverse CDF. For example,

$$F_X(\tilde x) = \frac 14 \implies \tilde x = F^{-1}_X(1/4)$$

and $\tilde x$ is the 1st quartile (or the 1st 4-quantile).

Moving now to a random sample of realizations of size $n$, we will collect in the interval defined by the first empirical quartile the smallest $\lceil n/4 \rceil$ realizations in it (because the sample is i.i.d. and so each realization emerged with equal probability).

We also have the concept of order-statistics : say $X_{(5)}$ or better $X_{(5:n)}$ is the fifth largest value counting from the smallest upwards, in a sample of size $n$.

We can see that sample order-statistics can be linked to the concept of quantiles: as said the smallest $\lceil n/4\rceil$ realizations will be included in the (area defined by the) 1st quartile. But then, the order-statistic $X_{\lceil n/4\rceil:n}$ is the largest realization included in it. And as such, it is the empirical cut-off point, the 1st quartile proper.
In symbols

$$X_{\lceil n/4\rceil:n} = \hat F^{-1}_X(1/4)$$

Note the appearance of $1/4$ in both sides. Generalize this to a value $0<q<1$ and write

$$X_{\lceil nq\rceil:n} = \hat F^{-1}_X(q)$$

Note that $q$ is an arbitrary number in $(0,1)$, not some "true probability"

Now we do have

$$\hat F_X(\tilde x) = Y_n(\tilde x) = \hat p \implies \tilde x = \hat F_X^{-1}[Y_n(\tilde x)] =\hat F_X^{-1}(\hat p) $$

and I believe the connection is now clear.