Confidence Interval for Population Quantile

Recall that for a continuous random variable $X$ with cdf F, the median $\nu$, satisfies $P(X \leq \nu)=F(\nu)=\frac{1}{2}$. More generally, let $\xi_p$ denote the population quantile of order $p$. That is $\xi_p$ satisfies

$$P(X \leq \xi_p)=F(\xi_p)=p.$$ (8.17)

[Clearly, the median is a special case of this where $\nu=\xi_{.5}$.]

A confidence interval for a population quantile can be obtained using order statistics as follows. Suppose we take a sample of size $n$ from the given population and write the ordered sample as $Y_1, \ldots, Y_n$. Then for integers $r$, $s$, ( $1 \leq r < s \leq n$) the probability $P(Y_r < \xi_p < Y_s)$ can be determined. It is equal to

$$\begin{eqnarray*} \mbox{P(\underline{at least} r but \underline{not more than} s... ...}^{s-1}{n \choose x} p^x(1-p)^{n-x}\\ &=& \gamma \mbox {, say}. \end{eqnarray*}$$

Alternatively, for a given value of $\gamma$, we'd like to be able to determine $r$ and $s$. Substituting observed values $y_r$ and $y_s$ for $Y_r$ and $Y_s$, the $100\gamma$% CI for $\xi_p$ is ($y_r, \, y_s$). Note that we can't usually choose $r$, $s$ to give the conventional $.9$, $.95$, $.99$, values for $\gamma$. Note also that the above has assumed no particular form for $F$, and is, in that sense, a distribution-free property.

Example 8..10
Let $Y_1 < Y_2 < Y_3 < Y_4 < Y_5$ denote the order statistics of a random sample of size $5$ from a distribution of the continuous type. Compute

$$(a) \, P(Y_1 < \xi_{.5} < Y_5); \quad (b) \, P(Y_1 < \xi_{.25}< Y_3).$$

(a)

 

$$\begin{eqnarray*} P(Y_1 < \xi_{.5} < Y_5) &=& \sum_{x=1}^4 {5 \choose x} (.5)^x... ...\choose 5}(.5)^5(.5)^0\right]\\ &=&1 - \frac{2}{32}\\ &=&.9375 \end{eqnarray*}$$

(b)

 

$$\begin{eqnarray*} P(Y_1 < \xi_{.25} < Y_3) &=& \sum_{x=1}^2{5 \choose x} (.25^x... ... \choose 1}(.25)(.75^4 + {5 \choose 2}(.25)^2(.75)^3\\ &=& .66. \end{eqnarray*}$$