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6.4.1 Quantiles

Often interest is in the values of a continuous random variable which are not exceeded with a given probability, such values are termed quantiles with $x_{p}$ the $100p\%$ quantile defined by

F_{X}(x_{p})=p.

Certain quantiles are of special interest:

Median:

the median is the middle of the distribution in the sense that half the values of the variable (in probability) are less than the median, and half are more. The median is the $50\%$ quantile, $x_{0.5}$ , so that $F(x_{0.5})=0.5$ . As a measure of location, the median has the advantage over the expectation of existing for all distributions.

Quartiles:

the quartiles split the distribution into four equally likely regions, $x_{0.25}$ the lower quartile, $x_{0.5}$ the median and $x_{0.75}$ the upper quartile.

{\rm P}(X<x_{0.25})={\rm P}(x_{0.25}<X<x_{0.5})={\rm P}(x_{0.5}<X<x_{0.75})={% \rm P}(X>x_{0.75})=0.25.

This is illustrated on Figure 6.4.

Inter-quartile range:

the difference in values of quartiles provides a measure of the variability of a random variable (measured in the units of the variable) that does not require the evaluation of the standard deviation (which can be infinite). The inter-quartile range is

x_{0.75}-x_{0.25}.

Figure 6.4: The cdf and pdf for a continuous random variable $X$ and the three quartiles $x_{0.25}$ , $x_{0.50}$ and $x_{0.75}$ . Note that the quartiles split the area under pdf into four equally sized regions.

Example 6.7.

A possible model for the claim sizes received by an insurance company is a random variable with cdf

\displaystyle F_{X}(x)=1-\exp(-\lambda x),

for some ${\lambda>0}$ . The company is legally obliged to pay the smallest ${99\%}$ of claims without requiring re-insurance support. What claim size must the company be able to pay without re-insurance support?

Solution.

The company must pay without support when a claim ${X}$ is less than ${x_{0.99}}$ , where

	$\displaystyle 0.99={\rm P}(X\leq x_{0.99})=F_{X}(x_{0.99})$
	$\displaystyle 1-\exp(-\lambda x_{0.99})=0.99$
	$\displaystyle\exp(-\lambda x_{0.99})=1-0.99$
	$\displaystyle-\lambda x_{0.99}=\log(1-0.99)$
	$\displaystyle x_{0.99}=-\frac{1}{\lambda}\log(0.01).$

Exercise 6.8.

It is considered suitable to model the annual maximum sea level by an extreme value distribution

F_{X}(x)=\exp[-\exp\{-(x-\alpha)/\beta\}],

for $\beta>0$ . The sea flood defence needs to be built to withstand a flood of the size which occurs in any year with probability $0.01$ (i.e. once on average every 100 years). Evaluate the required height of the flood defence.

Solution.

We seek $x$ such that ${\rm P}(X>x)=0.01$ .
Equivalently $F_{X}(x)=0.99$
We thus solve

	$\displaystyle\exp[-\exp\{-(x-\alpha)/\beta\}]=0.99$
	$\displaystyle-(x-\alpha)/\beta=\log[-\log\{0.99\}]$

Therefore

x=\alpha-\beta\log\{-\log(0.99)\}.