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2.5 The Quantile function

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 xp the 100p% quantile (ideally) defined by

FX(xp)=p.

Unnumbered Figure: Link

For a continuous random variable, X, FX(x) has an inverse FX-1(p):={x:F(x)=p}, for all values of p between 0 and 1. Typically, though not always, this inverse is unique except, perhaps at p=0 and p=1 (to see why, look again at Example 2.2.2). We therefore extend the definition to

FX-1(p)=inf{x|FX(x)p},

where 0<p1.

Roughly speaking, the quantile function of y is the smallest x such that FX(x)y.

In Example 2.2.2 FX-1(1)=2.

The above definition also applies to discrete random variables, although we will not discuss this in Math230.

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, x0.5, so that F(x0.5)=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, x0.25 the lower quartile, x0.5 the median and x0.75 the upper quartile.

𝖯(X<x0.25)=𝖯(x0.25<X<x0.5)=𝖯(x0.5<X<x0.75)=𝖯(X>x0.75)=0.25.

This is illustrated on Figure 2.4 (First Link, Second Link).

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

x0.75-x0.25.
Figure 2.4: First Link, Second Link, Caption: The cdf and pdf for a continuous random variable X and the three quartiles x0.25, x0.50 and x0.75. Note that the quartiles split the area under pdf into four equally sized regions.
Example 2.5.1.

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

FX(x)=exp[-exp{-(x-α)/β}],

for β>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 need 𝖯(X>x)=0.01 or, equivalently FX(x)=0.99.

  1. exp[-exp(-(x-α)/β)]=0.99

  2. -(x-α)/β=log[-log{0.99}]

so that

x0.99=α-βlog(-log(0.99)).