Relationship between f(x) and F(x)

We showed earlier that integration and differentiation relate the probability density function, \(f(x)\), of a continuous random variable, \(X\), to its cumulative distribution function, \(F(x)\).

\[ F(x) = \int_0^x f(u) \;du \spaced{and} f(x)= F'(x) \]

This sometimes allows us to derive the distribution of a random variable that is defined as a function of one or more others.

  1. Find the cumulative distribution function of \(X\), \(F(x) = P(X \le x)\).
  2. Differentiate it to get the probability density function, \(f(x)= F'(x)\)

Example: Square root of an exponential variable

Consider a random variable, \(X\), with an exponential distribution

\[ X \;\;\sim\;\; \ExponDistn(\lambda) \]

What is the distribution of \(Y = \sqrt{X}\)?

The cumulative distribution function of \(Y\) is

\[ F_Y(y) \;=\; P(Y \le y) \;=\; P(\sqrt{X} \le y) \;=\; P(X \le y^2) \;=\; F_X(y^2) \]

Since we know that the exponential distribution's cumulative distribution function is \(F_X(x) = 1 - e^{\large -\lambda x}\),

\[ F_Y(y) \;\;=\;\; 1 - e^{ -\lambda y^2} \]

It's pdf is therefore

\[ f_Y(y) \;=\; F'(y) \;=\; 2\lambda y \; e^{ -\lambda y^2} \]

This can be recognised as the probability density function of a \(\WeibullDistn(\alpha=2,\;\sqrt{\lambda})\) distribution.