14
Sep 23

## The magic of the distribution function

Let $X$ be a random variable. The function $F_{X}\left( x\right) =P\left( X\leq x\right) ,$ where $x$ runs over real numbers, is called a distribution function of $X.$ In statistics, many formulas are derived with the help of $F_{X}\left( x\right) .$ The motivation and properties are given here.

Oftentimes, working with the distribution function is an intermediate step to obtain a density $f_{X}$ using the link

$F_{X}\left( x\right) =\int_{-\infty }^{x}f_{X}\left( t\right) dt.$

A series of exercises below show just how useful the distribution function is.

Exercise 1. Let $Y$ be a linear transformation of $X,$ that is, $Y=\sigma X+\mu ,$ where $\sigma >0$ and $\mu \in R.$ Find the link between $F_{X}$ and $F_{Y}.$ Find the link between $f_{X}$ and $f_{Y}.$

The solution is here.

The more general case of a nonlinear transformation can also be handled:

Exercise 2. Let $Y=g\left( X\right)$ where $g$ is a deterministic function. Suppose that $g$ is strictly monotone and differentiable. Then $g^{-1}$ exists. Find the link between $F_{X}$ and $F_{Y}.$ Find the link between $f_{X}$ and $f_{Y}.$

Solution. The result differs depending on whether $g$ is increasing or decreasing. Let's assume the latter, so that $x_{1}\leq x_{2}$ is equivalent to $g\left( x_{1}\right) \geq g\left( x_{2}\right) .$ Also for simplicity suppose that $P\left( X=c\right) =0$ for any $c\in R.$ Then

$F_{Y}\left( y\right) =P\left( g\left( X\right) \leq y\right) =P\left( X\geq g^{-1}\left( y\right) \right) =1-P\left( X\leq g^{-1}\left( y\right) \right)=1-F_{X}\left( g^{-1}\left( y\right) \right) .$

Differentiation of this equation produces

$f_{Y}\left( y\right)=-f_{X}\left( g^{-1}\left( y\right) \right) \left( g^{-1}\left( y\right) \right) ^{\prime }=f_{X}\left( g^{-1}\left( y\right) \right) \left\vert\left( g^{-1}\left( y\right) \right) ^{\prime }\right\vert$

(the derivative of $g^{-1}$ is negative).

For an example when $g$ is not invertible see the post about the chi-squared distribution.

Exercise 3. Suppose $T=X/Y$ where $X$ and $Y$ are independent, have densities $f_{X},f_{Y}$ and $Y>0.$ What are the distribution function and density of $T?$

Solution. By independence the joint density $f_{X,Y}$ equals $f_{X}f_{Y},$ so

$F_{T}\left( t\right) =P\left( T\leq t\right) =P\left( X\leq tY\right) = \underset{x\leq ty}{\int \int }f_{X}\left( x\right) f_{Y}\left( y\right) dxdy$

(converting a double integral to an iterated integral and remembering that $f_{Y}$ is zero on the left half-axis)

$=\int_{0}^{\infty }\left( \int_{-\infty }^{ty}f_{X}\left( x\right) dx\right) f_{Y}\left( y\right)dy=\int_{0}^{\infty }F_{X}\left( ty\right) f_{Y}\left( y\right) dy.$

Now by the Leibniz integral rule

(1) $f_{T}\left( t\right) =\int_{0}^{\infty }f_{X}\left( ty\right) yf_{Y}\left( y\right) dy.$

A different method is indicated in Activity 4.11, p.207 of J.Abdey, Guide ST2133.

Exercise 4. Let $X,Y$ be two independent random variables with densities $f_{X},f_{Y}$. Find $F_{X+Y}$ and $f_{X+Y}.$

See this post.

Exercise 5. Let $X,Y$ be two independent random variables. Find $F_{\max \left\{ X_{1},X_{2}\right\} }$ and $F_{\min \left\{ X_{1},X_{2}\right\} }.$

Solution. The inequality $\max \left\{ X_{1},X_{2}\right\} \leq x$ holds if and only if both $X_{1}\leq x$ and $X_{2}\leq x$ hold. This means that the event $\left\{ \max \left\{ X_{1},X_{2}\right\} \leq x\right\}$ coincides with the event $\left\{ X_{1}\leq x\right\} \cap \left\{ X_{2}\leq x\right\}.$ It follows by independence that

(2) $F_{\max \left\{ X_{1},X_{2}\right\} }\left( x\right) =P\left( \max \left\{ X_{1},X_{2}\right\} \leq x\right) =P\left( \left\{ X_{1}\leq x\right\} \cap \left\{ X_{2}\leq x\right\} \right)$

$=P(X_{1}\leq x)P\left( X_{2}\leq x\right) =F_{X_{1}}\left( x\right) F_{X_{2}}\left( x\right) .$

For $\min \left\{ X_{1},X_{2}\right\}$ we need one more trick, namely, pass to the complementary event by writing

$F_{\min \left\{ X_{1},X_{2}\right\} }\left(x\right) =P\left( \min \left\{ X_{1},X_{2}\right\} \leq x\right) =1-P\left(\min \left\{ X_{1},X_{2}\right\} >x\right) .$

Now we can use the fact that the event $\left\{ \min \left\{ X_{1},X_{2}\right\} >x\right\}$ coincides with the event $\left\{ X_{1}>x\right\} \cap \left\{ X_{2}>x\right\} .$ Hence, by independence

(3) $F_{\min \left\{ X_{1},X_{2}\right\} }\left( x\right) =1-P\left( \left\{X_{1}>x\right\} \cap \left\{ X_{2}>x\right\} \right) =1-P\left(X_{1}>x\right) P\left( X_{2}>x\right)$

$=1-\left[ 1-P\left( X_{1}\leq x\right) \right] \left[ 1-P\left( X_{2}\leq x\right) \right] =1-\left( 1-F_{X_{1}}\left( x\right) \right) \left(1-F_{X_{2}}\left( x\right) \right) .$

Equations (2) and (3) can be differentiated to obtain the links in terms of densities.