Cumulative Distribution Functions (c. d. f.)

The cumulative distribution function of a continuous random variable extends the discrete definition in the natural way, replacing the sum by an integral. If $X$ is a continuous random variable with p.d.f. $f(x)$ on $(a,b)$, its c.d.f. is $F(x) = \int_a^x f(t)\,dt$ for $a<x<b$. While a discrete c.d.f. is a step function, a continuous c.d.f. is always a non-decreasing continuous function.

Worked examples.

Example 3. For $f(x)=3x^2$, $0<x<1$ (Example 1 of section 7.4.1), find $F(x)$.

$F(x) = \int_0^x 3t^2\,dt = \left[t^3\right]_0^x = x^3$.

Example 4. For $f(x)=x^3/4$, $0<x<4$, find $F(x)$.

$F(x) = \int_0^x \dfrac{t^3}{4}\,dt = \left[\dfrac{t^4}{16}\right]_0^x = \dfrac{x^4}{16}$.

Example 5. $X$ has p.d.f. $f(x) = x+1$ for $-1<x<0$ and $f(x) = 1-x$ for $0\le x<1$. Find the c.d.f. $F(x)$.

The density is defined piecewise, so $F$ is built piecewise too. For $-1<x\le0$: $F(x) = \int_{-1}^x (t+1)\,dt = \left[\dfrac{t^2}{2}+t\right]_{-1}^x = \dfrac{x^2}{2}+x-\left(\dfrac12-1\right) = \dfrac{x^2}{2}+x+\dfrac12 = \dfrac{(x+1)^2}{2}$; in particular $F(0)=1/2$. For $0<x<1$: $F(x) = F(0) + \int_0^x (1-t)\,dt = \dfrac12 + \left[t-\dfrac{t^2}{2}\right]_0^x = \dfrac12+x-\dfrac{x^2}{2}$. Collecting every piece: $F(x)=0$ for $x\le-1$; $F(x)=\dfrac{(x+1)^2}{2}$ for $-1<x\le0$; $F(x)=\dfrac12+x-\dfrac{x^2}{2}$ for $0<x<1$; $F(x)=1$ for $x\ge1$. The way $F$ accumulates is illustrated in Fig. 7.1, which shows the triangular density $f$ itself: $F(x)$ is the area under this triangle from $t=-1$ up to $t=x$ (the shaded region), growing from $0$ at $x=-1$, reaching exactly $1/2$ at $x=0$ (half of the triangle's total area of $1$), and climbing to $1$ at $x=1$ once the whole triangle is covered — matching the piecewise formula just derived. …