Cumulative Distribution Functions

The cumulative distribution function (CDF) is one of the most versatile tools in statistics. It provides a complete description of the distribution and allows you to calculate probabilities, find percentiles, and connect the PDF and CDF through differentiation and integration.

---

Definition

The cumulative distribution function of a random variable $X$X is:

$F(x) = P(X \leq x) = \int_{-\infty}^x f(t) \, dt$F(x)=P(X≤x)=∫−∞x​f(t)dt

Property	Detail
$F(-\infty) = 0$F(−∞)=0	No probability below the minimum
$F(\infty) = 1$F(∞)=1	All probability is accounted for
$F$F is non-decreasing	If $a < b$a<b, then $F(a) \leq F(b)$F(a)≤F(b)
$F$F is continuous (for continuous $X$X)	No jumps in the CDF

---

Relationship Between PDF and CDF

The PDF and CDF are linked by:

$F(x) = \int_{-\infty}^x f(t) \, dt \quad \text{and} \quad f(x) = F'(x)$F(x)=∫−∞x​f(t)dtandf(x)=F′(x)

In other words:

• Integrate the PDF to get the CDF
• Differentiate the CDF to get the PDF

Exam Tip: If you are given the CDF and asked for the PDF, differentiate. If given the PDF and asked for the CDF, integrate. This is a very common exam question format.

---

Worked Example: From PDF to CDF

Let $f(x) = \frac{3}{8}x^2$f(x)=83​x2 for $0 \leq x \leq 2$0≤x≤2.

For $x < 0$x<0: $F(x) = 0$F(x)=0.

For $0 \leq x \leq 2$0≤x≤2:

$F(x) = \int_0^x \frac{3}{8}t^2 \, dt = \frac{3}{8} \times \frac{t^3}{3}\Big|_0^x = \frac{x^3}{8}$F(x)=∫0x​83​t2dt=83​×3t3​​0x​=8x3​

For $x > 2$x>2: $F(x) = 1$F(x)=1.

So:

$F(x) = \begin{cases} 0 & x < 0 \\ \frac{x^3}{8} & 0 \leq x \leq 2 \\ 1 & x > 2 \end{cases}$F(x)=⎩⎨⎧​08x3​1​x<00≤x≤2x>2​

Verification: $F(0) = 0$F(0)=0 and $F(2) = \frac{8}{8} = 1$F(2)=88​=1. The CDF is continuous.

---

Worked Example: From CDF to PDF

Given:

$F(x) = \begin{cases} 0 & x < 1 \\ \frac{(x-1)^2}{4} & 1 \leq x \leq 3 \\ 1 & x > 3 \end{cases}$F(x)=⎩⎨⎧​04(x−1)2​1​x<11≤x≤3x>3​

Find the PDF:

$f(x) = F'(x) = \frac{2(x-1)}{4} = \frac{x-1}{2} \quad \text{for } 1 \leq x \leq 3$f(x)=F′(x)=42(x−1)​=2x−1​for 1≤x≤3

Check: $\int_1^3 \frac{x-1}{2} \, dx = \frac{1}{2}\left[\frac{(x-1)^2}{2}\right]_1^3 = \frac{1}{2} \times 2 = 1$∫13​2x−1​dx=21​[2(x−1)2​]13​=21​×2=1. Valid.

---

Using the CDF to Find Probabilities

The CDF provides a direct way to calculate probabilities:

Probability	Formula
$P(X \leq a)$P(X≤a)	$F(a)$F(a)
$P(X > a)$P(X>a)	$1 - F(a)$1−F(a)
$P(a \leq X \leq b)$P(a≤X≤b)	$F(b) - F(a)$F(b)−F(a)