TABLE OF CONTENTS

Probability II

VOL. IMAR 2026

CH1 — Part 1

Cumulative Distribution Function & Continuous Random Variables

§1 Cumulative Distribution Function

Let $X$ denote any random variable. The cumulative distribution function (cdf) of $X$ , denoted by $F_X(x)$ , is defined as the probability that the variable takes a value less than or equal to $x$ :

F_X(x) = P(X \leq x), \quad -\infty < x < \infty

Properties of the CDF (Click to explore):

Asymptotic Limits
$F_X(0.00) = 0.5000$
Approach -∞Approach +∞
As $x$ increases, probability mass accumulates toward an absolute ceiling of 1. As $x$ decreases, it vanishes entirely toward 0.
Right Continuity Jump
$F_X(1.00) = 0.75$
Approach from LeftApproach from Right
At exactly $x = 1$ , the value $F_X(x)$ is $0.75$ . Note the solid circle belongs to the right-hand segment.
Monotonic Accumulation
$P(-1.5 < X \leq 1.5) = 0.635$
Position x₁
Position x₂
Notice that as $x_2$ moves right, the difference $F(x_2) - F(x_1)$ never becomes negative. The probability of landing in a larger interval can only increase or stay constant.

Example 1

Binomial Case

Let $X \sim \text{Bin}(2, \tfrac{1}{2})$ . Find $F_X(x)$ .

Cumulative Distribution Function (CDF)

x = 0.50

Probability Mass Function (PMF)

0.25

X = 0

X = 1

X = 2

Current Probability

P(X ≤ 0.50) = 0.25

F_X(x) = \sum_{x_i \leq x} f_X(x_i)

§2 Continuous Random Variables

A random variable $X$ is said to be continuous if its cdf $F_X(x)$ is a continuous function for all $x \in \mathbb{R}$ .

Key Fact

For a continuous r.v., $P(X = x) = 0$ for any real number $x$ .

Formal Proof

$P(X = x) = F_X(x) - F_X(x^-)$

Since $F_X$ is continuous, $F_X(x) = F_X(x^-)$ , so the difference is zero.

§3 Probability Density Function

If $F_X(x)$ is the cdf of a continuous r.v. $X$ , its derivative is called the probability density function (pdf):

f_X(x) = \frac{d}{dx} F_X(x)

Which implies the Fundamental Theorem relationship:

F_X(x) = \int_{-\infty}^{x} f_X(t)\,dt

$f_X(x) \geq 0$ for all $x$
$\int_{-\infty}^{\infty} f_X(x)\,dx = 1$ (Total area = 1)

Probability Density

f_X(t)

Cumulative Distribution

F_X(x)

Adjust Value

x

x = 0.00

-303

Integration

P(X \le 0.00) = \int_{-\infty}^{0.00} f_X(t) \, dt

Cumulative Value & Confidence

F_X(0.00) = 0.5000

Confidence: 50.0%

Example 2

The Uniform Case

Suppose we are given the following CDF and need to find the underlying density: $F_X(x) = \begin{cases} 0 & x < 0 \\ x & 0 \leq x < 1 \\ 1 & x \geq 1 \end{cases}$

Cumulative

F_X(x)

Continuous

$F_X(x)$ is continuous everywhere

Density

f_X(x)

Discontinuous

$f_X(x)$ is discontinuous at $x=0,1$

Remark — Continuity Contrast

While the CDF of a continuous r.v. must be continuous everywhere, its PDF can have jump discontinuities. In the Uniform(0,1) case, the PDF jumps at $x=0$ and $x=1$ , yet the CDF remains continuous throughout.