TABLE OF CONTENTS

Probability II

VOL. IMAR 2026

CH1 — Part 1

§1 — Definition

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 $x$ :

Definition 1.1— CDF

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)

Solution

The probability mass function (pmf) for $X \sim \text{Bin}(2, 1/2)$ is:

f_X(0) = 1/4, \quad f_X(1) = 1/2, \quad f_X(2) = 1/4

Summing these piecewise gives the step function:

F_X(x) = \begin{cases} 0 & x < 0 \\ 1/4 & 0 \leq x < 1 \\ 3/4 & 1 \leq x < 2 \\ 1 & x \geq 2 \end{cases}

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 — Definition

Continuous Random Variables

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

Select a variable below to explore its continuous nature:

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 — Definition

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):

Definition 1.2— 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}

Solution

Differentiating piecewise gives:

f_X(x) = \frac{d}{dx} F_X(x) = \begin{cases} 1 & 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}

This result confirms that $X$ follows a Uniform(0,1) distribution.

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.

Next Stage

Where does the normalization constant come from?

In the next lecture, we'll learn how to find the constant $k$ that ensures the total area under a density curve is exactly one.

Continue to Part II

TABLE OF CONTENTS

Back to Archive

Probability II

VOL. IMAR 2026

CH1 — Part 1

§1 — Definition

Cumulative Distribution Function

Definition 1.1— CDF

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)

Solution

The probability mass function (pmf) for $X \sim \text{Bin}(2, 1/2)$ is:

f_X(0) = 1/4, \quad f_X(1) = 1/2, \quad f_X(2) = 1/4

Summing these piecewise gives the step function:

F_X(x) = \begin{cases} 0 & x < 0 \\ 1/4 & 0 \leq x < 1 \\ 3/4 & 1 \leq x < 2 \\ 1 & x \geq 2 \end{cases}

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 — Definition

Continuous Random Variables

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

Select a variable below to explore its continuous nature:

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 — Definition

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):

Definition 1.2— 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}

Solution

Differentiating piecewise gives:

f_X(x) = \frac{d}{dx} F_X(x) = \begin{cases} 1 & 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}

This result confirms that $X$ follows a Uniform(0,1) distribution.

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.

Next Stage

Where does the normalization constant come from?

In the next lecture, we'll learn how to find the constant $k$ that ensures the total area under a density curve is exactly one.

Continue to Part II

Cumulative Distribution Function

Asymptotic Limits

Right Continuity Jump

Monotonic Accumulation

Binomial Case

Continuous Random Variables

Key Fact

Formal Proof

Probability Density Function

The Uniform Case

Where does the normalization constant come from?

Cumulative Distribution Function

Asymptotic Limits

Right Continuity Jump

Monotonic Accumulation

Binomial Case

Continuous Random Variables

Key Fact

Formal Proof

Probability Density Function

The Uniform Case

Where does the normalization constant come from?