TABLE OF CONTENTS

Probability II

VOL. IMAR 2026

CH1 — Part 2

Example 3

Normalisation & Derivation

If $X$ is a random variable having the pdf of the form: $f_X(x) = \begin{cases} cx & 0 < x < 2 \\ 0 & \text{otherwise} \end{cases}$

Example 3 (a)

Find c

Find the constant

c

that makes

f_X(x)

a valid pdf.

Solution

For $f_X(x)$ to be a valid pdf, the total area must be 1:

\int_{-\infty}^{\infty} f_X(x)\,dx = 1 \implies \int_0^2 cx\,dx = 1

c\left[ \frac{x^2}{2} \right]_0^2 = 1 \implies 2c = 1 \implies c = 1/2

So:

f_X(x) = \begin{cases} \tfrac{1}{2}x & 0 < x < 2 \\ 0 & \text{otherwise} \end{cases}

PDF Normalization

Total Area0.60

Control Constant

c

c = 0.30

3 (b)

Deriving the Piecewise CDF

To find $F_X(x)$ , we integrate the piecewise density function $f_X(t)$ from $-\infty$ to $x$ across all three mathematical regions.

Mathematical Derivation Engine

Use the slider to see how the integral accumulates over each mathematical region.

— f(x): linear ramp from 0 to 1 on [0, 2].

— F(x): parabola 1/4x² on [0, 2].

Case 1: -0.5 ≤ x < 0

F_X(x) = \int_{-\infty}^{x} 0 \, dt = 0

F_X(-0.50) = 0

Density Function f(t)

Cumulative Tracing F(x)

Evaluation

F(-0.50) = 0.000

x = -0.5evaluation x = -0.50x = 2.5

Complete Definition Result

F_X(x) = \begin{cases} \color{#9b2c1c} 0 & x < 0 \\ \tfrac{1}{4}x^2 & 0 \leq x < 2 \\ 1 & x \geq 2 \end{cases}

Example 4

Area Calculation

Now that we have the PDF and CDF for Example 3, we can calculate the probability that $X$ falls between 1 and 2. This interactive visualization shows both the Integration Method (area under the PDF) and the Subtraction Method (CDF evaluation).

Example 4

P(1 < X < 2)

Find

P(1 < X < 2)

Solution

We can solve this using either the integral of the PDF or the subtraction of the CDF:

P(1 < X < 2) = \int_1^2 \frac{1}{2}x\,dx = 3/4

Or using CDF subtraction:

F_X(2) - F_X(1) = 1 - 1/4 = 3/4

Probability as Area

a = 1.00

b = 2.00

Bound A

Bound B

Integration Method

P(1.0 \leq X \leq 2.0) = \int_{1.0}^{2.0} f_X(x)\,dx \approx 0.7500

CDF Subtraction Method

F_X(2.0) - F_X(1.0) = 1.0000 - 0.2500 = 0.7500

Probability as Area

Remark — Area Interpretation

For a continuous random variable, the probability $P(a \leq X \leq b)$ is exactly the area under the density curve between $a$ and $b$ .

P(a \leq X \leq b) = \int_a^b f_X(x)\,dx = F_X(b) - F_X(a)

Interval Equivalence

Because $P(X=a)=0$ and $P(X=b)=0$ for continuous variables, all the following probabilities are equal:

P(a < X < b)

P(a \leq X < b)

P(a < X \leq b)

P(a \leq X \leq b)

Probability as Area

a = -1.00

b = 1.00

Bound A

Bound B

Integration Method

P(-1.0 \leq X \leq 1.0) = \int_{-1.0}^{1.0} f_X(x)\,dx \approx 0.6827

CDF Subtraction Method

F_X(1.0) - F_X(-1.0) = 0.8413 - 0.1587 = 0.6827

Example 5

The Tent Function

This example demonstrates the Fundamental Theorem of Calculus in its piecewise form. Notice how the continuity of $F_X(x)$ is preserved at the boundaries.

The "tent" function is a triangular density symmetric around $x=1$ : $f_X(x) = \begin{cases} x & 0 < x < 1 \\ 2-x & 1 < x < 2 \\ 0 & \text{otherwise} \end{cases}$

Example 5

Comprehensive Analysis

Find (a)

F_X(x)

, (b)

P(X \leq 0.2)

, (c)

P(X \leq 1.5)

, (d)

P(0.2 < X < 1.5)

, (e)

P(0.1 < X \leq 0.9)

, (f)

P(X > 6)

Solution

Final CDF

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

P(X \leq 0.2)

0.02

P(X \leq 1.5)

0.875

P(0.2 < X < 1.5)

0.855

P(0.1 < X \leq 0.9)

0.40

P(X > 6)

Interactive Visualisation

Building the Tent CDF

0 ≤ x < 1

Density

f(x) = x

Cumulative

F(x) = x^2/2

1 ≤ x < 2

Density

f(x) = 2 - x

Cumulative

F(x) = 1 - \frac{(2-x)^2}{2}

Direct Manipulation Active

Current Value

x = -0.20

Density Function Trace

Cumulative Distribution Trace

x = -0.20

Calculus Computation

Integration Result

F_X(-0.20) = \int_{-\infty}^{-0.20} f_X(t)\,dt = 0.0000

Region 0Live evaluation

Exercises

Exercise 1

Let

f_X(x) = 2

for

1 \leq x \leq 1.5

, and

0

otherwise. Is this a valid pdf?

Step-by-Step Derivation

Step 1: Area Visualization

Identify the region for the PDF

f_X(x) = 2

[1, 1.5]

Exercise 2

f_X(x) = cx^3(1-x)

for

0 \leq x \leq 1

.
(a) Find c.
(b) Find

P(X \leq 1/2)

.
(c) Find

P(X \geq 1/2)

Step-by-Step Derivation

Step 1: (a) Find c

Calculate the constant

c

such that

\int_0^1 cx^3(1-x)\,dx = 1

Example 5

The Tent Function

This example demonstrates the Fundamental Theorem of Calculus in its piecewise form. Notice how the continuity of $F_X(x)$ is preserved at the boundaries.

The "tent" function is a triangular density symmetric around $x=1$ : $f_X(x) = \begin{cases} x & 0 < x < 1 \\ 2-x & 1 < x < 2 \\ 0 & \text{otherwise} \end{cases}$

Example 5

Comprehensive Analysis

Find (a)

F_X(x)

, (b)

P(X \leq 0.2)

, (c)

P(X \leq 1.5)

, (d)

P(0.2 < X < 1.5)

, (e)

P(0.1 < X \leq 0.9)

, (f)

P(X > 6)

Solution

Final CDF

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

P(X \leq 0.2)

0.02

P(X \leq 1.5)

0.875

P(0.2 < X < 1.5)

0.855

P(0.1 < X \leq 0.9)

0.40

P(X > 6)

Exercises

Exercise 1

Let

f_X(x) = 2

for

1 \leq x \leq 1.5

, and

0

otherwise. Is this a valid pdf?

Step-by-Step Derivation

Step 1: Area Visualization

Identify the region for the PDF

f_X(x) = 2

[1, 1.5]

Exercise 2

f_X(x) = cx^3(1-x)

for

0 \leq x \leq 1

.
(a) Find c.
(b) Find

P(X \leq 1/2)

.
(c) Find

P(X \geq 1/2)

Step-by-Step Derivation

Step 1: (a) Find c

Calculate the constant

c

such that

\int_0^1 cx^3(1-x)\,dx = 1