Continuous Random Variables, CDF, and PDF – Study Notes

Continuous Random Variables

Definition and Examples

A continuous random variable is a variable that can take on an uncountably infinite number of possible values, typically over an interval or the entire real line. Examples include measurements such as weight, height, and time. These variables are defined over sets such as R (the real numbers), a single interval [a, b], or a union of disjoint intervals.

• Key Point 1: Continuous random variables are not countable; their values form a continuum.

• Key Point 2: Common examples include physical measurements and durations.

• Example: The exact time a bus arrives at a stop is a continuous random variable.

Cumulative Distribution Function (CDF)

Definition and Properties

The cumulative distribution function (CDF) of a random variable X, denoted as FX(x), gives the probability that X will take a value less than or equal to x. It is defined as:

$F_X(x) = P(X \leq x) = P[-\infty < X \leq x]$

• Key Point 1: The CDF is a non-decreasing function.

• Key Point 2: $\lim_{x \to -\infty} F_X(x) = 0$

• Key Point 3: $\lim_{x \to \infty} F_X(x) = 1$

• Key Point 4: $P(a < X \leq b) = F_X(b) - F_X(a)$

Observations about Continuous Random Variables

Probability at a Point and Interval Probabilities

• Key Point 1: For a single point a, $P[X = a] = 0$ for continuous random variables.

• Key Point 2: The probability over an interval does not depend on whether endpoints are included: $P(a < X < b) = P(a < X \leq b) = P(a \leq X < b) = P(a \leq X \leq b)$

Probability Density Function (PDF)

Definition and Relationship to CDF

The probability density function (PDF) of a continuous random variable X is the derivative of its CDF, provided the CDF is differentiable. The PDF, denoted as fX(x), satisfies:

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

Conversely, the CDF can be obtained by integrating the PDF:

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

• Key Point 1: The PDF describes the relative likelihood for the random variable to take on a given value.

• Key Point 2: The area under the PDF curve over an interval gives the probability that the variable falls within that interval.

Properties of the PDF

• Property 1: $f_X(x) \geq 0$ for all x.

• Property 2: $\int_{-\infty}^{\infty} f_X(x) dx = 1$

• Property 3: $P[a \leq X \leq b] = \int_{a}^{b} f_X(x) dx$

Worked Examples

Example 3.11 & 3.12: Verifying a PDF and Calculating Probabilities

Suppose the error in reaction temperature X (in °C) has the PDF:

$f(x) = \begin{cases} \frac{x^2}{3}, & -1 < x < 2 \\ 0, & \text{elsewhere} \end{cases}$

• 1. Verify that f(x) is a density function: Check that $f(x) \geq 0$ and $\int_{-1}^{2} \frac{x^2}{3} dx = 1$.

• 2. Find $P(0 < X \leq 1)$: $P(0 < X \leq 1) = \int_{0}^{1} \frac{x^2}{3} dx$

• 3. Find F(x): $F(x) = \int_{-1}^{x} \frac{t^2}{3} dt$ for $-1 < x < 2$

• 4. Use F(x) to evaluate $P(0 < X \leq 1)$: $F(1) - F(0)$

Example 3.13: Application to Bidding

The Department of Energy estimates the PDF for the winning bid Y as:

$f(y) = \begin{cases} \frac{5}{8b}, & \frac{2}{5}b \leq y \leq 2b \\ 0, & \text{elsewhere} \end{cases}$

• Find F(y): Integrate f(y) over the support interval.

• Find $P(Y < b)$: $F(b)$ gives the probability the winning bid is less than the DOE’s estimate.

Practice Exercises

Exercise 3.12: CDF and Probability Calculations

Given the CDF for T, the number of years to maturity for a bond:

t	F(t)
t < 1	0
1 ≤ t < 3	1/4
3 ≤ t < 5	1/2
5 ≤ t < 7	3/4
t ≥ 7	1

• (a) $P(T = 5)$: For continuous variables, $P(T = 5) = 0$.

• (b) $P(T > 3)$: $1 - F(3) = 1 - 1/2 = 1/2$

• (c) $P(1.4 < T < 6)$: $F(6) - F(1.4) = 3/4 - 1/4 = 1/2$

• (d) $P(T \leq 5 \mid T \geq 2)$: $\frac{F(5) - F(2)}{1 - F(2)} = \frac{3/4 - 1/4}{1 - 1/4} = \frac{1/2}{3/4} = 2/3$

Exercise 3.21: Finding k and Calculating Probabilities

Given the PDF:

$f(x) = \begin{cases} k\sqrt{x}, & 0 < x < 1 \\ 0, & \text{elsewhere} \end{cases}$

• (a) Evaluate k: Set $\int_{0}^{1} k\sqrt{x} dx = 1$ and solve for k.

• (b) Find F(x) and $P(0.3 < X < 0.6)$: $F(x) = \int_{0}^{x} k\sqrt{t} dt$; $P(0.3 < X < 0.6) = F(0.6) - F(0.3)$

Additional info: These exercises reinforce the calculation of probabilities using both the PDF and CDF, and illustrate the properties of continuous random variables in practical contexts.