Continuous Probability Distributions: Density, Uniform, and Normal Models

Definition of Continuous Random Variables: A continuous random variable is defined by its ability to assume an uncountable number of possible values within a given interval.
Probability Density Function (PDF): The behavior of a continuous random variable is described by its PDF, denoted as $f(x)$ .
- Non-negativity: The function must be non-negative for all values within its support: $f(x) ≥ 0$ for all $x ∈ [a, b]$ .
- Boundary Conditions: $f(x) = 0$ for all values of $x$ outside the support interval $[a, b]$ .
- Support: The interval $[a, b]$ is specifically referred to as the support of the density function $f(x)$ .
- Total Probability Requirement: The total area under the density curve must equal exactly 1. Mathematically, this is expressed through the integral of the function over its range: $\int_{-\infty}^{\infty} f(x)\,dx = 1$ Or, restricted to its support: $\int_a^b f(x)\,dx = 1$

Definition: A uniform distribution occurs when a continuous random variable is equally likely to take any value within a specified range from $a$ to $b$ .
Mathematical Representation: The PDF of a uniform distribution is a constant value expressed as: $f(x) = \begin{cases} \frac{1}{b-a} & \text{if } x ∈ [a, b] \\ 0 & \text{otherwise} \end{cases}$
Properties Check:
- $f(x) ≥ 0$ for all $x$ .
- $\int_a^b f(x)\,dx = 1$ .
Practical Example: Daily Gasoline Sales:
- Scenario: Daily gasoline sales at a service station follow a uniform distribution with a minimum ( $a$ ) of $2000\,gallons$ and a maximum ( $b$ ) of $5000\,gallons$ .
- PDF Calculation: The density value is $f(x) = \frac{1}{5000 - 2000} = \frac{1}{3000}$ .
- Probability Calculation 1: Find the probability that sales fall between $2500$ and $3000\,gallons$ ( $Pr(2500 ≤ X ≤ 3000)$ ). This is calculated as the area of the rectangle: $\text{Area} = (3000 - 2500) \times \frac{1}{3000} = \frac{500}{3000} = 0.1667$
- Probability Calculation 2: Find the probability that the station sells at least $4000\,gallons$ (Pr(X > 4000)). $\text{Area} = (5000 - 4000) \times \frac{1}{3000} = \frac{1000}{3000} = 0.3333$

Notation and Definition: A random variable $X$ that is normally distributed is denoted as $X \sim N(μ, σ^2)$ , where $μ$ is the mean and $σ^2$ is the variance.
Probability Density Function: The mathematical formula for the normal density curve is: $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left[-\frac{1}{2}\left(\frac{x-μ}{σ}\right)^2\right]$
- Range: The variable exists on the interval -∞ < x < ∞.
- Shape and Symmetry: The distribution is bell-shaped and perfectly symmetric around its mean ( $μ$ ).
Visualizing Parameters:
- Effect of the Mean: Increasing the mean ( $μ$ ) shifts the entire curve to the right along the horizontal axis while maintaining its shape. For example, comparing a curve with $μ = 1$ to one with $μ = 2$ (keeping variance constant).
- Effect of Standard Deviation: Increasing the standard deviation ( $σ$ ) "flattens" the curve, spreading it out over a wider range. Conversely, a smaller $σ$ (e.g., $σ = 0.5$ vs. $σ = 1$ or $σ = 3$ ) results in a taller, narrower peak centered at the mean.

The Z-Transformation: Any normal distribution can be transformed into the standard normal distribution using the formula: $Z = \frac{X - μ}{σ}$
Properties of Z:
- Expected Value: $E(Z) = 0$ .
- Variance: $V(Z) = 1$ .
- Distribution: $Z \sim N(0, 1)$ .
Standard Normal Density Function: The PDF for the variable $Z$ simplifies to: $f(z) = \frac{1}{\sqrt{2π}} \exp\left(-\frac{z^2}{2}\right)$
- Range: -∞ < z < ∞.
Practical Example: Normal Distribution Demand:
- Context: At a gasoline station, daily demand for regular gasoline is normally distributed with $μ = 1000\,gallons$ and $σ = 100\,gallons$ .
- Current Inventory: The manager has exactly $1100\,gallons$ in storage.
- Problem: Find the probability that the current stock is sufficient to satisfy today's demand until the next delivery at close of business (Pr(X < 1100)).
- Calculation via Standardization: Pr(X < 1100) = Pr\left(\frac{X - 1000}{100} < \frac{1100 - 1000}{100}\right) Pr(Z < 1)

Case 1: Left-Tail Probability: To find the probability of $Z$ being less than a specific negative value, such as Pr(Z < -1.52), look up the value directly in the table.
- Result: Pr(Z < -1.52) = 0.0643.
Case 2: Right-Tail Probability: To find the probability of $Z$ being greater than a value, use the complement rule: Pr(Z > 1.8) = 1 - Pr(Z < 1.8) $1 - 0.9641 = 0.0359$
Case 3: Interval Probability: To find the probability between two values, subtract the cumulative probability of the lower bound from the cumulative probability of the upper bound: Pr(-1.3 < Z < 2.1) = Pr(Z < 2.1) - Pr(Z < -1.3) $0.9821 - 0.0968 = 0.8853$

Conceptual Link: In finance, risk is frequently measured by variance and standard deviation. The following example demonstrates how a higher standard deviation increases the probability of poor outcomes (e.g., losing money).
Investment Return Scenario: Consider an investment where the return is normally distributed with a mean ( $μ$ ) of $10\%$ .
Comparison 1: Lower Volatility ( $σ = 5\%$ ):
- Problem: Find the probability of losing money (Pr(X < 0)).
- Calculation: Pr(X < 0) = Pr\left(\frac{X - 10}{5} < \frac{0 - 10}{5}\right) = Pr(Z < -2).
- Result: $0.0228$ (or $2.28\%$ probability of loss).
Comparison 2: Higher Volatility ( $σ = 10\%$ ):
- Problem: Find the probability of losing money (Pr(X < 0)).
- Calculation: Pr(X < 0) = Pr\left(\frac{X - 10}{10} < \frac{0 - 10}{10}\right) = Pr(Z < -1).
- Result: $0.1587$ (or $15.87\%$ probability of loss).

Definition: A quantile is a point $z_α$ such that the probability of the random variable exceeding that point is exactly $α$ .
- Relationship: Pr(Z > z_α) = α.
Specific Example: Find the value of $z_α$ for $α = 0.025$ .
- Consulting the standard normal table for the area corresponding to a tail of $0.025$ :
- Pr(Z > 1.96) = 0.025
- Therefore, $z_{0.025} = 1.96$ .