Properties of Probability Density Functions

$P(a \leq X \leq b) = \int_{a}^{b} f(x) dx$

This integral effectively sums up the "density" across the interval to give us a total probability.

The "Zero Probability" Paradox

A common point of confusion for students is why $P(a \leq X \leq a) = 0$. Mathematically, the integral from a point to itself has a width of zero:

$\int_{a}^{a} f(x) dx = 0$

Because the probability of a single point is zero, it does not matter whether you use strict inequalities ($<$) or inclusive inequalities ($\leq$). In a continuous distribution:

$P(a < X < b) = P(a \leq X \leq b)$

Exam Tip: If a question asks for the probability of a continuous variable being "exactly" a certain value, the answer is always 0. Always focus on the interval provided.

Fundamental Properties of a PDF

For a mathematical function to be considered a valid Probability Density Function, it must satisfy two strict criteria:

1. Non-negativity: The function must never be negative. For all possible values of $x$, the density must be zero or greater: $f(x) \geq 0$
2. Total Area: The total area under the entire curve, from negative infinity to positive infinity, must equal exactly 1. This represents the certainty that the variable will take some value within the possible range: $\int_{-\infty}^{\infty} f(x) dx = 1$

If a function is not defined for a certain range (for example, tree ages cannot be negative), the PDF is assumed to be zero in those regions.

Special Case: The Uniform Distribution

One of the simplest types of continuous distributions is the Uniform Distribution. This occurs when the probability is spread perfectly evenly across an interval from $a$ to $b$.

Imagine a spinning top that can come to rest at any angle between $0^\circ$ and $360^\circ$ with equal likelihood. Because the total area must be 1 and the shape is a rectangle, the height (the PDF value) must be constant. The formula for a Uniform PDF is:

$f(x) = \frac{1}{b - a} \text{ for } a \leq x \leq b$

In the spinning top example, the PDF would be $f(x) = \frac{1}{360}$ for any angle between 0 and 360.

Worked Example: Verifying a PDF

Let's test if a function meets the criteria of a PDF. Suppose we are given: $f(x) = 1.5x^2 \text{ for } -1 < x < 1, \text{ and } 0 \text{ elsewhere}$

Step 1: Check Non-negativity Since $x^2$ is always non-negative and $1.5$ is positive, $f(x) \geq 0$ for all $x$. Property 1 is satisfied.

Step 2: Check Total Area We integrate the function over its defined range: $\int_{-1}^{1} 1.5x^2 dx = \left[ \frac{1.5x^3}{3} \right]_{-1}^{1} = \left[ 0.5x^3 \right]_{-1}^{1}$ Substituting the limits: $(0.5(1)^3) - (0.5(-1)^3) = 0.5 - (-0.5) = 1$ Since the total area is 1, Property 2 is satisfied. This function is a valid PDF.