Probability Definitions and Distribution Functions

Exam Tip: When working with discrete variables, always ensure that your individual probabilities are non-negative and that their total sum equals $1$. If they do not, you have likely missed an outcome or made a calculation error.

Continuous Random Variables

Continuous random variables involve values that can take any value within a given range or interval. Unlike discrete counts, these are often measurements, such as temperature, pressure, the angle of a spinning top, or the age of trees in a forest.

Because there are an infinite number of possible values between any two points, we cannot assign a probability to a single, exact point. Instead, we use a Probability Density Function (PDF), denoted as $f(x)$, to describe the distribution.

The Probability Density Function (PDF)

The PDF represents how densely a population is distributed around specific values. While the height of the function $f(x)$ at a certain point indicates the relative likelihood of the variable falling near that value, the function itself does not represent a probability. Instead, probability is found by calculating the area under the curve between two points.

For a function to be a valid PDF, it must satisfy two essential properties:

1. Non-negativity: The value of the PDF is never negative ($f(x) \ge 0$).
2. Total Area: The total area under the full curve must equal $1$: $\int_{-\infty}^{\infty} f(x) \, dx = 1$

Calculating Probability in Continuous Distributions

To find the probability that a variable $X$ falls between two values, $a$ and $b$, we calculate the definite integral of the PDF: $P(a \le X \le b) = \int_{a}^{b} f(x) \, dx$

An important consequence of this is that the probability of a continuous variable taking on one specific, exact value is zero: $P(a \le X \le a) = \int_{a}^{a} f(x) \, dx = 0$

Worked Example: Exponential Distribution

Consider the diameter of copper wires produced in a plant. The thickness $X$ follows an exponential distribution with the following PDF: $f(x) = \begin{cases} 20e^{-20(x-12.5)} & x \ge 12.5 \\ 0 & x < 12.5 \end{cases}$

Question: What proportion of wires must be discarded if thicknesses greater than $12.6\text{ mm}$ are unacceptable?

To solve this, we find the area under the curve from $12.6$ to infinity: $\begin{aligned} P(X > 12.6) &= \int_{12.6}^{\infty} 20e^{-20(x-12.5)} \, dx \\ &= \left[ -e^{-20(x-12.5)} \right]_{12.6}^{\infty} \\ &= 0 - (-e^{-20(12.6-12.5)}) \\ &= e^{-20(0.1)} \\ &= e^{-2} \approx 0.135 \end{aligned}$

Thus, approximately $13.5%$ of the wires would be discarded.