IB AA: Continuous Random Variables

Continuous random variables form the mathematical backbone for modeling uncountably infinite outcomes, such as time, distance, or temperature. Mastering them is crucial for IB Analysis & Approaches, as they bridge calculus with real-world probability and are a frequent exam topic requiring both conceptual understanding and computational skill.

Probability Density Functions: The Foundation

A probability density function (PDF), denoted $f(x)$, describes the relative likelihood of a continuous random variable $X$ taking on a specific value. Unlike discrete probability, where $P(X=x)$ has meaning, for continuous variables, the probability at a single point is zero. Instead, probability is defined over intervals as the area under the PDF curve. Every valid PDF must satisfy two key properties. First, it must be non-negative for all $x$: $f(x) \geq 0$. Second, the total area under the curve must equal 1, representing certainty: $$\int_{-\infty}^{\infty} f(x) \, dx = 1.$$

The process of ensuring the second property holds is called normalization. You will often be given a function proportional to a PDF, such as $g(x)=kx$ for $0 \leq x \leq 2$, and must find the constant $k$ that makes it a valid PDF. You do this by setting the integral equal to 1: $\int_{0}^{2} kx \, dx = 1$. Solving gives $k[ \frac{x^2}{2} ]_{0}^{2} = 2k = 1$, so $k=\frac{1}{2}$. Think of the PDF as a smooth histogram; its height shows density, not probability, and the total area of all bars sums to 1.

Cumulative Distribution Functions: From Density to Probability

The cumulative distribution function (CDF), denoted $F(x)$, gives the probability that $X$ is less than or equal to a specific value: $F(x) = P(X \leq x)$. It is computed directly from the PDF by integration: $$F(x) = \int_{-\infty}^{x} f(t) \, dt.$$ The CDF is a non-decreasing function that ranges from 0 to 1, with $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to \infty} F(x) = 1$.

To find probabilities for an interval $a < X \leq b$, you use the CDF: $P(a < X \leq b) = F(b) - F(a)$. The fundamental theorem of calculus shows the inverse relationship: the PDF is the derivative of the CDF, $f(x) = F'(x)$, wherever the derivative exists. For a worked example, consider the PDF $f(x)= \frac{1}{2}x$ for $0 \leq x \leq 2$. Its CDF is $F(x) = \int_{0}^{x} \frac{1}{2}t \, dt = \frac{x^2}{4}$ for $0 \leq x \leq 2$. To find $P(1 < X < 1.5)$, compute $F(1.5) - F(1) = \frac{(1.5)^2}{4} - \frac{1^2}{4} = 0.5625 - 0.25 = 0.3125$.

Expected Value and Variance: Measuring Center and Spread

The expected value (mean) of a continuous random variable, denoted $E(X)$ or $\mu$, is the long-run average outcome, weighted by probability density. It is calculated by integrating $x$ times the PDF over all possible values: $$E(X) = \int_{-\infty}^{\infty} x f(x) \, dx.$$ Variance, denoted $Var(X)$ or $\sigma^2$, measures the spread or dispersion around the mean, defined as the expected value of the squared deviation: $Var(X) = E[(X-\mu)^2]$. A more computational formula is $Var(X) = E(X^2) - [E(X)]^2$, where $E(X^2) = \int_{-\infty}^{\infty} x^2 f(x) \, dx$. The standard deviation is simply the square root of the variance: $\sigma = \sqrt{Var(X)}$.

Let's compute these for the PDF $f(x)= \frac{1}{2}x$ on $[0,2]$. First, the mean: $$E(X) = \int_{0}^{2} x \cdot \frac{1}{2}x \, dx = \frac{1}{2} \int_{0}^{2} x^2 \, dx = \frac{1}{2} \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}.$$ Next, find $E(X^2)$: $$E(X^2) = \int_{0}^{2} x^2 \cdot \frac{1}{2}x \, dx = \frac{1}{2} \int_{0}^{2} x^3 \, dx = \frac{1}{2} \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{1}{2} \cdot 4 = 2.$$ Thus, variance is $Var(X) = E(X^2) - [E(X)]^2 = 2 - \left( \frac{4}{3} \right)^2 = 2 - \frac{16}{9} = \frac{2}{9}$.

Median and Mode: Other Measures of Central Tendency

For continuous variables, the median is the value $m$ such that half the probability lies below it and half above, satisfying $P(X \leq m) = 0.5$. In terms of the CDF, you solve $F(m) = 0.5$. Using our example CDF $F(x)=\frac{x^2}{4}$, set $\frac{m^2}{4} = 0.5$, so $m^2 = 2$ and $m = \sqrt{2} \approx 1.414$. The mode is the value at which the PDF $f(x)$ achieves its maximum. It represents the most likely outcome in a density sense. For $f(x)=\frac{1}{2}x$ on $[0,2]$, the function increases linearly, so the maximum is at the right endpoint: $x=2$. In symmetric distributions like the normal, the mean, median, and mode coincide.

Common Continuous Distributions and Real-World Modeling

The uniform distribution is the simplest continuous model, where all intervals of equal length have the same probability. If $X$ is uniform on $[a, b]$, its PDF is constant: $f(x) = \frac{1}{b-a}$ for $a \leq x \leq b$. Its CDF is a straight line: $F(x) = \frac{x-a}{b-a}$. The mean is the midpoint $E(X) = \frac{a+b}{2}$, and variance is $Var(X) = \frac{(b-a)^2}{12}$. Uniform distributions model scenarios like random number generation or waiting for a bus that arrives at fixed intervals.

Other essential continuous distributions include the exponential distribution for modeling waiting times or decay processes, and the normal distribution for bell-shaped data like heights or test scores. Connecting these to real-world modeling involves identifying the key characteristics of a phenomenon—such as whether it is memoryless (exponential) or symmetric (normal)—and selecting the appropriate PDF. For instance, the time between customer arrivals at a store might follow an exponential distribution, while errors in a physical measurement often follow a normal distribution. The power of continuous random variables lies in using integration over these PDFs to make precise probabilistic predictions about complex, measurable events.

Common Pitfalls

1. Treating PDF value as a probability: A common error is interpreting $f(a)$ as $P(X=a)$. Remember, for continuous variables, $P(X=a)=0$. Probability is always an area under the PDF curve, so you must integrate over an interval.

1. Misapplying integration limits: When computing CDFs or expected values, ensure your limits match the support of the PDF. For a PDF defined only on $[0, 2]$, integrals for $F(x)$ should run from 0 to $x$, not from $-\infty$. Similarly, $E(X)$ integrates from 0 to 2.

1. Forgetting to normalize: If given a function like $g(x)=c(3-x)$ on an interval, you must first find $c$ by setting $\int g(x) dx = 1$ to obtain the valid PDF $f(x)$. Skipping this step leads to incorrect probabilities and expectations.

1. Incorrectly finding the median: The median satisfies $F(m)=0.5$, not $f(m)=0.5$. Confusing the PDF and CDF here will yield the wrong value. Always use the CDF equation to solve for the median.

Summary

• A probability density function (PDF) $f(x)$ must be non-negative and integrate to 1 over all space; it defines probabilities via area under the curve.
• The cumulative distribution function (CDF) $F(x)$ is found by integrating the PDF and gives $P(X \leq x)$; probabilities for intervals are differences in CDF values.
• Expected value and variance are calculated through integration: $E(X) = \int x f(x) dx$ and $Var(X) = E(X^2) - [E(X)]^2$.
• The median is the solution to $F(m)=0.5$, and the mode is the value maximizing $f(x)$.
• The uniform distribution has a constant PDF and is a foundational model; other distributions like exponential and normal extend modeling to various real-world scenarios.
• Always remember that for continuous random variables, probability is area, not height, and proper integration techniques are essential for accurate computation.