Continuous random variables: VCE Math Methods Unit 4

What this dot point is asking

VCAA wants you to define a continuous random variable through its probability density function (pdf), compute probabilities as definite integrals, and compute expected value, variance and standard deviation as integrals. The dot point is the integration application that bridges Unit 4 calculus to Unit 4 statistics.

What is a continuous random variable

A continuous random variable $X$ takes values in a continuum (an interval of real numbers), not a finite or countable set. Examples: the time a customer waits in a queue, the length of a manufactured part, the maximum temperature on a given day.

Because $X$ has uncountably many possible values, the probability that $X$ takes any specific single value is zero. Probabilities are computed for intervals, not for points.

The probability density function

A continuous random variable $X$ is described by its probability density function $f(x)$, with the property:

A valid pdf must satisfy two conditions:

1. Non-negative. $f(x) \geq 0$ for all $x$.
2. Integrates to 1. $\int_{-\infty}^{\infty} f(x) \, dx = 1$.

If $f$ is given as zero outside a finite interval $[a, b]$ (the most common Unit 4 case), condition 2 becomes $\int_{a}^{b} f(x) \, dx = 1$. The interval $[a, b]$ is called the support of the random variable.

Finding a normalising constant

A typical Paper 2 question gives $f(x) = k g(x)$ on some interval and asks for the value of $k$ that makes $f$ a valid pdf.

Procedure: Set $\int_{a}^{b} k g(x) \, dx = 1$. Compute the integral; solve for $k$.

Example. If $f(x) = k x (2 - x)$ on $[0, 2]$, normalising gives $k = \frac{3}{4}$ (worked above).

Computing probabilities

For a continuous random variable with pdf $f$:

The endpoints' open or closed nature does not matter: $P(X = a) = 0$ for any single value, so:

For probabilities on one side only:

For an $X$ with support $[a, b]$, these reduce to integrals over the appropriate sub-interval.

The cumulative distribution function

The cumulative distribution function (cdf) of $X$ is:

Properties of the cdf:

• $F$ is non-decreasing.
• $F(-\infty) = 0$ and $F(\infty) = 1$.
• $P(a \leq X \leq b) = F(b) - F(a)$.

The cdf is the antiderivative of the pdf (with appropriate constant), so $F'(x) = f(x)$ where $f$ is continuous. This is the fundamental theorem applied to probability.

Expected value (mean)

The expected value of a continuous random variable $X$ is:

For $X$ with support $[a, b]$:

Interpretation: $E(X)$ is the centre of mass of the pdf, the long-run average of many independent observations of $X$.

Linearity of expectation

$E(aX + b) = a E(X) + b$ for constants $a, b$.

$E(X + Y) = E(X) + E(Y)$ for any two random variables on the same sample space.

Expectation of a function

For a function $g$:

In particular, $E(X^2) = \int_{a}^{b} x^2 f(x) \, dx$.

Variance and standard deviation

The variance of $X$ is:

The right-hand identity is the working formula; compute $E(X^2)$ and $E(X)$ separately, then subtract.

The standard deviation is $\sigma = \sqrt{\text{Var}(X)}$.

Properties

$\text{Var}(aX + b) = a^2 \text{Var}(X)$. Adding a constant does not change variance; multiplying by a constant scales variance by the constant squared.

Worked variance

For the pdf $f(x) = \frac{x}{8}$ on $[0, 4]$ (worked above), $E(X) = \frac{8}{3}$, $E(X^2) = 8$, $\text{Var}(X) = \frac{8}{9}$, $\sigma = \frac{2 \sqrt{2}}{3}$.

Median and quartiles

The median of a continuous random variable $X$ is the value $m$ such that $P(X \leq m) = \frac{1}{2}$. Find by solving:

The first and third quartiles satisfy similar equations with $\frac{1}{4}$ and $\frac{3}{4}$.

Median and quartiles do not, in general, equal the mean. For symmetric pdfs (rectangle, isosceles triangle, normal) the median equals the mean. For skewed pdfs they differ.

The uniform distribution

The simplest continuous random variable. $X \sim U(a, b)$ has pdf:

Properties:

• $E(X) = \frac{a + b}{2}$ (midpoint)
• $\text{Var}(X) = \frac{(b - a)^2}{12}$
• Median = $E(X)$ (symmetric)
• $P(c \leq X \leq d) = \frac{d - c}{b - a}$ for $a \leq c \leq d \leq b$

The uniform is the "no information" distribution: all values in $[a, b]$ are equally likely.

Examples in context

Example 1. Bus waiting time. Wait time $X$ (minutes) at a stop is uniform on $[0, 10]$, so $f(x) = \frac{1}{10}$ for $0 \le x \le 10$. The probability of waiting at most $3$ minutes is $P(X \le 3) = \int_0^3 \frac{1}{10}\,dx = \frac{3}{10} = 0.3$. The mean wait is $E(X) = \frac{0 + 10}{2} = 5$ minutes, and the variance is $\frac{(10 - 0)^2}{12} = \frac{100}{12} = 8.33$.

Example 2. Normalising a lifetime density. A component's lifetime $X$ (years) is modelled by $f(x) = kx(2 - x)$ on $[0, 2]$, zero elsewhere. The normalisation condition requires $\int_0^2 kx(2 - x)\,dx = 1$. Evaluating, $k\left[x^2 - \frac{x^3}{3}\right]_0^2 = k\left(4 - \frac{8}{3}\right) = \frac{4k}{3} = 1$, so $k = \frac{3}{4}$. The density is symmetric about $x = 1$, so the mean lifetime is $E(X) = 1$ year.

Try this

Q1. A pdf is $f(x) = kx$ on $[0, 4]$, zero elsewhere. Find $k$. [2 marks]

• Cue. $\int_0^4 kx\,dx = 8k = 1$, so $k = \frac{1}{8}$.

Q2. For $f(x) = \frac{1}{8}x$ on $[0, 4]$, find $P(1 \le X \le 3)$. [2 marks]

• Cue. $\left[\frac{x^2}{16}\right]_1^3 = \frac{9}{16} - \frac{1}{16} = \frac{1}{2}$.

Q3. For $f(x) = \frac{1}{8}x$ on $[0, 4]$, find the median $m$. [3 marks]

• Cue. $\int_0^m \frac{x}{8}\,dx = \frac{m^2}{16} = \frac{1}{2} \Rightarrow m^2 = 8 \Rightarrow m = 2\sqrt{2} \approx 2.83$.