Continuous Random Variables

This lesson introduces continuous random variables — variables that can take any value in an interval (or the whole real line). Unlike a discrete variable, a continuous variable assigns zero probability to any single value; instead, probability is area under a probability density function (PDF), computed by integration. This is the calculus-meets-statistics heart of 7367/3S, and it sets up the expectation, CDF and named-distribution work that follows.

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1. Where this sits in AQA 7367

This belongs to the Paper 3 Statistics option (7367/3S) (AO weighting AO1 40% / AO2 25% / AO3 35%). Setting up and evaluating the defining integrals is AO1; finding unknown constants and probabilities across piecewise boundaries is AO3; interpreting "density is not probability" is AO2. The prerequisite is A-Level Mathematics integration (definite integrals, areas) and the discrete-variable language from Lesson 1 — every formula here is the integral analogue of a discrete sum.

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2. Core theory: from sums to integrals

Why $P(X = x) = 0$P(X=x)=0

For a discrete variable, $P(X = x)$P(X=x) can be positive. For a continuous variable,

$P(X = x) = 0 \quad\text{for every single value } x,$P(X=x)=0for every single value x,

because a single point has zero width, and area over zero width is zero. Probability is assigned to intervals:

$P(a \leq X \leq b) = \int_a^b f(x)\,dx,$P(a≤X≤b)=∫ab​f(x)dx,

where $f$f is the PDF — the continuous analogue of the probability mass function, with $\int$∫ replacing $\sum$∑.

Because endpoints carry no probability, for a continuous $X$X all four are equal: $P(a \leq X \leq b) = P(a < X < b) = P(a \leq X < b) = P(a < X \leq b)$P(a≤X≤b)=P(a<X<b)=P(a≤X<b)=P(a<X≤b).

Conditions for a valid PDF

Property	Requirement
Non-negativity	$f(x) \geq 0$f(x)≥0 for all $x$x
Normalisation	$\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$∫−∞∞​f(x)dx=1

Crucially, $f(x)$f(x) is a density, not a probability — it may exceed 1. Only the area is a probability, and the total area is 1. For example $f(x) = 3$f(x)=3 on $[0, \tfrac13]$[0,31​] is a valid PDF: $f = 3 > 1$f=3>1, but $\int_0^{1/3} 3\,dx = 1$∫01/3​3dx=1.

The continuous uniform (rectangular) distribution

The simplest model is $X \sim U(a,b)$X∼U(a,b) with constant density

$f(x) = \frac{1}{b - a}, \quad a \leq x \leq b,$f(x)=b−a1​,a≤x≤b,

giving, by the standard results,

Quantity	Value
$E(X)$E(X)	$\dfrac{a+b}{2}$2a+b​
$\text{Var}(X)$Var(X)	$\dfrac{(b-a)^2}{12}$12(b−a)2​

Probabilities are just length ratios: $P(c \leq X \leq d) = \dfrac{d-c}{b-a}$P(c≤X≤d)=b−ad−c​ for $a \leq c \leq d \leq b$a≤c≤d≤b.

Reading a density: area, shape and sketches

Because probability is area under the PDF, a quick sketch is one of the most powerful checks available in this topic. Three habits pay off repeatedly.

First, the total area is always 1, so the height of the density is constrained by the width of its support. A uniform density on an interval of width $2$2 must have height $\tfrac12$21​; a uniform density on an interval of width $\tfrac13$31​ must have height $3$3. This is why a density can legitimately exceed 1: a narrow support forces a tall curve to keep the area at 1. Whenever you find a constant $k$k, sanity-check that the resulting curve encloses unit area.

Second, the shape of the curve tells you where probability concentrates. For $f(x) = \tfrac38 x^2$f(x)=83​x2 on $[0,2]$[0,2], the density rises steeply toward $x = 2$x=2, so most of the probability sits in the right-hand part of the interval — which is exactly why $P(1 \leq X \leq 2) = \tfrac78$P(1≤X≤2)=87​ is so much larger than $P(0 \leq X \leq 1) = \tfrac18$P(0≤X≤1)=81​. A glance at the graph predicts that imbalance before any integration. For the triangular density of Example 3, the symmetric "tent" shape immediately tells you the variable is most likely to land near the middle, $x = 1$x=1, and that $P(0.5 \leq X \leq 1.5)$P(0.5≤X≤1.5) (the central region) should be a clear majority of the area — confirmed as $0.75$0.75.

Third, a sketch fixes the limits of integration. Drawing the support and shading the required region is the surest way to avoid the single most common error in this topic: integrating over the wrong range. For a probability such as $P(X > 1.5)$P(X>1.5) on a support $[0,2]$[0,2], the picture makes it obvious that the limits are $1.5$1.5 to $2$2, not $1.5$1.5 to $\infty$∞.

These ideas extend to modelling a real continuous quantity. Suppose the time $X$X (in minutes, $0 \leq X \leq 5$0≤X≤5) that a customer waits at a counter is modelled by a density that decreases linearly from a maximum at $x = 0$x=0 to zero at $x = 5$x=5 — capturing "short waits are common, long waits are rare". Such a density has the form $f(x) = c(5 - x)$f(x)=c(5−x); imposing $\int_0^5 c(5-x)\,dx = 1$∫05​c(5−x)dx=1 gives $c\left[5x - \tfrac{x^2}{2}\right]_0^5 = c(25 - 12.5) = 12.5\,c = 1$c[5x−2x2​]05​=c(25−12.5)=12.5c=1, so $c = \tfrac{2}{25} = 0.08$c=252​=0.08. The probability of waiting under one minute is then $P(X < 1) = 0.08\left[5x - \tfrac{x^2}{2}\right]_0^1 = 0.08(4.5) = 0.36$P(X<1)=0.08[5x−2x2​]01​=0.08(4.5)=0.36. Choosing a sensible algebraic form for a density to match a described behaviour, then fixing its constant by normalisation, is a recurring AO3 modelling skill.

Accumulating area: a first look at the CDF

Every probability you compute in this lesson is an integral of the density from the left up to some point, and that observation leads directly to the cumulative distribution function $F$F, studied in detail in the next lesson:

$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)\,dt.$F(x)=P(X≤x)=∫−∞x​f(t)dt.

$F(x)$F(x) is the running total of probability accumulated as you sweep from left to right. It starts at 0 (no area has accumulated below the support) and climbs to 1 (all the area, at and beyond the right end of the support). Two features follow immediately from "probability is area". First, $F$F is non-decreasing: moving the upper limit to the right can only add more (non-negative) area. Second, because $F$F is an integral of $f$f, the Fundamental Theorem of Calculus gives $f(x) = F'(x)$f(x)=F′(x) — the density is the rate at which probability accumulates. Where the density is tall, probability piles up quickly and $F$F rises steeply; where the density is zero (outside the support), $F$F is flat.

This perspective re-derives the interval formula painlessly. Since the area from $a$a to $b$b is the area up to $b$b minus the area up to $a$a,

$P(a \leq X \leq b) = \int_a^b f(x)\,dx = F(b) - F(a).$P(a≤X≤b)=∫ab​f(x)dx=F(b)−F(a).

For our worked density $f(x) = \tfrac38 x^2$f(x)=83​x2 on $[0,2]$[0,2], accumulating area from 0 gives $F(x) = \int_0^x \tfrac38 t^2\,dt = \tfrac{x^3}{8}$F(x)=∫0x​83​t2dt=8x3​ for $0 \leq x \leq 2$0≤x≤2 (and 0 below, 1 above). Then $P(1 \leq X \leq 2) = F(2) - F(1) = 1 - \tfrac18 = \tfrac78$P(1≤X≤2)=F(2)−F(1)=1−81​=87​, recovering the Example-1 result instantly — and the median will satisfy $F(m) = 0.5$F(m)=0.5, the cleanest possible route to it. Keeping the area picture in mind throughout the continuous-distribution lessons is the thread that connects the PDF, the CDF, probabilities, the median and the quartiles into a single coherent story.

A fuller piecewise example

Piecewise densities deserve a second, more searching example, because they concentrate several of the skills above. Consider

$f(x) = \begin{cases} kx & 0 \leq x \leq 2 \\ k(4 - x) & 2 < x \leq 4 \\ 0 & \text{otherwise,} \end{cases}$f(x)=⎩⎨⎧​kxk(4−x)0​0≤x≤22<x≤4otherwise,​

a symmetric "tent" peaking at $x = 2$x=2. To find $k$k, integrate each piece and set the total to 1:

$\int_0^2 kx\,dx + \int_2^4 k(4 - x)\,dx = k\left[\frac{x^2}{2}\right]_0^2 + k\left[4x - \frac{x^2}{2}\right]_2^4 = 2k + k\big[(16 - 8) - (8 - 2)\big] = 2k + 2k = 4k = 1,$∫02​kxdx+∫24​k(4−x)dx=k[2x2​]02​+k[4x−2x2​]24​=2k+k[(16−8)−(8−2)]=2k+2k=4k=1,

so $k = \tfrac14$k=41​. The geometry confirms it: the graph is a triangle of base $4$4 and height $k\cdot 2 = \tfrac12$k⋅2=21​, area $\tfrac12\cdot 4\cdot\tfrac12 = 1$21​⋅4⋅21​=1. Now compute $P(1 \leq X \leq 3)$P(1≤X≤3), a region straddling the join at $x = 2$x=2, so the integral must be split:

$$\begin{aligned} P(1 \leq X \leq 3) &= \int_1^2 \frac{x}{4}\,dx + \int_2^3 \frac{4 - x}{4}\,dx \\ &= \frac14\left[\frac{x^2}{2}\right]_1^2 + \frac14\left[4x - \frac{x^2}{2}\right]_2^3 \\ &= \frac14\left(2 - \tfrac12\right) + \frac14\big[(12 - 4.5) - (8 - 2)\big] = \frac14(1.5) + \frac14(1.5) = 0.75. \end{aligned}$$P(1≤X≤3)​=∫12​4x​dx+∫23​44−x​dx=41​[2x2​]12​+41​[4x−2x2​]23​=41​(2−21​)+41​[(12−4.5)−(8−2)]=41​(1.5)+41​(1.5)=0.75.​

The symmetry of the tent about $x = 2$x=2 provides an instant check: the central band $[1,3]$[1,3] is symmetric about the peak, the two half-integrals are equal ($0.375$0.375 each), and the result $0.75$0.75 is the natural "central three-quarters". Recognising symmetry both shortcuts the arithmetic and guards against error — if your two pieces had come out unequal, you would know at once that a slip had crept in. Piecewise densities reward neatness: label each piece, integrate within its own limits, and split any probability that crosses a boundary.

Densities on an infinite support

So far every support has been a bounded interval, but a continuous variable can be defined on an unbounded range — for instance $[0, \infty)$[0,∞) — provided the area is still finite and equal to 1. This requires the density to decay fast enough as $x \to \infty$x→∞, and it brings improper integrals into play. Consider

$f(x) = k e^{-2x} \quad \text{for } x \geq 0, \qquad f(x) = 0 \text{ otherwise.}$f(x)=ke−2xfor x≥0,f(x)=0 otherwise.

For normalisation,

$\int_0^{\infty} k e^{-2x}\,dx = k\left[-\tfrac12 e^{-2x}\right]_0^{\infty} = k\left(0 - \left(-\tfrac12\right)\right) = \frac{k}{2} = 1 \implies k = 2,$∫0∞​ke−2xdx=k[−21​e−2x]0∞​=k(0−(−21​))=2k​=1⟹k=2,

where the upper limit uses $e^{-2x} \to 0$e−2x→0 as $x \to \infty$x→∞. This is the exponential distribution $\text{Exp}(2)$Exp(2), a continuous model you will study in its own right; here it simply illustrates that the PDF rules carry over unchanged to infinite supports — non-negativity holds ($2e^{-2x} > 0$2e−2x>0) and the (improper) integral equals 1. Probabilities follow the same area principle:

$P(X > 1) = \int_1^{\infty} 2e^{-2x}\,dx = \left[-e^{-2x}\right]_1^{\infty} = e^{-2} = 0.135, \qquad P(X \leq 0.5) = \left[-e^{-2x}\right]_0^{0.5} = 1 - e^{-1} = 0.632.$P(X>1)=∫1∞​2e−2xdx=[−e−2x]1∞​=e−2=0.135,P(X≤0.5)=[−e−2x]00.5​=1−e−1=0.632.

The key practical point is that the limit at infinity must be handled by recognising that the exponential term vanishes — writing the upper boundary as a limit and evaluating it. Densities on $(-\infty, \infty)$(−∞,∞), such as the normal, work the same way, with the area in both tails contributing. Far from being exotic, infinite-support densities are central to statistics: they model waiting times, lifetimes and measurement errors, and they obey precisely the framework — non-negative density, unit area, probability by integration — established in this lesson.

The uniform distribution in context

The continuous uniform distribution deserves a worked application, because its "all values equally likely" structure makes it the natural model for rounding and for a value known only to lie in an interval. Suppose a train is equally likely to arrive at any moment in the 15-minute window after its scheduled time; let $X$X be the delay in minutes, so $X \sim U(0, 15)$X∼U(0,15) with constant density $f(x) = \tfrac{1}{15}$f(x)=151​ on $[0, 15]$[0,15].