Continuous random variables

40 min

When every value in an interval is possible

The time a pupil takes to walk to school, the mass of a bag of maize flour, the lifetime of a light bulb: these quantities can take any value in an interval, not just whole numbers. The chance of exactly 12.000… minutes is zero, so a table of single values no longer works. Instead, probability is spread along the number line by a curve, the density, and the probability of an interval is the area under that curve. The integrals you learned in S5 and S6 do the work.

Definition

A random variable $X$ is continuous if there is a function $f$ , its probability density function (pdf), such that

f(x) \ge 0 \text{ for all } x , \qquad \int_{-\infty}^{+\infty} f(x)\,dx = 1 , \qquad P(a \le X \le b) = \int_a^b f(x)\,dx .

Its cumulative distribution function is $F(x) = P(X \le x) = \int_{-\infty}^{x} f(t)\,dt$ , and $f (x) = F^{'} (x)$ wherever $f$ is continuous. The median $m$ satisfies $F(m) = \frac{1}{2}$ .

Example: if $X$ is equally likely to be anywhere in $[0, 20]$ , then $f(x) = \frac{1}{20}$ on that interval and 0 outside.

Important

Single values have probability zero. $P(X = c) = \int_c^c f(x)\,dx = 0$ . So for a continuous $X$ , $P(a \lt X \lt b) = P(a \le X \le b) = F(b) - F(a)$ : strict and non-strict inequalities give the same answer. A density value $f (x)$ is not a probability, and it may be greater than 1.

Mean and variance replace sums by integrals:

E(X) = \mu = \int_{-\infty}^{+\infty} x f(x)\,dx , \qquad \operatorname{Var}(X) = \int_{-\infty}^{+\infty} (x - \mu)^2 f(x)\,dx = E(X^2) - \mu^2 .

The rules $E (a X + b) = a E (X) + b$ and $\operatorname{Var}(aX + b) = a^2\operatorname{Var}(X)$ still hold.

A probability density. Probabilities are areas under the curve: the shaded area is

P(X \le 0.5) = \frac{5}{32}

. Drawn from the formula of worked example 1.

Worked example 1: finding the constant, a probability, the mean and the variance

$X$ has pdf $f (x) = k x (2 - x)$ for $0 \le x \le 2$ , and $f (x) = 0$ otherwise. Find $k$ , $P(X \le 0.5)$ , $E (X)$ and $\operatorname{Var}(X)$ .

Solution. $\int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$ , so $k \cdot \frac{4}{3} = 1$ and $k = \frac{3}{4}$ .

P(X \le 0.5) = \frac{3}{4}\left[x^2 - \frac{x^3}{3}\right]_0^{0.5} = \frac{3}{4}\left(\frac{1}{4} - \frac{1}{24}\right) = \frac{3}{4} \times \frac{5}{24} = \frac{5}{32} \approx 0.156 .

The graph is symmetric about $x = 1$ , so $E (X) = 1$ . (Check: $\frac{3}{4}\int_0^2 (2x^2 - x^3)\,dx = \frac{3}{4}\left(\frac{16}{3} - 4\right) = 1$ .) Next, $E(X^2) = \frac{3}{4}\int_0^2 (2x^3 - x^4)\,dx = \frac{3}{4}\left(8 - \frac{32}{5}\right) = \frac{6}{5}$ , so $\operatorname{Var}(X) = \frac{6}{5} - 1 = \frac{1}{5}$ and $\sigma \approx 0.447$ .

Worked example 2: the cdf and the median

$X$ has pdf $f(x) = \frac{3}{8}x^2$ for $0 \le x \le 2$ , and 0 otherwise. Find $F$ , the median, $P(X \gt 1)$ and $E (X)$ .

Solution. For $0 \le x \le 2$ , $F(x) = \int_0^x \frac{3}{8}t^2\,dt = \frac{x^3}{8}$ ; $F (x) = 0$ for $x \lt 0$ and $F (x) = 1$ for $x \gt 2$ .

Median: $\frac{m^3}{8} = \frac{1}{2}$ , so $m^{3} = 4$ and $m = \sqrt[3]{4} \approx 1.587$ . $P(X \gt 1) = 1 - F(1) = 1 - \frac{1}{8} = \frac{7}{8}$ . $E(X) = \int_0^2 \frac{3}{8}x^3\,dx = \frac{3}{8} \times 4 = 1.5$ . The mean (1.5) is less than the median (1.587) because the density is heavier on the right and has a long tail to the left.

Worked example 3: a uniform waiting time

A bus passes a stop every 20 minutes, and a passenger arrives at a random moment, so the waiting time $W$ is uniform on $[0, 20]$ . Find $P(W \gt 15)$ , $E (W)$ and the standard deviation of $W$ .

Solution. $f(w) = \frac{1}{20}$ on $[0, 20]$ . $P(W \gt 15) = \frac{20 - 15}{20} = \frac{1}{4}$ . $E(W) = \int_0^{20} \frac{w}{20}\,dw = 10$ minutes. $E(W^2) = \int_0^{20} \frac{w^2}{20}\,dw = \frac{400}{3}$ , so $\operatorname{Var}(W) = \frac{400}{3} - 100 = \frac{100}{3}$ and $\sigma = \frac{20}{\sqrt{12}} \approx 5.77$ minutes.

In general, the uniform distribution on $[a, b]$ has $f(x) = \frac{1}{b - a}$ , mean $\frac{a + b}{2}$ and variance $\frac{(b - a)^2}{12}$ .

Warning

Integrate over the whole support. If $f$ is defined in pieces, $F (x)$ must add the full areas of all the pieces to the left of $x$ .

$f (x)$ is not $P (X = x)$ . In worked example 2, $f (2) = 1.5$ , which is not a probability. Only areas under $f$ are probabilities.

Tip

Use symmetry before integrating. If the graph of $f$ is symmetric about $x = c$ , then $E (X) = c$ and the median is also $c$ , as in worked example 1.

3Blue1Brown: Why “probability of 0” does not mean “impossible” (about 10 min). Watch how a density spreads probability over an interval so that every single value has probability zero, while areas, not heights, give probabilities.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks you to explain.

$f (x) = k x$ for $0 \le x \le 4$ , and 0 otherwise, is a pdf. Find $k$ and $P(1 \le X \le 3)$ .
$X$ has pdf $f (x) = 2 x$ on $[0, 1]$ . Find $E (X)$ , $\operatorname{Var}(X)$ and the median.
$X$ is uniform on $[2, 8]$ . Find $P(X \gt 6.5)$ , $E (X)$ and $\operatorname{Var}(X)$ .
The time $T$ (in minutes) between two calls to a help line has cdf $F(t) = 1 - e^{-0.5t}$ for $t \ge 0$ . Find the pdf, $P(T \gt 2)$ and the median.
$f (x) = k (4 - x^{2})$ for $-2 \le x \le 2$ , and 0 otherwise. Find $k$ , $P(X \gt 1)$ , $E (X)$ and $\operatorname{Var}(X)$ .
Show that $f(x) = \frac{1}{x^2}$ for $x \ge 1$ (and 0 otherwise) is a pdf, and find $P(X \gt 4)$ . Does $X$ have a mean?
Explain why, for a continuous random variable, $P(X \lt 3) = P(X \le 3)$ , while for a discrete one they can differ.

Show answersHide answers

$8 k = 1$ , $k = \frac{1}{8}$ ; $P(1 \le X \le 3) = \frac{9 - 1}{16} = \frac{1}{2}$ .
$E(X) = \frac{2}{3}$ , $E(X^2) = \frac{1}{2}$ , $\operatorname{Var}(X) = \frac{1}{18}$ ; median $m^2 = \frac{1}{2}$ , $m = \frac{1}{\sqrt{2}} \approx 0.707$ .
$\frac{1.5}{6} = 0.25$ ; $E (X) = 5$ ; $\operatorname{Var}(X) = \frac{36}{12} = 3$ .
$f(t) = 0.5e^{-0.5t}$ ; $P(T \gt 2) = e^{-1} \approx 0.368$ ; median $2\ln 2 \approx 1.39$ minutes.
$k = \frac{3}{32}$ ; $P(X \gt 1) = \frac{3}{32}\left(4 - \frac{7}{3}\right) = \frac{5}{32}$ ; $E (X) = 0$ (symmetry); $\operatorname{Var}(X) = E(X^2) = 0.8$ .
$\int_1^\infty x^{-2}\,dx = 1$ ; $P(X \gt 4) = \frac{1}{4}$ ; $\int_1^\infty x \cdot x^{-2}\,dx = \int_1^\infty \frac{dx}{x}$ diverges, so there is no mean.
They differ by $P (X = 3)$ , which is $\int_3^3 f = 0$ for a continuous variable but can be positive for a discrete one.

Sources and credits

Written by Ganzaa: the explanations, the diagram (plotted from the density of worked example 1), the worked examples, the practice items and the quiz questions. Every integral was checked numerically.
Video: Why “probability of 0” does not mean “impossible”, 3Blue1Brown (Grant Sanderson), embedded from YouTube.

1When every value in an interval is possible

When every value in an interval is possible