Use probability density functions and cumulative distribution functions to find probabilities, medians, modes, means and variances of continuous random variables

What this dot point is asking

NESA wants you to work with continuous random variables defined by a probability density function (pdf). You must find unknown constants by enforcing total probability $= 1$, compute probabilities as definite integrals, and find the mean, variance, median and mode using integrals.

The answer

Probability density functions

A continuous random variable $X$ is described by a probability density function $f$ satisfying

• IMATH_9 for all $x$,
• IMATH_11 .

In Maths Advanced, $f$ is non-zero only on a finite interval $[a, b]$ called the support, and the total integral is taken over that interval.

For a continuous random variable, $P(X = c) = 0$ for any single value $c$. Probabilities live on intervals only.

Probabilities as integrals

For any interval $[c, d]$ inside the support,

Because single points have zero probability, $P(c \le X \le d) = P(c < X < d)$. The strict and non-strict inequalities give the same value.

Cumulative distribution function

The cumulative distribution function (cdf) is

Useful properties:

• IMATH_18 is non-decreasing, with $F(-\infty) = 0$ and $F(\infty) = 1$.
• IMATH_21 .
• Where $f$ is continuous, $F'(x) = f(x)$.

Mean, variance, median, mode

The mean (expected value) is

The variance is

where $E(X^2) = \int x^2 f(x) \, dx$. The standard deviation is $\sigma = \sqrt{\text{Var}(X)}$.

The median $m$ splits the distribution in half:

The mode is the value of $x$ where $f$ is largest. If $f$ is differentiable on the interior of the support, the mode is at a critical point of $f$ (or at an endpoint if $f$ is monotone there).

Worked examples

Finding a constant and a probability

$f(x) = c(1 - x^2)$ for $-1 \le x \le 1$, $0$ elsewhere. Find $c$, then $P(X > 0)$.

$\int_{-1}^{1} c(1 - x^2) \, dx = c \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = c \left( \frac{2}{3} - \left(-\frac{2}{3}\right) \right) = \frac{4 c}{3} = 1$, so $c = \frac{3}{4}$.

By symmetry of $1 - x^2$ about $x = 0$, $P(X > 0) = \frac{1}{2}$.

Computing the mean and variance

For $f(x) = \frac{x}{8}$ on $[0, 4]$:

$E(X) = \int_0^4 x \cdot \frac{x}{8} \, dx = \frac{1}{8} \cdot \frac{64}{3} = \frac{8}{3}$.

$E(X^2) = \int_0^4 x^2 \cdot \frac{x}{8} \, dx = \frac{1}{8} \cdot \frac{256}{4} = 8$.

$\text{Var}(X) = 8 - \left(\frac{8}{3}\right)^2 = 8 - \frac{64}{9} = \frac{8}{9}$. So $\sigma = \frac{2\sqrt{2}}{3}$.

Cdf from a pdf

For $f(x) = \frac{x}{8}$ on $[0, 4]$, the cdf is

So $P(X \le 2) = F(2) = \frac{4}{16} = \frac{1}{4}$.

Median and mode

For $f(x) = \frac{x}{8}$ on $[0, 4]$, the median $m$ solves $\frac{m^2}{16} = \frac{1}{2}$, so $m^2 = 8$ and $m = 2 \sqrt{2} \approx 2.83$.

The mode is at $x = 4$, the right endpoint, because $f$ is increasing on $[0, 4]$.

Common traps

Treating $P(X = c) > 0$. For a continuous random variable, single points have zero probability. The pdf value $f(c)$ is a density, not a probability.

Forgetting to enforce total probability. When a pdf has an unknown constant, the first step is always $\int f = 1$.

Confusing pdf and cdf. $f$ is the density (can exceed $1$), $F$ is the cumulative probability (always between $0$ and $1$). $f = F'$ where $F$ is differentiable.

Integrating $x f(x)$ over the wrong range. When computing $E(X)$, integrate only over the support. Outside the support, $f = 0$ contributes nothing.

Picking an interior critical point that is actually a minimum. The mode is the maximum of $f$. If $f$ is monotone, the mode is at an endpoint of the support.

In one sentence

A continuous random variable is described by a pdf $f$ with $\int f = 1$; probabilities come from integrating $f$ over intervals, the cdf is $F(x) = \int_{-\infty}^{x} f$, and the mean, variance, median and mode are computed by integrals or by solving $F(m) = \frac{1}{2}$.