4.14 Discrete and continuous random variablesIB Maths: Analysis and Approaches HL: Revision notes

For a discrete random variable $X$: $E(X) = \sum x\,P(X = x), \qquad E(X^2) = \sum x^2\,P(X = x),$ $\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = E\big((X - \mu)^2\big).$ The standard deviation is $\sqrt{\mathrm{Var}(X)}$.

Example: $P(X = 0, 1, 2, 3) = 0.1, 0.3, 0.4, 0.2$ gives $E(X) = 1.7$, $E(X^2) = 3.7$ and $\mathrm{Var}(X) = 3.7 - 2.89 = 0.81$.

If $a$ and $b$ are constants: $E(aX + b) = aE(X) + b, \qquad \mathrm{Var}(aX + b) = a^2\,\mathrm{Var}(X).$ Adding $b$ shifts every value, so it changes the mean but not the spread. Multiplying by $a$ scales the standard deviation by $|a|$ and the variance by $a^2$.

A game is fair if the player's expected gain is 0. If a player pays 8 dollars and wins 5 dollars per point, the gain is $Y = 5X - 8$, so $E(Y) = 5(1.7) - 8 = 0.5 \neq 0$: not fair, it favours the player.

A continuous random variable is described by a probability density function (pdf) $f(x)$ with $f(x) \ge 0 \text{ for all } x, \qquad \int_{-\infty}^{\infty} f(x)\,dx = 1.$ Probabilities are areas: $P(a < X < b) = \int_a^b f(x)\,dx$. For any single value, $P(X = a) = 0$, so $<$ and $\le$ give the same probability.

A pdf may be piecewise: integrate each piece over its own interval and add. For $f(x) = kx$ on $[0, 2]$ and $\frac{k(6 - x)}{2}$ on $(2, 6]$: $2k + 4k = 1$, so $k = \frac{1}{6}$.

The mode is the value of $x$ where $f(x)$ is greatest. Find it by differentiating and checking for a maximum, and remember to compare with the endpoints. For $f(x) = \frac{4}{27}x^2(3 - x)$: $6x - 3x^2 = 0$ gives the mode $x = 2$.

The median $m$ satisfies $\int_{-\infty}^{m} f(x)\,dx = \frac{1}{2}.$ For a piecewise pdf, first find which piece contains $m$ by working out the probability of the first piece. If the equation gives two roots, reject the one outside the domain.

Replace sums by integrals: $E(X) = \int_{-\infty}^{\infty} x f(x)\,dx, \qquad E(X^2) = \int_{-\infty}^{\infty} x^2 f(x)\,dx, \qquad \mathrm{Var}(X) = E(X^2) - [E(X)]^2.$ In practice you only integrate over the interval where $f(x) \neq 0$. For $f(t) = \frac{t}{8}$ on $[0, 4]$: $E(T) = \frac{8}{3}$, $E(T^2) = 8$, $\mathrm{Var}(T) = \frac{8}{9}$.

The linear transformation rules hold for continuous variables too: a charge $C = 3 + 2.5T$ has $E(C) = \frac{29}{3}$ and $\mathrm{Var}(C) = 6.25\times\frac{8}{9}$.