Expectation and variance of discrete random variablesEdexcel International A Level Maths: Revision notes

The expectation (mean) of a discrete random variable is the long-run average value, found by weighting each value by its probability: $E(X)=\sum x\,P(X=x).$ Example: $P(X=0)=0.2$, $P(X=1)=0.3$, $P(X=2)=0.4$, $P(X=3)=0.1$ gives $E(X)=0+0.3+0.8+0.3=1.4$. $E(X)$ need not be a possible value of $X$. If $X$ takes values symmetrically about a centre, the mean is at the centre.

$E(X^2)=\sum x^2P(X=x)$: square each value first, then weight by its probability. The variance measures spread: $\mathrm{Var}(X)=E(X^2)-[E(X)]^2.$ The standard deviation is $\sqrt{\mathrm{Var}(X)}$. Variance cannot be negative. Example: with the distribution above, $E(X^2)=0+0.3+1.6+0.9=2.8$, so $\mathrm{Var}(X)=2.8-1.4^2=0.84$. A fully worked layout (a table of $x$, $P(X=x)$, $xP$, $x^2P$) keeps this tidy. If you are given $\mathrm{Var}(X)$ and $E(X)$, rearrange to find $E(X^2)=\mathrm{Var}(X)+[E(X)]^2$.

For constants $a$ and $b$: $E(aX+b)=aE(X)+b,\qquad \mathrm{Var}(aX+b)=a^2\mathrm{Var}(X).$

• The mean is shifted and scaled in the same way as the values.
• The constant $b$ has no effect on the spread, because adding a constant moves every value by the same amount.
• The multiplier $a$ is squared in the variance. Example: $E(W)=10$, $\mathrm{Var}(W)=6$ and $V=3W-4$: $E(V)=3(10)-4=26$ and $\mathrm{Var}(V)=9\times6=54$. For a negative multiplier, e.g. $5-2X$: $\mathrm{Var}(5-2X)=(-2)^2\mathrm{Var}(X)=4\mathrm{Var}(X)$.

Often the distribution contains unknowns. Write one equation from $\sum p=1$ and another from a given mean or variance, then solve the simultaneous equations. Example: $P(X=1)=0.1$, $P(X=2)=a$, $P(X=3)=0.4$, $P(X=4)=b$ with $E(X)=2.9$. From $\sum p=1$: $a+b=0.5$. From $E(X)$: $0.1+2a+1.2+4b=2.9$, so $2a+4b=1.6$. Solving gives $a=0.2$, $b=0.3$. Then $E(X^2)=0.1+0.8+3.6+4.8=9.3$ and $\mathrm{Var}(X)=9.3-2.9^2=0.89$.

In a game with prize $X$ and an entry fee $f$, the expected net gain is $E(X)-f$. The game is fair if $E(X)=f$. A positive expected gain favours the player; a negative one favours the organiser. Example: a die roll pays 8 pounds for a 6, 3 pounds for a 4 or 5, and nothing otherwise: $E(X)=\frac73$. At a fee of 2 pounds the expected net gain is $\frac13$ pounds, about 33p per game. Remember that $E(X)$ is a long-run average: any single game gives only 0, 3 or 8.