Mean and variance of discrete distributionsEdexcel A-Level Further Maths: Revision notes

A discrete random variable $X$ takes separate values $x$ with probabilities $P(X=x)$ that sum to 1. The expected value or mean is $E(X)=\mu=\sum xP(X=x).$ It is the long-run average of many observations, not necessarily a value $X$ can take. Example: $P(X=1)=0.5$, $P(X=2)=0.3$, $P(X=3)=0.2$ gives $E(X)=0.5+0.6+0.6=1.7$. Always weight each value by its probability; the plain average of the values ($2$ here) is wrong unless all probabilities are equal.

The variance measures spread: $\text{Var}(X)=\sigma^2=E(X^2)-\mu^2=\sum x^2P(X=x)-\mu^2.$ Find $E(X^2)$ by squaring each value before weighting. Continuing the example: $E(X^2)=0.5+1.2+1.8=3.5$, so $\text{Var}(X)=3.5-1.7^2=0.61$. The standard deviation is $\sigma=\sqrt{0.61}=0.781$. A variance cannot be negative; if you get a negative answer, you have made an arithmetic error or forgotten to square.

For any function $g$, $E(g(X))=\sum g(x)P(X=x).$ Apply $g$ to each value, then weight by the probability. For $g(x)=x^2$ this is $E(X^2)$. In the example, $E(2X^2-1)=2(3.5)-1=6$. In general $E(g(X))\neq g(E(X))$: for instance $E(X^2)=3.5$ but $[E(X)]^2=2.89$, and the difference is exactly the variance. The special case $g(x)=ax+b$ gives $E(aX+b)=aE(X)+b$. Costs, profits and penalties are often functions of a random variable, so $E(g(X))$ gives the expected cost or profit.

Probabilities are often given by a formula with an unknown constant. Use $\sum P(X=x)=1$ to find it. Example: $P(X=x)=kx^2$ for $x=1,2,3$. Then $k(1+4+9)=1$, so $k=\frac1{14}$. $E(X)=\frac{1^3+2^3+3^3}{14}=\frac{36}{14}=\frac{18}{7}$ and $E(X^2)=\frac{1+16+81}{14}=7$. Then $\text{Var}(X)=7-\left(\frac{18}{7}\right)^2=\frac{19}{49}$. Check every probability lies between $0$ and $1$ once $k$ is found.

A game's profit is a function of the outcome, so $E(\text{profit})$ is found by weighting each profit by its probability. A game is fair if the expected profit is zero. Example: a game costs £3 and pays £20 with probability $0.05$, £6 with probability $0.2$ and nothing otherwise. The profit is $17$, $3$ or $-3$, so $E=0.85+0.6-2.25=-0.8$: the player loses 80p per game on average. Interpret answers in context, with units. The variance tells you how much results vary around the average, which matters when the expected value is small.

A model for a real situation should reproduce the observed mean and variance. To assess it, calculate $E(X)$ and $\text{Var}(X)$ for the model and compare with the sample mean and variance (or standard deviation) from data. If they are close, the model may be suitable. If the model's variance is much smaller than the data's, it underestimates spread and so underestimates the chance of extreme values; costs calculated with $E(g(X))$ from such a model are too low. Also check whether the model allows every value seen in the data (a model with a maximum of 3 cannot describe a sample containing 5). Always state a conclusion about suitability and say why.