Functions of a DRVAQA A-Level Further Maths: Revision notes

If $X$ is a DRV and $g$ is a function, then $Y=g(X)$ is also a DRV. It takes the values $g(x_i)$, with the probability of each value equal to $P(X=x_i)$ (if two values of $x$ give the same $g(x)$, add their probabilities). Examples of $g$ in this topic: $X^2$, $5X^3$, $18X^{-3}$ and $6X^{-1}$. A table with rows $x$, $g(x)$ and $P(X=x)$ keeps the working clear.

$E\big(g(X)\big)=\sum g(x_i)\,p_i.$ Apply $g$ to each value, multiply by its probability and add. You do not need the distribution of $g(X)$ itself. Example: $X=1,2,3$ with probabilities $0.2,0.5,0.3$: $E(X^2)=0.2+2+2.7=4.9$ and $E(5X^3)=5(0.2+4+8.1)=61.5$. Constant multiples come out: $E(18X^{-3})=18E(X^{-3})$.

In general $E(g(X))\ne g(E(X))$. For $g(x)=x^2$, $E(X^2)=\mathrm{Var}(X)+[E(X)]^2\ge[E(X)]^2$. Example: $X=2,3,4$ with probabilities $0.25,0.5,0.25$: $E(X)=3$, so $[E(X)]^2=9$, but $E(X^2)=1+4.5+4=9.5$. The gap is $\mathrm{Var}(X)=0.5$. The equality $E(g(X))=g(E(X))$ holds only when $g$ is linear, $g(x)=ax+b$.

Treat $Y=g(X)$ as a variable and use $\mathrm{Var}(Y)=E(Y^2)-[E(Y)]^2$, where $E(Y^2)=\sum[g(x_i)]^2p_i$. A constant multiple scales the variance: $\mathrm{Var}(kg(X))=k^2\mathrm{Var}(g(X))$. Example: $X=1,2,4$ with probabilities $0.4,0.4,0.2$. $E(X^{-1})=0.65$, $E(X^{-2})=0.5125$, so $\mathrm{Var}(X^{-1})=0.5125-0.4225=0.09$ and $\mathrm{Var}(6X^{-1})=36\times0.09=3.24$. For $\mathrm{Var}(5X^3)$ you need $E(X^6)$: $\mathrm{Var}(5X^3)=25\left[E(X^6)-(E(X^3))^2\right]$.

1. Write out the distribution as a table (find unknown constants from $\sum p=1$).
2. Add rows for $g(x)$ and, for variance, $[g(x)]^2$.
3. Compute $E(g(X))$ and $E([g(X)]^2)$.
4. Use $\mathrm{Var}=E(\cdot^2)-[E(\cdot)]^2$ and the scaling rules $E(aY+b)=aE(Y)+b$, $\mathrm{Var}(aY+b)=a^2\mathrm{Var}(Y)$.
5. Round only at the end. Context questions: the area $A=X^2$ of a square tile with random side $X$, and a cost $C=0.4A+1$ are functions of $X$; find the mean cost by combining the two rules.