Linear functions of a DRVAQA A-Level Further Maths: Revision notes

A linear function of a random variable has the form $Y=aX+b$. The constant $a$ scales every value and $b$ shifts every value; the probabilities are unchanged. Typical uses: converting units (${}^\circ\mathrm{C}\to{}^\circ\mathrm{F}$), profit $=$ price $\times$ number sold $-$ fixed cost, and standardising a variable. Example: if $X=1,2,3,4$ with probabilities $0.3,0.4,0.2,0.1$, then $Y=5-2X$ takes the values $3,1,-1,-3$ with the same probabilities.

$E(aX+b)=aE(X)+b.$ The mean is scaled and shifted in exactly the same way as the values. The proof uses the definition: $E(aX+b)=\sum(ax_i+b)p_i=a\sum x_ip_i+b\sum p_i=aE(X)+b$, since $\sum p_i=1$. Example: $E(X)=5$, so $E(3X+2)=3(5)+2=17$ and $E(2-4X)=2-4(5)=-18$.

$\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X).$ A shift by $b$ does not change the spread, so $b$ disappears. Scaling by $a$ multiplies the spread by $|a|$, so the variance is multiplied by $a^2$. Hence the standard deviation is $|a|\times$ the standard deviation of $X$. Because $a^2\ge0$, a negative $a$ never gives a negative variance: $\mathrm{Var}(2-4X)=(-4)^2\mathrm{Var}(X)=16\mathrm{Var}(X)$. Example: $\mathrm{Var}(X)=4$, so $\mathrm{Var}(3X+2)=9\times4=36$ and the standard deviation is $6$.

A café sells $X$ meals with $E(X)=8.5$ and $\mathrm{Var}(X)=2.25$; profit $T=6X-20$. $E(T)=6(8.5)-20=31$, $\mathrm{Var}(T)=6^2(2.25)=81$ and the standard deviation is $9$. You can always check with the distribution itself: list the new values $y=ax+b$ with the same probabilities, then compute $E(Y)=\sum yp$ and $\mathrm{Var}(Y)=E(Y^2)-[E(Y)]^2$. The results must match the formulas. To standardise, choose $a$ and $b$ so that $E(Y)=0$ and $\mathrm{Var}(Y)=1$: $a=\frac1\sigma$ and $b=-\frac\mu\sigma$.