4.14 Linear combinations of random variables and unbiased estimatesIB Maths: Applications and Interpretation HL: Revision notes

If $X$ is a random variable and $a$, $b$ are constants then $E(aX+b)=aE(X)+b.$ The mean is shifted and scaled in the same way as each value. For the variance, $Var(aX+b)=a^2Var(X)$: adding $b$ does not change the spread and multiplying by $a$ multiplies the variance by $a^2$ (you will not be asked to recall this variance formula for a single variable, but it explains the results for combinations). Example: $E(X)=120$ and income $I=15X+200$ gives $E(I)=15(120)+200=2000$.

For any random variables $X_1,\ldots,X_n$ and constants $a_1,\ldots,a_n$, $E(a_1X_1+\cdots+a_nX_n)=a_1E(X_1)+\cdots+a_nE(X_n).$ This holds whether or not the variables are independent. If the variables are independent, $Var(a_1X_1+\cdots+a_nX_n)=a_1^2Var(X_1)+\cdots+a_n^2Var(X_n).$ So for independent $X$ and $Y$: $Var(X+Y)=Var(X)+Var(Y)$ and, because $(-1)^2=1$, $Var(X-Y)=Var(X)+Var(Y)$. Variances add even when the variables are subtracted. Example: $E(X)=120$, $Var(X)=64$, $E(Y)=90$, $Var(Y)=36$. Then $E(X+Y)=210$ and $Var(X-Y)=100$.

If $X_1$ and $X_2$ are two independent copies of $X$ (for example two separate drivers) then $E(X_1+X_2)=2E(X)$ and $Var(X_1+X_2)=2Var(X)$. But for one variable doubled, $E(2X)=2E(X)$ while $Var(2X)=4Var(X)$. The means agree but the variances differ: in $X_1+X_2$ a high value of one can be offset by a low value of the other, whereas $2X$ doubles one value. Check which situation the question describes: separate independent items or one item counted several times.

A sample is used to estimate the properties of a population. An estimate is unbiased if its expected value equals the parameter being estimated: on average it is neither too high nor too low. The sample mean is an unbiased estimate of the population mean: $\bar x=\frac{\sum x_i}{n}\quad(\text{or }\bar x=\frac{\sum f_ix_i}{n}\text{ for a frequency table}),$ so $E(\bar X)=\mu$. (The proof is not examined.) Example: masses $58, 61, 63, 60, 57, 65$ give $\bar x=\frac{364}{6}=60.7$ g as an unbiased estimate of $\mu$.

The variance of the sample, dividing by $n$, $s_n^2=\frac{\sum f_i(x_i-\bar x)^2}{n},$ systematically underestimates $\sigma^2$. The unbiased estimate divides by $n-1$: $s_{n-1}^2=\frac{n}{n-1}s_n^2=\frac{\sum f_i(x_i-\bar x)^2}{n-1},\quad n=\sum f_i.$ On most GDCs, $\sigma_x$ is $s_n$ and $s_x$ is $s_{n-1}$ (check your model). Square the standard deviation to get the variance. Example: for the six eggs, $s_n^2=7.56$, so $s_{n-1}^2=\frac{6}{5}(7.56)=9.07$ g$^2$. For a frequency table with $n=40$ and $s_n^2=1.634$: $s_{n-1}^2=\frac{40}{39}(1.634)=1.68$.

Once $\bar x$ and $s_{n-1}^2$ are found, treat them as $\mu$ and $\sigma^2$ in the linear combination formulas. Example (parcels per driver): $\bar x=10.375$, $s_{n-1}^2=1.676$. For payment $P=6X+40$: $E(P)=6(10.375)+40=102.25$ AED. For two independent drivers: $E(X_1+X_2)=20.75$ and $Var(X_1+X_2)=2(1.676)=3.35$. State clearly that these are estimates. Keep unrounded values until the final answer and give 3 significant figures (money to 2 decimal places).