6.2 - Means and variances of combinations | Cambridge A-Level Mathematics

Multiplying one random measurement by four and adding four independent measurements can give the same expected value but different spreads. Learn how shifts, scales, sums and differences affect expectation and variance, and distinguish repeated observations from repeated use of the same observation.

Scaling and shifting

Suppose a random quantity has mean $10$ and standard deviation $2$ . Adding $5$ to every outcome moves its centre to $15$ , but leaves every distance from that centre unchanged. Multiplying every outcome by $3$ multiplies those distances by $3$ , so its variance is multiplied by $9$ .

This is why a scale change affects variance differently from a shift. If $Y=aX+b$ , its deviation from its own mean is

Y-\mathrm E(Y)=aX+b-\{a\mathrm E(X)+b\}=a\{X-\mathrm E(X)\}.

Squaring the deviation introduces $a^2$ , while the shift $b$ cancels.

Scaling and shifting a random variable

For constants $a,b$ and a random variable with finite mean and variance,

\begin{aligned} \mathrm E(aX+b)&=a\mathrm E(X)+b\\ \operatorname{Var}(aX+b)&=a^2\operatorname{Var}(X)\\ \operatorname{SD}(aX+b)&=|a|\operatorname{SD}(X). \end{aligned}

Not in the formula booklet: learn these transformation rules.

A negative scale reflects the outcomes and changes the sign of the mean's contribution. It does not make variance or standard deviation negative. If $a=0$ , the result is the constant $b$ , with variance $0$ .

Worked example 1: a positive scale

Question. Given $\mathrm E(X)=10$ and $\operatorname{Var}(X)=4$ , find the mean, variance and standard deviation of $Y=3X+5$ .

Plan. Apply the scale and shift to the mean, but square only the scale for the variance.

\begin{aligned} \mathrm E(Y)&=3(10)+5=35\\ \operatorname{Var}(Y)&=3^2(4)=36\\ \operatorname{SD}(Y)&=\sqrt{36}=6 \end{aligned}

Answer. Mean $35$ , variance $36$ , standard deviation $6$ .

Check. For a concrete variable with the stated moments, take equally likely values $8,12$ . They become $29,41$ : their mean is $35$ and their deviations are $-6,6$ , giving variance $36$ . The general deviation argument above shows why the rules do not depend on this particular distribution.

Change only the sign of the scale. The centre changes, while the spread stays the same.

Worked example 2: a negative scale

Question. For the same $X$ , find the mean, variance and standard deviation of $W=-3X+5$ .

Plan. Retain the negative coefficient in the mean and square it in the variance.

\begin{aligned} \mathrm E(W)&=-3(10)+5=-25\\ \operatorname{Var}(W)&=(-3)^2(4)=36\\ \operatorname{SD}(W)&=6 \end{aligned}

Answer. Mean $-25$ , variance $36$ , standard deviation $6$ .

Check. The equally likely values $8,12$ would become $-19,-31$ , centred at $-25$ with deviations $6,-6$ . Reflecting those deviations has not changed their squares.

These rules describe moments; they do not by themselves tell you the distribution's name. A mean and variance alone do not establish that a variable is normal or Poisson.

Watch out

$\operatorname{Var}(3X+5)$ is $9\operatorname{Var}(X)$ , not $3\operatorname{Var}(X)+5$ . A fixed addition changes location, not spread.

Use the negative scale carefully in this fresh transformation.

Combining two variables

Suppose two independently measured quantities fluctuate around their own means. Adding them can reinforce or partly offset their deviations on a particular occasion. Subtracting them can do the same. Over repeated observations, both a sum and a difference have variability from both measurements.

Expectation adds linearly even when the variables are dependent. For variance, independence gives a simple sum of the squared-scale contributions.

Mean and variance of a linear combination

For constants $a,b,c$ ,

\mathrm E(aX+bY+c)=a\mathrm E(X)+b\mathrm E(Y)+c.

If $X$ and $Y$ are independent,

\operatorname{Var}(aX+bY+c) =a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y).

The expectation rule does not require independence. The variance rule is guaranteed under the stated independence assumption.
Not in the formula booklet: learn these combination rules.

In particular, for independent variables, both $X+Y$ and $X-Y$ have variance $\operatorname{Var}(X)+\operatorname{Var}(Y)$ . The minus sign is squared; it does not remove uncertainty.

Worked example 3: a weighted total

Question. Independent random variables $X,Y$ have means $8,5$ and variances $9,4$ , respectively. Find the mean and variance of $T=2X+3Y$ .

Plan. Combine the means with the signed coefficients and the variances with their squares.

\begin{aligned} \mathrm E(T)&=2(8)+3(5)=31\\ \operatorname{Var}(T)&=2^2(9)+3^2(4)=36+36=72 \end{aligned}

Answer. Mean $31$ , variance $72$ .

Check. One concrete independent pair with these moments takes $X=5,11$ and $Y=3,7$ , each equally likely. The four equally likely totals are $19,31,31,43$ , with mean $31$ and variance $(144+0+0+144)/4=72$ .

Keep the variables and change only the sign of the $Y$ term. Its contribution to the mean changes sign; its contribution to variance does not.

Worked example 4: a weighted difference

Question. For the same independent variables, find the mean and variance of $D=2X-3Y$ .

Plan. Use coefficient $-3$ in the expectation and $(-3)^2$ in the variance.

\begin{aligned} \mathrm E(D)&=2(8)-3(5)=1\\ \operatorname{Var}(D)&=2^2(9)+(-3)^2(4)=72 \end{aligned}

Answer. Mean $1$ , variance $72$ .

Check. Using the same concrete four outcomes gives differences $1,-11,13,1$ . Their mean is $1$ , and their squared deviations again average to $(0+144+144+0)/4=72$ .

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6.2 - Means and variances of combinations

Scaling and shifting

Combining two variables

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