Mathematics 9709 · Probability & Statistics 2

The two rules of combining

Real quantities are built by scaling and adding random pieces (a triathlon time = swim + cycle + run). The headline lesson: means behave the obvious way; variances do not. Adding a constant slides the whole spread along unchanged; scaling multiplies variance by the coefficient squared; and combining two independent variables always ADDS the variances, even for a difference.

Mean is honest; variance is sneaky

Picture a die relabelled $+3$ : every score moves three to the right, so the mean moves +3, but the numbers sit exactly as far apart as before, so the spread is identical. That one picture is the whole rule: a constant $+b$ shifts the mean and leaves the variance untouched.

Adding 3 to every value shifts the centre from 2 to 5 but keeps the shape (and so the variance) identical: a +b moves the mean, never the spread.

$E(aX+b)=aE(X)+b,\quad \operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)$
$E(aX\pm bY)=aE(X)\pm bE(Y),\quad \operatorname{Var}(aX\pm bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$

The $+b$ vanishes from the variance (a constant has no spread); the $a$ comes out squared; and for independent $X,Y$ you add the variances even when the variables are subtracted, because two sources of randomness never cancel each other's wobble.

+b moves the mean, not the variance; a comes out a²; and Var(X − Y) = Var(X) + Var(Y), never minus.

Worked example

Mean and variance straight from a table

$X$ takes 2, 3, 5, 7 with probabilities 0.4, 0.1, 0.3, 0.2. Read off the two summary numbers before doing anything else:

$E(X)=\textstyle\sum xP=4,\qquad \operatorname{Var}(X)=\textstyle\sum x^2P-E(X)^2=19.8-16=3.8$

Answer

E(X)=4,\ \operatorname{Var}(X)=3.8

Worked example

Scale a single variable

Same $X$ ( $E=4,\ \operatorname{Var}=3.8$ ). For $2X+3$ , the mean follows the obvious rule and the variance squares the multiplier:

$E(2X+3)=2(4)+3=11,\qquad \operatorname{Var}(2X+3)=2^2(3.8)=15.2$

Answer

E=11,\ \operatorname{Var}=15.2

The $+3$ bumps the mean but is invisible to the variance.

Worked example

The rules work on named distributions too

$X\sim B(8,0.3)$ , so quote the binomial mean and variance, then apply the linear rules:

$E(X)=np=2.4$ , $\operatorname{Var}(X)=npq=8(0.3)(0.7)=1.68$ .
$E(4X-1)=4(2.4)-1=8.6$ .
$\operatorname{Var}(4X-1)=4^2(1.68)=26.88$ (the $-1$ is invisible to variance).

Answer

E(4X-1)=8.6,\ \operatorname{Var}(4X-1)=26.88

Combining two independent variables: add the variances

Sum or difference, the means combine the natural way; the variances always add. State “since independent”, square every coefficient, then add.

Worked example

Sum and difference from two tables

Independent $X$ (2, 3, 5 with 0.2, 0.3, 0.5) and $Y$ (0, 1 with 0.6, 0.4) give $E(X)=3.8,\ \operatorname{Var}(X)=1.56$ and $E(Y)=0.4,\ \operatorname{Var}(Y)=0.24$ .

$S=X+Y$ : $E(S)=3.8+0.4=4.2$ , $\operatorname{Var}(S)=1.56+0.24=1.8$ .
$T=X-Y$ : $E(T)=3.8-0.4=3.4$ , $\operatorname{Var}(T)=1.56+0.24=1.8$ .
The two variances are identical: subtracting does not subtract spread.

Answer

\operatorname{Var}(X+Y)=\operatorname{Var}(X-Y)=1.8

Worked example

A difference adds variances

$X\sim B(12,0.2)$ ( $E=2.4,\ \operatorname{Var}=1.92$ ), $Y\sim B(2,0.8)$ ( $E=1.6,\ \operatorname{Var}=0.32$ ), independent. For $W=X-Y$ :

$E(W)=2.4-1.6=0.8,\qquad \operatorname{Var}(W)=1.92+0.32=2.24$

Answer

E(W)=0.8,\ \operatorname{Var}(W)=2.24

Means combine the obvious way: $E(aX\pm bY)=aE(X)\pm bE(Y)$ .
State “since independent”.
Variances: $a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$ , coefficients SQUARED, always ADD.
A $+b$ constant moves the mean, not the variance.

The signature trap: 2X is not X₁ + X₂

$2X$ means “roll once and double it”; $X_1+X_2$ means “roll twice and add”. Same mean, but very different spread, because two rolls let highs and lows partly cancel.

Worked example

When is 2X = X₁ + X₂?

Xing's fair die scores 1, 1, 2, 2, 2, 4, giving $E(X)=2$ , $\operatorname{Var}(X)=1$ . Compare doubling one roll with adding two independent rolls:

Double one roll: $E(2X)=2(2)=4$ , $\operatorname{Var}(2X)=2^2(1)=4$ .
Add two rolls: $E(X_1+X_2)=2+2=4$ , $\operatorname{Var}(X_1+X_2)=1+1=2$ .
Means match ( $4=4$ ) but variances do not: $4\ne2$ .

Answer

E(2X)=E(X_1+X_2)=4,\ \text{but }\operatorname{Var}(2X)=4\ne\operatorname{Var}(X_1+X_2)=2

“Total of $n$ ” scales variance by $n$ ; “ $n$ times one” scales it by $n^2$ .

“Total of n” (X₁+…+Xₙ, Var ×n) ≠ “n times one” (nX, Var ×n²). Same mean, different variance.

Combination	Mean	Variance
$aX+b$	$aE(X)+b$	$a^2\operatorname{Var}(X)$
$aX\pm bY$	$aE(X)\pm bE(Y)$	$a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$
sum of n copies	$nE(X)$	$n\operatorname{Var}(X)$
$nX$	$nE(X)$	$n^2\operatorname{Var}(X)$

Your turn— tap to reveal the worked answer (rules)

Independent $X\sim B(20,0.3)$ and $Y\sim B(12,0.4)$ . Find $E(2X-Y)$ and $\operatorname{Var}(2X-Y)$ .

$E(X)=6,\ \operatorname{Var}(X)=4.2$ ; $E(Y)=4.8,\ \operatorname{Var}(Y)=2.88$ . So $E=2(6)-4.8$ , $\operatorname{Var}=2^2(4.2)+2.88$ .

Answer

E(2X-Y)=7.2,\ \operatorname{Var}(2X-Y)=19.68

Examiner note

Write “since independent, variances add”, square every coefficient, and keep variance (not SD) in the

N(\cdot,\cdot)

slot. (9709/62/M/J/23 Q5) (9709/62/M/J/22 Q6)

Common mistake

Subtracting variances for a difference (never), forgetting to square the multiplier (

\operatorname{Var}(3X)=9\operatorname{Var}(X)

), adding

b^2

into the variance, and confusing

2X

with

X_1+X_2

. State “since independent” before adding variances.

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The two rules of combining

Mean is honest; variance is sneaky

Adding 3 to every value shifts the centre from 2 to 5 but keeps the shape (and so the variance) identical: a +b moves the mean, never the spread.

$E(aX+b)=aE(X)+b,\quad \operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)$
$E(aX\pm bY)=aE(X)\pm bE(Y),\quad \operatorname{Var}(aX\pm bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$

+b moves the mean, not the variance; a comes out a²; and Var(X − Y) = Var(X) + Var(Y), never minus.

Worked example

Mean and variance straight from a table

$X$ takes 2, 3, 5, 7 with probabilities 0.4, 0.1, 0.3, 0.2. Read off the two summary numbers before doing anything else:

$E(X)=\textstyle\sum xP=4,\qquad \operatorname{Var}(X)=\textstyle\sum x^2P-E(X)^2=19.8-16=3.8$

Answer

E(X)=4,\ \operatorname{Var}(X)=3.8

Worked example

Scale a single variable

Same $X$ ( $E=4,\ \operatorname{Var}=3.8$ ). For $2X+3$ , the mean follows the obvious rule and the variance squares the multiplier:

$E(2X+3)=2(4)+3=11,\qquad \operatorname{Var}(2X+3)=2^2(3.8)=15.2$

Answer

E=11,\ \operatorname{Var}=15.2

The $+3$ bumps the mean but is invisible to the variance.

Worked example

The rules work on named distributions too

$X\sim B(8,0.3)$ , so quote the binomial mean and variance, then apply the linear rules:

$E(X)=np=2.4$ , $\operatorname{Var}(X)=npq=8(0.3)(0.7)=1.68$ .
$E(4X-1)=4(2.4)-1=8.6$ .
$\operatorname{Var}(4X-1)=4^2(1.68)=26.88$ (the $-1$ is invisible to variance).

Answer

E(4X-1)=8.6,\ \operatorname{Var}(4X-1)=26.88

Combining two independent variables: add the variances

Sum or difference, the means combine the natural way; the variances always add. State “since independent”, square every coefficient, then add.

Worked example

Sum and difference from two tables

Independent $X$ (2, 3, 5 with 0.2, 0.3, 0.5) and $Y$ (0, 1 with 0.6, 0.4) give $E(X)=3.8,\ \operatorname{Var}(X)=1.56$ and $E(Y)=0.4,\ \operatorname{Var}(Y)=0.24$ .

$S=X+Y$ : $E(S)=3.8+0.4=4.2$ , $\operatorname{Var}(S)=1.56+0.24=1.8$ .
$T=X-Y$ : $E(T)=3.8-0.4=3.4$ , $\operatorname{Var}(T)=1.56+0.24=1.8$ .
The two variances are identical: subtracting does not subtract spread.

Answer

\operatorname{Var}(X+Y)=\operatorname{Var}(X-Y)=1.8

Worked example

A difference adds variances

$X\sim B(12,0.2)$ ( $E=2.4,\ \operatorname{Var}=1.92$ ), $Y\sim B(2,0.8)$ ( $E=1.6,\ \operatorname{Var}=0.32$ ), independent. For $W=X-Y$ :

$E(W)=2.4-1.6=0.8,\qquad \operatorname{Var}(W)=1.92+0.32=2.24$

Answer

E(W)=0.8,\ \operatorname{Var}(W)=2.24

Means combine the obvious way: $E(aX\pm bY)=aE(X)\pm bE(Y)$ .
State “since independent”.
Variances: $a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$ , coefficients SQUARED, always ADD.
A $+b$ constant moves the mean, not the variance.

The signature trap: 2X is not X₁ + X₂

$2X$ means “roll once and double it”; $X_1+X_2$ means “roll twice and add”. Same mean, but very different spread, because two rolls let highs and lows partly cancel.

Worked example

When is 2X = X₁ + X₂?

Xing's fair die scores 1, 1, 2, 2, 2, 4, giving $E(X)=2$ , $\operatorname{Var}(X)=1$ . Compare doubling one roll with adding two independent rolls:

Double one roll: $E(2X)=2(2)=4$ , $\operatorname{Var}(2X)=2^2(1)=4$ .
Add two rolls: $E(X_1+X_2)=2+2=4$ , $\operatorname{Var}(X_1+X_2)=1+1=2$ .
Means match ( $4=4$ ) but variances do not: $4\ne2$ .

Answer

E(2X)=E(X_1+X_2)=4,\ \text{but }\operatorname{Var}(2X)=4\ne\operatorname{Var}(X_1+X_2)=2

“Total of $n$ ” scales variance by $n$ ; “ $n$ times one” scales it by $n^2$ .

“Total of n” (X₁+…+Xₙ, Var ×n) ≠ “n times one” (nX, Var ×n²). Same mean, different variance.

Combination	Mean	Variance
$aX+b$	$aE(X)+b$	$a^2\operatorname{Var}(X)$
$aX\pm bY$	$aE(X)\pm bE(Y)$	$a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$
sum of n copies	$nE(X)$	$n\operatorname{Var}(X)$
$nX$	$nE(X)$	$n^2\operatorname{Var}(X)$

Your turn— tap to reveal the worked answer (rules)

Independent $X\sim B(20,0.3)$ and $Y\sim B(12,0.4)$ . Find $E(2X-Y)$ and $\operatorname{Var}(2X-Y)$ .

$E(X)=6,\ \operatorname{Var}(X)=4.2$ ; $E(Y)=4.8,\ \operatorname{Var}(Y)=2.88$ . So $E=2(6)-4.8$ , $\operatorname{Var}=2^2(4.2)+2.88$ .

Answer

E(2X-Y)=7.2,\ \operatorname{Var}(2X-Y)=19.68

Examiner note

Write “since independent, variances add”, square every coefficient, and keep variance (not SD) in the

N(\cdot,\cdot)

slot. (9709/62/M/J/23 Q5) (9709/62/M/J/22 Q6)

Common mistake

Subtracting variances for a difference (never), forgetting to square the multiplier (

\operatorname{Var}(3X)=9\operatorname{Var}(X)

), adding

b^2

into the variance, and confusing

2X

with

X_1+X_2

. State “since independent” before adding variances.