Mathematics 9709 · Probability & Statistics 1

Expectation & variance

Imagine running the experiment many times and averaging the values you get.
That long-run average is $E(X)=\sum xp$ .
Each value is weighted by how often it happens.
$\operatorname{Var}(X)$ measures how spread out the values are.
It uses the same engine as Chapter 3: mean of squares minus square of mean.

E(X) = Σxp, Var(X) = Σx²p − [E(X)]²

$E(X)=\sum xp,\qquad \operatorname{Var}(X)=\sum x^2p-\big[E(X)\big]^2,\qquad \text{SD}=\sqrt{\operatorname{Var}}$

E(X) is the balance point of the distribution; the variance measures how far the mass sits from it. Get E(X) first, square it, subtract last.

E(X) is the balance point of the distribution; the variance measures how far the bars spread from it.

Picture the distribution as weights on a ruler: $E(X)$ is the point where it balances. Drag the mass and watch the fulcrum slide, and the variance grow as mass moves to the extremes.

centrespread

E(X) = Σxp = 2.00·Var(X) = Σx²p − [E(X)]² = 1.06

Worked example

Start simple: equal-probability values

$T\in\{1,3,6,10\}$ , each equally likely so $p=\tfrac14$ . With equal probabilities $E(T)$ is just the plain average of the values:

$E(T)=\tfrac14(1+3+6+10)=5,\qquad \sum t^2p=\tfrac14(1+9+36+100)=36.5$

$\operatorname{Var}(T)=36.5-5^2=11.5$

Answer

E(T)=5,\quad \operatorname{Var}(T)=11.5

Worked example

Next rung: a clean decimal table (E, Var, SD)

$x=1,2,3,4$ with $p=0.1,0.2,0.3,0.4$ . $E(X)=0.1+0.4+0.9+1.6=3$ ; $\sum x^2p=0.1+0.8+2.7+6.4=10$ .

$\operatorname{Var}(X)=10-3^2=1$

Answer

E(X)=3,\ \operatorname{Var}(X)=1,\ \text{SD}=1

Worked example

Fractional table: factor out the common denominator

$x=0,5,15,20$ with $p=\tfrac{1}{12},\tfrac{3}{12},\tfrac{5}{12},\tfrac{3}{12}$ . Pull $\tfrac{1}{12}$ outside so you add whole numbers:

$E(X)=\tfrac{1}{12}(0+15+75+60)=\tfrac{150}{12}=12.5$ .
$\sum x^2p=\tfrac{1}{12}(0+75+1125+1200)=\tfrac{2400}{12}=200$ .
$\operatorname{Var}(X)=200-12.5^2=43.75$ , so $\text{SD}=\sqrt{43.75}=6.61$ (3 s.f.).

Answer

E(X)=12.5,\ \operatorname{Var}(X)=43.75,\ \text{SD}\approx6.61

Worked example

Two unknowns: Σp = 1 AND E(X) together

$X\in\{0,1,2,3\}$ with $p=0.4,\ p,\ r,\ 0.15$ and $E(X)=1.1$ . Two letters need two equations.

$\sum p=1$ : $0.4+p+r+0.15=1\Rightarrow p+r=0.45$ .
$E(X)=p+2r+0.45=1.1\Rightarrow p+2r=0.65$ .
Subtract: $r=0.20$ , then $p=0.25$ .

Answer

p=0.25,\quad r=0.20

Whenever a table has two letters, hunt for a second fact — almost always the given E(X). (9709/52/O/N/23 Q1)

Worked example

Expected value in context: a decision tool

An investment's percentage profit over 3 years has this distribution:

Profit %	P
1, 5, 10, 15, 20	0.05, 0.10, 0.50, 0.20, 0.05
30, 40, 45, 50	0.04, 0.03, 0.02, 0.01

Expected percentage profit: $E(X)=\sum xp=13.35\%$ .
On a $\$40\,000$ stake that is $\tfrac{13.35}{100}\times 40\,000=\$5340$ expected return.

Answer

E(X)=13.35\%\Rightarrow\$5340\text{ expected}

E(X) is what you compare against alternatives (a savings rate, another venture). Expected frequency over N trials is N×P; an expected total is N×E(X).

$E(X)=\sum xp$ — value×prob, add. Keep it exact.
Square it: park $[E(X)]^2$ .
$\sum x^2p$ — square the value, ×prob, add.
$\operatorname{Var}=\sum x^2p-[E(X)]^2$ (the subtraction is the mark); SD = √Var if asked.

Asked for	Use	The mark / slip
E(X), mean	$\sum xp$	value×prob, not prob alone
Var(X)	$\sum x^2p-[E(X)]^2$	don't stop at Σx²p
SD	$\sqrt{\operatorname{Var}}$	root at the very end
expected total over N	$N\,E(X)$	a total, not a probability

Examiner note

Show the

\sum x^2p

line and the

-[E(X)]^2

step separately; if both E and Var are asked, label which is which, and keep

E(X)

exact through the Var line. (9709/52/M/J/24 Q5) (9709/52/M/J/23 Q1)

Why minus the square of the mean?

$\sum x^2p$ measures spread from 0; you want spread from the centre $E(X)$ . Subtracting $[E(X)]^2$ shifts the origin to the mean — the same $\overline{x^2}-\bar x^2$ engine as Chapter 3.

Common mistake

Subtracting

E(X)

instead of

[E(X)]^2

; squaring the probability instead of the value (

x^2p

, not

xp^2

); and rounding

E(X)

early. A negative variance always means a slip.

Common mistake

Giving two numbers for E and Var without saying which is which — the accuracy mark needs them identified.

Your turn— tap to reveal the worked answer (expected frequency)

For the clean table above ( $E(X)=3$ ), what is the expected total over 50 independent trials?

Expected total $=N\times E(X)=50\times3$ .

Answer

150

PreviousBuilding a probability distribution NextThe binomial distribution

Expectation & variance

Imagine running the experiment many times and averaging the values you get.
That long-run average is $E(X)=\sum xp$ .
Each value is weighted by how often it happens.
$\operatorname{Var}(X)$ measures how spread out the values are.
It uses the same engine as Chapter 3: mean of squares minus square of mean.

E(X) = Σxp, Var(X) = Σx²p − [E(X)]²

$E(X)=\sum xp,\qquad \operatorname{Var}(X)=\sum x^2p-\big[E(X)\big]^2,\qquad \text{SD}=\sqrt{\operatorname{Var}}$

E(X) is the balance point of the distribution; the variance measures how far the mass sits from it. Get E(X) first, square it, subtract last.

E(X) is the balance point of the distribution; the variance measures how far the bars spread from it.

Picture the distribution as weights on a ruler: $E(X)$ is the point where it balances. Drag the mass and watch the fulcrum slide, and the variance grow as mass moves to the extremes.

centrespread

E(X) = Σxp = 2.00·Var(X) = Σx²p − [E(X)]² = 1.06

Worked example

Start simple: equal-probability values

$T\in\{1,3,6,10\}$ , each equally likely so $p=\tfrac14$ . With equal probabilities $E(T)$ is just the plain average of the values:

$E(T)=\tfrac14(1+3+6+10)=5,\qquad \sum t^2p=\tfrac14(1+9+36+100)=36.5$

$\operatorname{Var}(T)=36.5-5^2=11.5$

Answer

E(T)=5,\quad \operatorname{Var}(T)=11.5

Worked example

Next rung: a clean decimal table (E, Var, SD)

$x=1,2,3,4$ with $p=0.1,0.2,0.3,0.4$ . $E(X)=0.1+0.4+0.9+1.6=3$ ; $\sum x^2p=0.1+0.8+2.7+6.4=10$ .

$\operatorname{Var}(X)=10-3^2=1$

Answer

E(X)=3,\ \operatorname{Var}(X)=1,\ \text{SD}=1

Worked example

Fractional table: factor out the common denominator

$x=0,5,15,20$ with $p=\tfrac{1}{12},\tfrac{3}{12},\tfrac{5}{12},\tfrac{3}{12}$ . Pull $\tfrac{1}{12}$ outside so you add whole numbers:

$E(X)=\tfrac{1}{12}(0+15+75+60)=\tfrac{150}{12}=12.5$ .
$\sum x^2p=\tfrac{1}{12}(0+75+1125+1200)=\tfrac{2400}{12}=200$ .
$\operatorname{Var}(X)=200-12.5^2=43.75$ , so $\text{SD}=\sqrt{43.75}=6.61$ (3 s.f.).

Answer

E(X)=12.5,\ \operatorname{Var}(X)=43.75,\ \text{SD}\approx6.61

Worked example

Two unknowns: Σp = 1 AND E(X) together

$X\in\{0,1,2,3\}$ with $p=0.4,\ p,\ r,\ 0.15$ and $E(X)=1.1$ . Two letters need two equations.

$\sum p=1$ : $0.4+p+r+0.15=1\Rightarrow p+r=0.45$ .
$E(X)=p+2r+0.45=1.1\Rightarrow p+2r=0.65$ .
Subtract: $r=0.20$ , then $p=0.25$ .

Answer

p=0.25,\quad r=0.20

Whenever a table has two letters, hunt for a second fact — almost always the given E(X). (9709/52/O/N/23 Q1)

Worked example

Expected value in context: a decision tool

An investment's percentage profit over 3 years has this distribution:

Profit %	P
1, 5, 10, 15, 20	0.05, 0.10, 0.50, 0.20, 0.05
30, 40, 45, 50	0.04, 0.03, 0.02, 0.01

Expected percentage profit: $E(X)=\sum xp=13.35\%$ .
On a $\$40\,000$ stake that is $\tfrac{13.35}{100}\times 40\,000=\$5340$ expected return.

Answer

E(X)=13.35\%\Rightarrow\$5340\text{ expected}

E(X) is what you compare against alternatives (a savings rate, another venture). Expected frequency over N trials is N×P; an expected total is N×E(X).

$E(X)=\sum xp$ — value×prob, add. Keep it exact.
Square it: park $[E(X)]^2$ .
$\sum x^2p$ — square the value, ×prob, add.
$\operatorname{Var}=\sum x^2p-[E(X)]^2$ (the subtraction is the mark); SD = √Var if asked.

Asked for	Use	The mark / slip
E(X), mean	$\sum xp$	value×prob, not prob alone
Var(X)	$\sum x^2p-[E(X)]^2$	don't stop at Σx²p
SD	$\sqrt{\operatorname{Var}}$	root at the very end
expected total over N	$N\,E(X)$	a total, not a probability

Examiner note

Show the

\sum x^2p

line and the

-[E(X)]^2

step separately; if both E and Var are asked, label which is which, and keep

E(X)

exact through the Var line. (9709/52/M/J/24 Q5) (9709/52/M/J/23 Q1)

Why minus the square of the mean?

$\sum x^2p$ measures spread from 0; you want spread from the centre $E(X)$ . Subtracting $[E(X)]^2$ shifts the origin to the mean — the same $\overline{x^2}-\bar x^2$ engine as Chapter 3.

Common mistake

Subtracting

E(X)

instead of

[E(X)]^2

; squaring the probability instead of the value (

x^2p

, not

xp^2

); and rounding

E(X)

early. A negative variance always means a slip.

Common mistake

Giving two numbers for E and Var without saying which is which — the accuracy mark needs them identified.

Your turn— tap to reveal the worked answer (expected frequency)

For the clean table above ( $E(X)=3$ ), what is the expected total over 50 independent trials?

Expected total $=N\times E(X)=50\times3$ .

Answer

150