Expectation of a random variable | CAIE 9709 Probability & Statistics 1 Notes

Probability & Statistics 1

Mathematics9709

Multiply each value by its probability

E(X)=\sum xp

Symbols

$x$ = one possible value of X (context unit)
$p$ = probability of that value (none)
$E(X)$ = expectation or long-run mean (same as X)

Here $p=P(X=x)$ , so the same formula can be written as $E(X)=\sum xP(X=x)$ .

In many repetitions, a value with probability $p$ occurs about $Np$ times out of $N$ . That is why the mean uses probability weights.

Worked example

A four-value distribution

x	P(X = x)
0	0.1
1	0.3
2	0.4
3	0.2

x	0	1	2	3
P(X = x)	0.1	0.3	0.4	0.2

0(0.1)+1(0.3)=0.3

2(0.4)+3(0.2)=1.4

Answer

E(X)=0.3+1.4=1.7

Common mistake

Do not average the listed values unless they are equally likely. Match every value to the probability directly below it.

Expectation predicts an average, not the next result

$E(X)$ need not be a possible value. A long-run average of 1.7 is possible even when each result is an integer.
It has the same units as the random variable.
Across $N$ repetitions of the same model, the expected total is $N E(X)$ .

Worked example

Expected total over 200 games

If the expected score in one game is 1.7, then

Answer

\text{expected total}=200(1.7)=340

Key idea

Expected does not mean guaranteed. A short run can be far from the expectation; the interpretation is about repeated use of the model.

Two unknown probabilities need two equations

The total $\sum p=1$ gives one equation. A stated expectation can give the second.

Worked example

Use the total and E(X) together

For $X=0,1,2,3$ , the probabilities are $0.4,p,r,0.15$ and $E(X)=1.1$ . (9709/52/O/N/23 Q1)

Total: $p+r+0.55=1$ , so $p+r=0.45$ .
Expectation: $p+2r+0.45=1.1$ , so $p+2r=0.65$ .
Subtract, then substitute back.

Answer

r=0.20,\qquad p=0.25

Examiner note

Write both equations. A correct pair of final numbers alone can lose the method marks shown in the mark scheme.

Your turnIndependent check [3 marks]

$X=0,1,2$ has probabilities 0.2, 0.5 and 0.3. Find $E(X)$ and the expected total over 40 repetitions.

Show worked answer

$E(X)=0(0.2)+1(0.5)+2(0.3)=1.1$ .

Answer

E(X)=1.1

Answer

40E(X)=44

Multiply each value by its probability

E(X)=\sum xp

Symbols

$x$ = one possible value of X (context unit)
$p$ = probability of that value (none)
$E(X)$ = expectation or long-run mean (same as X)

Here $p=P(X=x)$ , so the same formula can be written as $E(X)=\sum xP(X=x)$ .

In many repetitions, a value with probability $p$ occurs about $Np$ times out of $N$ . That is why the mean uses probability weights.

Worked example

A four-value distribution

x	P(X = x)
0	0.1
1	0.3
2	0.4
3	0.2

x	0	1	2	3
P(X = x)	0.1	0.3	0.4	0.2

0(0.1)+1(0.3)=0.3

2(0.4)+3(0.2)=1.4

Answer

E(X)=0.3+1.4=1.7

Common mistake

Do not average the listed values unless they are equally likely. Match every value to the probability directly below it.

Expectation predicts an average, not the next result

$E(X)$ need not be a possible value. A long-run average of 1.7 is possible even when each result is an integer.
It has the same units as the random variable.
Across $N$ repetitions of the same model, the expected total is $N E(X)$ .

Worked example

Expected total over 200 games

If the expected score in one game is 1.7, then

Answer

\text{expected total}=200(1.7)=340

Key idea

Expected does not mean guaranteed. A short run can be far from the expectation; the interpretation is about repeated use of the model.

Two unknown probabilities need two equations

The total $\sum p=1$ gives one equation. A stated expectation can give the second.

Worked example

Use the total and E(X) together

For $X=0,1,2,3$ , the probabilities are $0.4,p,r,0.15$ and $E(X)=1.1$ . (9709/52/O/N/23 Q1)

Total: $p+r+0.55=1$ , so $p+r=0.45$ .
Expectation: $p+2r+0.45=1.1$ , so $p+2r=0.65$ .
Subtract, then substitute back.

Answer

r=0.20,\qquad p=0.25

Examiner note

Write both equations. A correct pair of final numbers alone can lose the method marks shown in the mark scheme.

Your turnIndependent check [3 marks]

$X=0,1,2$ has probabilities 0.2, 0.5 and 0.3. Find $E(X)$ and the expected total over 40 repetitions.

Show worked answer

$E(X)=0(0.2)+1(0.5)+2(0.3)=1.1$ .

Answer

E(X)=1.1

Answer

40E(X)=44