When dealing with a model, we use the letter for the mean.

We write

)( xXxP

or, more often, replacing p by , )( xXP

Instead of , we can also write E(X).

( Think of this as being what we expect to get on average ).

This notation comes from the idea of the mean being the Expected value of the r.v. X.

px

Mean of a Discrete Random Variable

S1 Reminder about Expectation

e.g. 1. A random variable X has the probability distribution

P 4

1

2

11 5 10x

(X = )x p

Find (a) the value of p and (b) the mean of X.

Solution:

(a) Since X is a discrete r.v., 1)( xXP

121

41 p

41p

(b) mean,( ) ( )E X xP X x 411

41 1051 2 4

21

54Col max

–35–32

2241

Rowmin21

BA

Mixed StrategiesUsing probabilities to determine the best mixed strategy

1) 22 games

maximin

minimax

Max(row/min) 2 ≠ Min(col max) 4 so not stable

Check to see if there is a stable solution

5–32

241

21 BA

Suppose player A chooses option 1 (row1) with probability p and option 2 (row 2) with probability 1– p

If B chooses option 2 then the expected win for A is: E(win) = 2p + 5(1 – p) = –3p + 5

The expected pay–off is deduced using E(X) = xP(X=x) from S1

If B chooses option 1 then the expected win for A is: E(win) = 4p – 3(1 – p) = 7p – 3

p

1–p

-1

-2

-3

-4

-5

1

2

3

4

5

0.2 0.4 0.6 0.8 1.00X->

|̂Y

A`s expected pay-off

p

B chooses 1B chooses 2

These 2 lines are plotted as shown and intersect at the highest minimum winnings when:

7p – 3 = –3p + 5

10p = 8

p = 0.8

E(win) = –3p + 5 E(win) = 7p – 3

So if A chooses option 1 with probability 0.8 then his expected win is:

E(win) = 70.8 – 3 = 2.6

E(win) = 7p – 3

If B chooses option 1 then

If A chooses option 1 with a value of less than 0.8 then:

If B chooses option 2 then E(win) = –3p + 5

E(win) = 7p – 3

E(win ) ≥ 2.6

E(win) ≤ 2.6

The expected win of 2.6 is the greatest that A can guarantee. The value of the game is 2.6. V(A) = 2.6

The same analysis can be carried out for player B but considering the losses.

Remember the matrix shows A`s gain so to find B`s gain change the signs.

5–32

241

21 BA

A`s gain

–532

–2–41

21 BA

B`s gain

Suppose player B chooses option 1(col 1) with probability q and option 2 (col 2) with probability 1 – q

–532

–2–41

21 BA

If A chooses option 2 then the expected win for B is:

The expected pay–off is deduced using E(X) = xP(X=x) from S1

If A chooses option 1 then the expected win for B is:

E(win) = –4q – 2(1 – q) = –2q – 2

E(win) = 3q – 5(1 – q) = 8q – 5

q 1–q

The intersection of these 2 lines occurs when:

–2q – 2 = 8q – 5

10q = 3

q = 0.3

If q = 0.3 then B expects to lose 2.6 as we expected from considering A`s winnings.

So to minimise his loses B should choose option 1 with probability 0.3 and option 2 with probability 0.7

E(win) = –2q – 2 = -2.6