Chapter 3
Game Theory
3.1
Two-player constant-sum games
Let A and B be two competitors (‘players’). Each has a set of available strategies. A
game consists of several rounds. In each round the players, without co-operating, choose
their strategies and obtain a certain payoff.
The payoff function may be represented by a table or matrix, in which case the game is
in strategic form. Usually we only show the payoff matrix for player A. In this chapter,
matrices are denoted by bold letters, e.g. A, to distinguish them from the players’ names.
Example (i)
A and B both have one black card and one red card. They each select one card, and display
them simultaneously. If the displayed cards are of the same colour then B pays A £1. If
they are of different colours then A pays B £1. The payoff matrix for player A is:
Player A
Black
Red
Player B
Black Red
1
−1
−1
1
or more simply A =
1 −1
−1
1
A two-player constant-sum game is a game in which, when the payoff to A is k, the
payoff to B is c − k. The sum of the payoffs to A and B has a constant value c.
When c = 0, so A gains exactly the amount that B loses, the game is a zero-sum game.
A game is solved by finding strategies which optimize the players’ payoffs.
If it is optimal to do the same thing every time, we say there is a solution in pure strategies.
If the players’ best course of action is to pick strategies at random with certain probabilities,
we call this a solution in mixed strategies.
Let A be the payoff matrix for A in a two-player constant-sum game between A and B.
Each row of A has a minimal entry. The largest of these is the maximin entry. Player A
can choose a strategy which gives at least this payoff, whatever strategy Player B uses.
Each column of A has a maximal entry. The smallest of these is the minimax entry.
Player B can choose a strategy so that Player A gains no more than this, whatever strategy
A uses.
25
Example (ii)
A and B share £50 every week. Each of them chooses one of the numbers 1, 2, 3. If their
numbers differ by 2, they get £25 each. If they differ by 1, the person with the larger
number gets £35 and the other gets £15. If the numbers are equal then A gets that number
times £10 and B gets the rest.
This game has constant sum £50. The payoff matrix for A is:
Player A
1
2
3
Column maximum
minimax
Player
1
2
10 15
35 20
25 35
35 35
B
3
25
15
30
30
30
Row minimum
10
15
25
maximin
25
The maximin is 25, so A can guarantee a payoff of at least £25 by choosing 3.
The minimax is 30, so B can guarantee a ‘loss’ to A of at most £30 (i.e. a payoff of at least
£20) by choosing 3.
A constant-sum game can be converted into a zero-sum game by subtracting half the constant sum (25 in the above example) from each entry of the matrix. (If A and B each get
£25 then neither ‘loses’ nor ‘wins’ from the other.) The resulting matrix represents A’s
payoff in the zero-sum game. In the example, the maximin and minimax would then be 0
and 5 respectively.
The maximin entry of the game matrix A is
min
max
min ars and the minimax entry is
r=1,...,m s=1,...,n
max ars .
s=1,...,n r=1,...,m
Thus by Proposition 2.3 we always have maximin ≤ minimax. By Proposition 2.4, they are
equal if and only if there is a pair (i, j) such that arj ≤ aij ≤ ais for all r, s, i.e. an entry
which is both maximal in its column and minimal in its row.
When there is such an entry, the payoff matrix is said to have a saddle-point at (Ai , Bj ).
This pair is also called a solution of the game in pure strategies. Neither player can
benefit by unilaterally changing from their saddle-point strategy. aij , the payoff to Player A
at the saddle point, is called the value of the game.
There may sometimes be more than one saddle point, but they will all yield the same value
for the game.
Example (iii)
Consider the game with this payoff matrix for Player A:
Player A
Column maximum
minimax
A1
A2
A3
Player B
B1 B2 B3
−1
2
0
1
1
3
0
2 −1
1
2
3
1
Row minimum
−1
1
−1
maximin
1
Minimax = maximin = 1, so the value of the game is 1 and there is a solution in pure
strategies, (A2 , B1 ).
26
In Example (iii), B would never play strategy B2 because B1 would always be at least as
good. Then A would never play A3 as this would be less good than A2 . Column 2 and row
3 are dominated, which means they can be deleted without affecting the solution.
Definitions
If every entry in row i is less than or equal to the entry in the same column in row j, then
row i is dominated by row j and can be deleted.
If every entry in column i is greater than or equal to the entry in the same row in column
j, then column i is dominated by column j and can be deleted.
(Note that we can delete a minimal row and a maximal column.)
Example (iv)
Consider the game with this payoff matrix for Player A:
Player B
B1 B2 B3 Row minimum maximin
A1
2 −2 −4
−4
Player A
A2 −2
2
2
−2
−2
4 −2
0
−2
A3
Column maximum
4
2
2
minimax
2
A would never play strategy A1 . Row 1 is dominated by Row 3, so delete Row 1.
Then B would never play strategy B3 , because B2 is at least as good and sometimes better.
Column 3 (after Row 1 has been deleted) is dominated by Column 2, so delete Column 3.
These deletions do not affect the fact that maximin = −2 and minimax = 2. There is no
solution in pure strategies.
When a unique pure-strategy solution exists, as in Example (iii), one way of finding it is to
delete dominated rows and columns until only one entry remains.
3.2
Solutions in mixed strategies
Most games do not have pure-strategy solutions, such as Example (iv) above in which it is
best for player A to play strategy A2 if B plays the minimax strategy B2 . However, then
it would be beter for B to play B1 , and so on. The game is unstable.
Let A and B have m and n available strategies respectively. Suppose A chooses strategy Ai
with probability pi , and B chooses strategy Bj with probability qj . Then
p1 + · · · + pm = 1
and
q1 + · · · + qn = 1.
Let p = (p1 · · · pm )t and q = (q1 · · · qn )t .
Each of p and q is a probability vector, also called a mixed strategy.
Let A = (aij ) be the m × n payoff matrix for A. Assuming theX
rounds to be independent,
the mean or expected payoff to A in any round is E(p, q) =
aij pi qj = pt Aq.
i,j
For example, the game in Example (i) has no solution in pure strategies. Suppose A chooses
a black card once in every three rounds, and B chooses black once in every four rounds,
these choices being random and independent. We say A is using the mixed strategy ( 13 , 32 )
1 1 −1
1 3
1
2
4
and B is using ( 4 , 4 ). The expected payoff to A is then 3 3
= £ 16
3
−1
1
4
On the other hand, if both players use the strategy ( 12 , 12 ) then the mean payoff is 0.
27
Let Pm , Pn be the sets of all probability vectors in Rm , Rn respectively. These sets are
compact so all continuous functions on them attain their least and greatest values.
If A chooses a mixed strategy p, the worst possible gain for A is the smallest value of pt Aq
as q varies in Pn . To ensure the ‘best worst outcome’ A should choose p = p∗ to solve the
problem
Maximize min pt Aq subject to p ∈ Pm .
q∈Pn
If B chooses a mixed strategy q, the worst possible outcome for B is that A gains the largest
value of pt Aq as p varies in Pm . Thus B should choose q = q∗ to solve the problem
Minimize max pt Aq subject to q ∈ Pn .
p∈Pm
Note that the two problems are duals of each other.
p∗ and q∗ are called optimal mixed strategies for A and B respectively. The pair (p∗ , q∗ )
is called a solution in mixed strategies.
In due course we shall prove:
Proposition 3.1 (Fundamental theorem of zero-sum games) For any two-player zerosum game, there exist mixed strategies p∗ , q∗ such that for all mixed strategies p, q we have
E(p, q∗ ) ≤ E(p∗ , q∗ ) ≤ E(p∗ , q).
The value of the game is defined to be v = E(p∗ , q∗ ). This is consistent with our earlier
definition in the pure-strategy case. It is the expected payoff to A when both players act
optimally. Without co-operating, neither can benefit by unilaterally deviating from their
optimal strategy.
Thus by using the mixed strategy p∗ , Player A can ensure an expected payoff of at least v
regardless of B’s strategy.
There is no way that A can ensure an expected payoff of more than v, for suppose there
were another mixed strategy p′ for A such that E(p′ , q) > v for all q. Then in particular
E(p′ , q∗ ) > v, which contradicts the property that E(p, q∗ ) ≤ v for all p.
Thus the best A can do is to ensure an expected payoff of v. Similarly, the best B can do
is to prevent A’s expected payoff from exceeding v.
The optimal mixed strategies p∗ , q∗ may not be unique, but every mixed-strategy solution
will yield the same value of the game.
A pure-strategy solution (Ai , Bj ) is a special case of a mixed-strategy solution, where
pi = qj = 1 and all other entries are 0.
Now consider the problem:
Minimize ct q subject to q ∈ Pn , where c is a fixed vector.
Let cj be the smallest entry of c. Let ej have 1 in position j and 0 elsewhere. Then ej ∈ Pn
and ct ej = cj .
For any q ∈ Pn , ct q = c1 q1 + · · · + cn qn ≥ cj q1 + · · · + cj qn = cj (q1 + · · · + qn ) = cj . Hence
the minimum of ct q is cj when q = ej .
To apply this to Player A’s problem, take c = At p to get min pt Aq = min pt Aej , which
q∈Pn
j=1,...,n
is the minimum of the expected payoffs to A when B plays a pure strategy.
A’s problem is to find p∗ to maximize this.
Similarly max pt Aq = max ei t Aq, the maximum of the expected payoffs to B when A
p∈Pm
i=1,...,m
plays a pure strategy. B’s problem is to find q∗ to minimize this.
28
3.3
The 2 × n case
Suppose player A has just two available strategies and B has n. Let A = (aij ) be the 2 × n
payoff matrix for A.
If there is a pure-strategy solution, it can easily be found by inspection. Otherwise, let A
play A1 , A2 with probabilities p, 1 − p respectively.
If B plays strategy Bj then A’s expected payoff is a1j p + a2j (1 − p) = (a1j − a2j )p + a2j .
We can graph each of these n linear payoff functions. A’s minimum payoff is maximized
at the highest point on their lower envelope, i.e. the highest point of the region that lies
below all the lines.
Example
If A =
3 2 4
2 4 1
, there is no solution in pure strategies.
The payoffs for A when B plays each of B1 , B2 , B3 are p + 2, −2p + 4, 3p + 1.
We must find the maximum value of the smallest of these as p varies between 0 and 1.
4
3.5
3
2.5
2
1.5
1
0
0.2
0.4
0.6
0.8
1
p
2
The highest point on the lower envelope occurs where p = , so A’s optimal mixed strategy
3
2 1
8
is
,
. The value of the game is . When A uses this strategy, B cannot choose a
3 3
3
strategy that will ensure a lower mean payoff to A.
a b
In the 2 × 2 case we can generalise this result. Let A =
, so there are two payoff
c d
functions (a − c)p + c and (b − d)p + d. Assuming the solution is where these intersect (i.e.
d−c
a−b
A has no solution in pure strategies), we get p =
, so 1 − p =
.
a−b−c+d
a−b−c+d
a−c
d−b
, 1−q =
.
Similarly B’s optimal mixed strategy is (q, 1 − q): q =
a−b−c+d
a−b−c+d
So long as these probabilities are all in the interval (0, 1), they give a valid optimal strategy.
Note that these results do not apply if the game matrix has a saddle point.
The value of the game is the payoff to A when these probabilities are used, which can be
ad − bc
shown to be
. Note that the numerator is det(A).
a−b−c+d
Proposition 3.2 Let A be the payoff matrix for Player A in a two-player zero-sum game in
which each player has two available strategies. If A doesnot have a saddle-point
then the op
d−c
d−b
a−b
a−c
timal mixed strategies are
,
,
for A and
a−b−c+d a−b−c+d
a−b−c+d a−b−c+d
ad − bc
.
for B. The value of the game is
a−b−c+d
29
3.4
Solution by linear programming
We have seen that A must find p to maximize the minimum payoff when B plays a pure
m
P
strategy. When B uses strategy Bj this payoff is pt Aej , which equals
aij pi .
i=1
To maximize the smallest of these A must choose pi ≥ 0 for i = 1, . . . , m, with
m
P
pi = 1,
i=1
to
(
m
X
maximize min
ai1 pi , . . . ,
m
X
ain pi
i=1
i=1
Similarly B’s problem is to choose qj ≥ 0 for j = 1, . . . , n with
!)
n
P
.
qj = 1 so as to
j=1
minimize




max 
n
X
a1j qj , . . . ,
j=1
n
X
j=1


amj qj  .

Let ars be the maximin entry of A and suppose this is positive. Then every entry in row
m
P
r of A must be positive. Taking p = er gives
aij pi = arj > 0 for j = 1, . . . , n, so for
i=1
m
m
P
P
ai1 pi , . . . ,
this p we have min
ain pi > 0. It follows that the maximum value of this
i=1
i=1
expression over all probability vectors p is positive.
m
m
P
P
ai1 pi , . . . ,
In A’s problem let u = min
ain pi . Assume, as above, that umax > 0.
i=1
The problem is:
i=1
maximize 
u
m
P


aij pi ≥ u for j = 1, . . . , n,

i=1
subject to
m
P


pi = 1

i=1
and
pi ≥ 0 for i = 1, . . . , m.
As umax > 0, we can divide through each constraint by u without losing any feasible
solutions or changing the optimal point. This gives
p1
a11
u
..
.
p1
a1n
u
p1
and
u
Now let xi =
p2
+ a21
u
..
.
p2
+ a2n
u
p2
+
u
+ ···
+ ···
+ ···
pm
+ am1
u
..
.
pm
+ amn
u
pm
+
u
≥
≥
=
1,
..
.
1,
1
.
u
pi
for i = 1, . . . , m.
u
1
p1 + · · · + pm
=
= x1 + · · · + xm , so player A’s
u
u
xi
optimal mixed strategy (p1 , . . . , pm ) is such that for i = 1, . . . , m,
pi =
x1 + · · · + xm
where (x1 , . . . , xm ) solves the linear programming problem
Maximizing u is equivalent to minimizing
30
minimize z = x1 + · · · + xm


 a11 x1 + · · ·
..
subject to
.


a1n x1 + · · ·
+ am1 xm ≥ 1,
..
..
.
.
+ amn xm ≥ 1,
(LP A)
and xi ≥ 0 for i = 1, . . . , m.
Similarly, letting w = max
n
P
a1j qj , . . . ,
j=1
n
P
amj qj , assumed positive, and yj =
j=1
j = 1, . . . , n, player B’s problem becomes
!
qj
for
w
maximize z = y1 + · · · + yn


 a11 y1 + · · ·
..
subject to
.


am1 y1 + · · ·
+
a1n yn
..
.
≤ 1,
..
.
(LP B)
+ amn yn ≤ 1,
and yj ≥ 0 for j = 1, . . . , n.
Letting 1 be a vector whose entries are all 1, the problems can be written as follows
LP A:
minimize 1t x subject to At x ≥ 1, x ≥ 0,
LP B:
maximize 1t y subject to Ay ≤ 1, y ≥ 0.
B’s problem is the dual of A’s problem. By the duality theorems of Linear Programming,
both problems have the same optimal objective values. Using complementary slackness, the
optimal solution of either problem can be constructed from that of the other.
Let p∗ , q∗ be the mixed strategies obtained from the solutions of (LPA) and (LPB).
By Proposition 2.4 (p∗ , q∗ ) satisfies the saddle-point condition E(p, q∗ ) ≤ E(p∗ , q∗ ) ≤
E(p∗ , q), which proves Proposition 3.1.
The maximum value of z in A’s problem equals the minimum value of z in B’s problem.
The reciprocal of this common optimal value is E(p∗ , q∗ ), the payoff to player A when both
players are using their optimal strategies, and is the value of the game, denoted by v.
The maximin entry of A is a value taken by the function whose maximum is v. Similarly
the minimax entry of A is a value taken by the function whose minimum is v. Thus v lies
between the maximin and minimax entries of A inclusive.
Proposition 3.3 Let A be the payoff matrix for player A in a two-player zero-sum game
between A and B. Suppose the maximin entry of A is positive. Then the optimal mixed
strategies p∗ and q∗ for A and B are respectively (x1 v, . . . , xm v) and (y1 v, . . . , yn v) where
1
is the optimal
(x1 , . . . , xm ) and (y1 , . . . , yn ) solve the problems (LPA) and (LPB), and
v
value of z in both problems. Furthermore, maximin ≤ v ≤ minimax.
In the above reasoning we assumed that the payoff matrix has a positive maximin entry.
If this is not the case we can add a suitable constant k to all entries to make it so. It can
be shown that this gives a game with the same optimal mixed strategies and value k more
than that of the original game. Subtracting k gives the value of the original game.
31
Example (i)
Consider the game with the following payoff matrix for A:
A1
A2
Player A
Column maximum
minimax
Player B
B1 B2 B3
3
2
4
4
3
1
4
3
4
3
Row minimum
2
1
Maximin > 0 so we do not need to modify the matrix.
maximin
2
2≤v≤3
The L.P. formulation of A’s problem is: Minimize x1 + x2
subject to 3x1 + 4x2 ≥ 1, 2x1 + 3x2 ≥ 1, 4x1 + x2 ≥ 1 and x1 ≥ 0, x2 ≥ 0.
(Note that each inequality comes from a column of the payoff matrix.)
1
0.8
0.6
x2
0.4
0.2
0.1
0.2
x1
0.3
0.4
0.5
1
2
The minimum value of x1 + x2 is , when x1 = x2 = . Hence the value of the game is
5
5
1
1 5
1
5
= . A plays strategies A1 and A2 with probability × = each.
x1 + x2
2
5 2
2
B’s problem is to: Maximize y1 + y2 + y3
subject to 3y1 + 2y2 + 4y3 ≤ 1, 4y1 + 3y2 + y3 ≤ 1 and y1 ≥ 0, y2 ≥ 0, y3 ≥ 0.
This can be solved by the Simplex algorithm or using Maple, as follows :
with(simplex): maximize(x+y+z, {3*x+2*y+4*z<=1, 4*x+3*y+z<=1}, NONNEGATIVE);
1
3
, y3 = .
to give the optimal solutions y1 = 0, y2 =
10
10
5
Again the value of the game to A is . B plays strategies B1 , B2 and B3 with probabilities
2
1
3
0, and respectively.
4
4
These results can also be obtained from those for A’s problem using the complementary
slackness conditions of Linear Programming. Write A’s constraints as 3x1 + 4x2 − x3 = 1,
2x1 + 3x2 − x4 = 1, 4x1 + x2 − x5 = 1 using surplus variables x3 , x4 , x5 ≥ 0 and and B’s
constraints as 3y1 + 2y2 + 4y3 + y4 = 1, 4y1 + 3y2 + y3 + y5 = 1 using slack variables
y4 , y5 ≥ 0. Then x1 y4 = x2 y5 = x3 y1 = x4 y2 = x5 y3 = 0.
2
At the optimal solution x3 = , so y1 = 0. Also x1 , x2 6= 0 so y4 = y5 = 0. Hence
5
3
1
2y2 + 4y3 = 1, 3y2 + y3 = 1, which we can solve to get y2 =
, y3 =
as before.
10
10
Note that column 1 is dominated in the table, so we could have deleted it. This is reflected
in the fact that the first constraint in A’s problem was not active, and that B never plays
strategy B1 . Solving the simplified L.P. problem would give the same solution.
32
Example (ii)
The game of Morra is played as follows. Each of two players holds up 1 or 2 fingers, and
simultaneously guesses how many fingers the other player is holding up. If only one player
guesses correctly, that player wins the sum of the number of fingers held up by both players.
Otherwise the game is a draw.
Let (i, j) be the strategy of holding up i fingers and guessing that the other player is holding
up j fingers. The payoff matrix for player A is:
(1, 1)
(1, 2)
(2, 1)
(2, 2)
Player A
Column maximum
minimax
(1, 1)
0
−2
3
0
3
Player B
(1, 2) (2, 1)
2
−3
0
0
0
0
−3
4
2
4
2
(2, 2)
0
3
−4
0
3
Row minimum
−3
−2
−4
−3
maximin
−2
By the symmetry of the situation it is clear that the value of the game is zero. To find the
optimal strategies we can add any constant which makes the value positive, such as 1, to
every entry. This is enough because increasing all the payoffs to A must increase the value
of the game. There would be nothing wrong with adding 5, to make each entry positive.
Player A
(1, 1)
(1, 2)
(2, 1)
(2, 2)
(1, 1)
1
−1
4
1
Player B
(1, 2) (2, 1)
3
−2
1
1
1
1
−2
5
(2, 2)
1
4
−3
1
Now A’s problem is equivalent to: Minimize z = x1 + x2 + x3 + x4
subject to
x1 − x2 + 4x3 + x4 ≥ 1,
3x1 + x2 + x3 − 2x4 ≥ 1,
−2x1 + x2 + x3 + 5x4 ≥ 1,
x1 + 4x2 − 3x3 + x4 ≥ 1,
and x1 , x2 , x3 , x4 ≥ 0.
Solving this, e.g. with Maple, gives an optimal point (not necessarily unique) as
x1 = 0, x2 = 0.6, x3 = 0.4, x4 = 0.
The value of the modified game is
is 1 − 1 = 0.
1
= 1, so the value of the original game
0 + 0.6 + 0.4 + 0
Since x1 + x2 + x3 + x4 = 1, we have p1 = 0, p2 = 0.6, p3 = 0.4, p4 = 0
An optimal strategy for A is to play (1, 2) with probability 0.6 and (2, 1) with probability
0.4. Again this may not be unique, but the value of the game is.
This example illustrates the following:
Proposition 3.4 If the payoff matrix for a two-player zero-sum game is skew-symmetric,
i.e. At = −A, then the value of the game is 0 and the optimal mixed strategies for both
players are the same. The game is then said to be symmetric.
33
3.5
Two-player general-sum games
In practice, many situations cannot be modelled as constant-sum games. For example,
consider the Prisoners’ Dilemma.
Two criminals are taken into custody. They are both offered the same deal:
If you confess and your accomplice does not, you will go to prison for 1 year and your
accomplice will go to prison for 15 years. If you both confess, you will both get 5 years in
prison. If neither of you confesses, you will both get 3 years.
We can write down a payoff matrix whose entries are ordered pairs (a, b) representing the
penalties (in years) to the two prisoners:
Prisoner A
Confess
Don’t Confess
Prisoner B
Confess
Don’t Confess
(−5, −5)
(−1, −15)
(−15, −1)
(−3, −3)
This is a non-constant sum game, i.e. a general-sum game or bimatrix game. The
matrix can be separated into payoff matrices for A and B and expressed as (A, B) where:
−5 −1
−5 −15
A=
, B=
.
−15 −3
−1 −3
If A and B use mixed strategies p and q then their payoff functions are pt Aq and pt Bq.
We do not generally transpose to get B’s matrix, as we always index the first player’s payoffs
by the rows and it is convenient for both matrices to have the same dimensions when they
are not square. However, if we regard B as the first player then B has payoff matrix Bt ;
the payoff function is then qt Bt p, which of course is the same as pt Bq.
An equilibrium pair is a pair of mixed strategies (p, q) for A and B such that when they
are used together, neither player can benefit by a unilateral change of mixed strategy. An
equilibrium pair is not necessarily unique.
A Nash equilibrium is an equilibrium pair of pure strategies. This can be identified from
the matrix of ordered pairs as follows: the first entry is maximal in its column and the
second entry is maximal in its row.
If each player has a pure strategy which gives the best payoff whichever strategy the other
player uses, then that pair of strategies is a dominant-strategy equilibrium.
In the Prisoners’ Dilemma, ‘Confess’ is a dominant strategy for both players. (Confess,
Confess) is both a dominant-strategy equilibrium and a Nash equilibrium.
It would benefit both prisoners to choose the strategy (Don’t confess, Don’t confess) and
get the payoffs (−3, −3). However, this is not an equilibrium pair, as each prisoner would
gain by changing from it (i.e. confessing and double-crossing the accomplice) so long as the
other does not change.
A player’s safety level is the maximum expected payoff that the player can be guaranteed.
Thus A’s safety level is the value of the game with matrix A and B’s safety level is the
value of the game with matrix Bt .
34
Example
A1
A2
A3
Player A
Player B
B1
B2
(15, 5) (10, 25)
(5, 10) (15, 30)
(25, 20) (5, 15)
In this general-sum game, the pure strategies (A2 , B2 ) and (A3 , B1 ) both give Nash equilibria, being of the form (column maximum, row maximum).
There is no dominant-strategy equilibrium, but if the payoffs from (A3 , B2 ) were changed to
(20, 25) then this would be one. Then the matrices A and Bt would each have a dominant
row. A and B would then have safety levels 20 and 25 respectively.
Equilibrium pairs in mixed strategies are generally not easy to find for general-sum games,
nor is it obvious that they must exist. We shall show later that there is always at least one
equilibrium pair for a general-sum two-player game. There need not always be a dominantstrategy or Nash equilibrium.
The payoffs in a game can be represented by a tree diagram rather than a table. This is
called the extensive form of the game. A game in strategic form is easily converted to
extensive form. The reverse is less trivial, especially when there is an element of chance.
Example
A coin is biased so that the probability of heads is 23 . The coin is tossed by Player A, who
does not show it to Player B. A makes a claim about the outcome, and B then guesses how
the coin fell. B wins £3 for a correct guess and nothing otherwise. If A’s claim was true
then A wins £3. In addition, A wins a further £6 if B guesses heads.
This game can be illustrated in extensive form as follows :
Heads
A
B
Tails
True
False
H
T
H
(9, 3)
(3, 0)
(6, 3)
True
T
H
(0, 0)
(9, 0)
False
T
(3, 3)
H
T
(6, 0)
(0, 3)
Since B does not have any information about how the coin fell or whether A’s claim is true,
the four points in the diagram at which B makes a decision are indistinguishable (from B’s
point of view) so they are grouped into an information set. Each of A’s decision points
is an information set in itself.
A diagram which incorporates the information set(s) is called a Kuhn tree.
We convert to strategic form by allocating the mean payoff to each strategy pair :
35
Player A’s claim
Player B’s guess
Heads
Tails
(9, 2)
(3, 1)
(6, 2)
(0, 1)
True
False
1
2
(9, 3) + (9, 0).
3
3
The two separate matrices in this example both have saddle points, from which we can see
that A and B have safety levels 3 and 2 respectively.
e.g. (9, 2) comes from
3.5.1
Equilibrium pairs in general-sum games
Proposition 3.5 (The Brouwer fixed-point theorem) Let X be a non-empty compact
convex subset of Rn and let f : X → X be a continuous mapping on X. Then f has a fixed
point in X, i.e. there is a point x0 ∈ X such that f(x0 ) = x0 .
This apparently simple theorem is easy to justify if X ⊂ R, but its proof in the general case
is far more complicated. An outline proof will be given in the lectures.
Proposition 3.6 Let c(x) =
n
X
i=1
ri xi where
n
X
ri = 1, 0 ≤ ri ≤ 1 and x1 , . . . , xn ∈ R.
i=1
(i)
If c(x) > k, then xi > k for at least one i.
(ii)
If c(x) < k, then xi < k for at least one i.
(iii)
If c(x) = k, then xi ≤ k for at least one i and xj ≥ k for at least one j.
Proof
(i)
Suppose xi ≤ k for i = 1, . . . , n. Then c(x) =
n
X
i=1
Hence if c(x) > k, we cannot have xi ≤ k for all i.
ri xi ≤
n
X
kri = k
i=1
n
X
ri = k(1) = k.
i=1
(ii) and (iii) are proved similarly.
Now consider a two-player general-sum game whose m × n payoff matrices for A and B are
A and B respectively, where A’s m available strategies are represented by the rows and B’s
n strategies are represented by the columns.
If A and B play the mixed strategies p and q, the expected payoffs to A and B are respectively pt Aq and pt Bq.
A pair of mixed strategies (p, q) is called an equilibrium pair if for all possible mixed
strategies (p0 , q0 ), we have
pt Aq ≥ p0 t Aq and pt Bq ≥ pt Bq0 ,
i.e. given that B plays q then A does best by playing p, and vice-versa.
Let ei have 1 in position i and 0 everywhere else, so this represents the ith pure strategy.
(To avoid additional notation, we use ei for vectors with both m and n entries.)
Let p, q be any pair of mixed strategies for A and B respectively.
For i = 1, . . . , m define Ci = max(0, ei t Aq − pt Aq).
For j = 1, . . . , n define Dj = max(0, pt Bej − pt Bq).
36
These are the increases, if any, in a player’s payoff as a result of switching to a pure strategy.
Note that Ci = 0 if and only if pt Aq ≥ ei t Aq and Dj = 0 if and only if pt Bq ≥ pt Bej .
 ′ 
 ′ 
p1
q1
p i + Ci
qj + Dj
 .. 
 .. 
′
′
′
′
Define pi =
, qj =
, p =  . , q =  . .
m
n
P
P
1+
1+
Ci
Dj
p′m
qn′
i=1
j=1
are mixed strategies, i.e. they are probability vectors, because p′i ≥ 0 for
m
m
P
P
1+
Ci
(pi + Ci )
m
X
i=1
i=1
′
=
= 1, and likewise for the qj′ .
pi =
all i and
m
m
P
P
i=1
1+
1+
Ci
Ci
Then
p′
and
q′
i=1
i=1
Proposition 3.7 (p′ , q′ ) = (p, q) if and only if (p, q) is an equilibrium pair.
Proof
(⇐) Suppose (p, q) is an equilibrium pair. Then by the definition, for each i, ei t Aq ≤ pt Aq
so Ci = 0. Similarly, Dj = 0 for all j. Thus p′ = p and q′ = q.
Hence (p′ , q′ ) = (p, q) if (p, q) is an equilibrium pair.
(⇒) Suppose (p, q) is not an equilibrium pair. Then either there exists p0 such that
p0 t Aq > pt Aq, or there exists q0 such that pt Bq0 > pt Bq. Assume the former (the
argument in the latter case is identical).
m
P
As p0 is a probability vector, p0 t Aq =
p0 i ei t Aq is a convex linear combination of the
i=1
ei t Aq. Thus if p0 t Aq > pt Aq, then by Proposition 3.6 (i), er t Aq > pt Aq for some r.
m
P
Hence for this r, Cr > 0, and it follows that
Ci > 0.
i=1
pt Aq itself is also a convex linear combination of the ei t Aq, so by Proposition 3.6 (iii),
there is some s such that es t Aq ≤ pt Aq and hence Cs = 0.
m
P
ps
, which is strictly less than ps because
Ci > 0.
Then p′s =
m
P
i=1
1+
Ci
i=1
Hence
p′
6= p as they differ at least in position s.
We have thus shown that if (p, q) is not an equilibrium pair then (p′ , q′ ) 6= (p, q), from
which it follows that if (p′ , q′ ) = (p, q) then (p, q) is an equilibrium pair.
It follows from the above result that a necessary and sufficient condition for (p, q) to be
an equilibrium pair is that Ci = 0 for i = 1, . . . , m and Dj = 0 for j = 1, . . . , n; that is,
pt Aq ≥ ei t Aq for all i and pt Bq ≥ pt Bej for all j.
Proposition 3.8 (Nash’s theorem) Every two-player general-sum game has at least one
equilibrium pair.
Proof Define a mapping f by f(p, q) = (p′ , q′ ). f is a continuous function on the set of
all mixed strategy pairs, which is a non-empty compact convex subset of Rm+n . By the
Brouwer fixed-point theorem, f has a fixed point, i.e. there exists a pair (p, q) such that
(p, q) = (p′ , q′ ). By Proposition 3.7 this is an equilibrium pair, so such a pair always
exists.
37
Using this result we can check whether a given mixed strategy is an equilibrium pair. It does
not in general provide a method for finding optimal strategies, though it may be possible
to do so in simple cases. There are also methods for finding equilibrium pairs by solving a
non-linear programming problem.
Note that an equilibrium pair need not be unique, nor need all equilibrium pairs yield the
same expected payoff.
Zero-sum games are a special case of this result with B = −A.
John Nash won a Nobel prize for Economics in 1994 for his work on Game Theory.
3.5.2
Example
Consider the two-player non-constant-sum game with payoff matrix
Player A
A1
A2
B1
(2, 5)
(5, 3)
Player B
B2
B3
(4, 2) (3, 6)
(2, 4) (4, 1)
2 3
, , 0 for B are an equilibrium pair.
for A and q =
Verify that the strategies p =
5 5
16
16 t
7
9
7
t
For these strategies, Aq =
, pB=
.
5
5
2
2
4
7
16
The payoffs to A and B are pt Aq = , pt Bq = .
5
2
9 7
We have C1 = C2 = D1 = D2 = D3 = 0; e.g. D3 = max 0, −
.
4 2
1 3
,
4 4
Hence (p′ , q′ ) = (p, q), so (p, q) is an equilibrium pair.
Neither player could improve their payoff by a unilateral change of mixed strategy, but there
may be other equilibrium pairs with different payoffs.
38
Exercises 3
1. Consider the game with the following payoff matrix for A. Delete any dominated rows
or columns and determine whether or not the reduced matrix has a solution in pure
strategies. If it does not, then determine the solution in mixed strategies.
Player A
A1
A2
A3
Player B
B1 B2 B3
1 −1
2
−1
1
3
−3 −2
4
2. Find the optimal strategies for A and B and the value in the following games:
(i)
Player A
A1
A2
Player B
B1 B2
2
2
1
3
(ii)
Player A
A1
A2
Player B
B1 B2 B3
4
1 −1
−1
3
6
3. Ann has two cards, numbered 2 and 3. Bill has two cards, numbered 1 and 4. They
both choose one card. If the sum of the numbers on the cards is odd, Ann pays Bill the
sum of the numbers. If the sum of the numbers is even, Bill pays Ann the difference
between the numbers.
(a) Explain briefly why this is a zero-sum game.
(b) Find the payoff matrix for Ann and state the minimax and maximin values.
(c) State how you can tell that the game has no solution in pure strategies.
(d) Find the expected payoff to Ann if she always chooses 2 and Bill is equally likely
to choose 1 or 4.
(e) Find the optimal strategy for each player and the value of the game. What is
the significance of the value?
(f) If Bill’s 4 is replaced by another 3 and all other conditions remain unchanged,
show that there is a solution in pure strategies and find the value of the game.
4. A and B both have three cards, numbered 0, 1 and 2. Each player chooses two cards
and displays them. If both players’ cards show the same two numbers, A pays B the
larger of the two values. Otherwise, B pays A the larger of the different values. For
example, if both show the pair (0, 1) then A pays B 1p, but if A shows (0, 1) and B
shows (1, 2) then B pays A 2p.
(a) Formulate this situation as a two-player zero-sum game and show that it has no
solution in pure strategies.
(b) Express the game as a linear programming problem, clearly stating the significance
of each variable that you introduce. Given that A’s optimal mixed strategy is to show
5 3 3
, 11 , 11 respectively, use complementary
(0,1), (0, 2) and (1, 2) with probabilities 11
slackness to show that B’s optimal strategy is the same.
5. A two-person game has payoff matrix
Player A
A1
A2
A3
39
Player B
B1 B2 B3
0 −1
2
1
0 −1
−2
1
0
(a) State the value of the game.
(b) Formulate the problem of finding the optimal strategies for A as a linear programming problem.
(c) Given that z is minimized when x1 = x3 = 0.125, x2 = 0.25, find the optimal
mixed strategy for A and deduce the optimal strategy for B.
6. Two armies are advancing on two cities. The first army consists of four regiments. The
second army has three regiments. For each city, the army that sends more regiments
to the city captures both the city and the opposing army’s regiments that are sent to
that city. If both armies send the same number of regiments to a city, then the battle
for the city results in a draw.
Each army scores 1 point per city captured and 1 point per enemy regiment captured.
Assume that each army wants to maximize the difference between its reward and its
opponent’s reward, and that all regiments are used. Formulate this situation as a
two-player zero-sum game, giving the payoff matrix for the first army.
Express the problem of finding that army’s optimal strategy in linear programming
form. Given that the value of the game lies strictly between the maximin and minimax,
explain why there is no need to add a constant to the entries of the payoff matrix.
Given that the L.P. problem has the solution (x1 , x2 , x3 , x4 , x5 ) = (2/7, 0, 1/14, 0, 2/7),
find the optimal mixed strategy for the first army and the value of the game.
7. A total of 90,000 customers go to either Safeways supermarket or Wholefoods supermarket. To attract customers, each supermarket gives away a free item which is
announced in the local newspaper on Monday. Neither supermarket knows which item
the other supermarket will give away this week. Safeways is considering giving away
a bottle of Pepsi or 2 litres of milk. Wholefoods is considering giving away a pound
of butter or 2 litres of orange juice.
Wholefoods
Butter Orange Juice
Pepsi 40,000
50,000
Safeways
Milk 60,000
30,000
From market research, it is known that for each possible choice of items the number of
shoppers who will shop at Safeways is as shown in the table below. Each supermarket
wants to maximize its expected number of customers during the current week.
Determine an optimal strategy for each store. Find and interpret the value of the
game. Also express the game in extensive form, using a Kuhn tree.
8. Two companies A and B each have to choose one medium for their advertising. There
are three possible media: Newspapers (N ), Radio (R) and Television (T ). It is estimated that, of the company’s potential customers, 40% read newspapers, 20% listen
to radio and 50% watch television.
A′ s
choice of medium
N
R
T
40
B’s choice of medium
N
R
T
20, 20 40, 20
40, 50
20, 40
50, 40
If the companies choose the same medium they will split the customers who are
reached through that medium equally, while if they choose different media then each
will get all the customers from that medium. The payoffs, as percentages of the total
customer base, are shown in the table.
(a) Complete the table. Explain why this is a non-constant sum game and write
down the separate payoff matrices for A and for B. Are there any dominant-strategy
equilibria?
(b) Show, with explanation, that there are two Nash equilibria. Verify that they both
yield the same total payoff.
(c) Based on the given data, describe what factors might prevent the companies from
choosing one of these Nash equilibria.
9. Contestants A and B in a game show start the last round with £400 and £500
respectively. Each must decide whether to pass or gamble, not knowing the choice
of the other. Any player who passes keeps the money they started with. If player A
gambles, he is equally likely to win an extra £200 or lose his entire £400. If player
B gambles, she is equally likely to win £200 or lose £200. Then the contestant with
the higher amount at the end wins a bonus of £200.
Draw the Kuhn tree. Convert the game into strategic form and find the safety level
for each player. Does the game have a dominant-strategy equilibrium or a Nash
equilibrium?
given by
10. Let T be the subset of Rm+n
+
T = {(p1 , . . . , pm , q1 , . . . , qn ) : p1 + · · · + pm = 1, q1 + · · · + qn = 1}
Prove that T is a convex set. (This was used in the proof of Nash’s Theorem.)
[Hint : Let (p, q) and (p′ , q′ ) be in T . Show that (1 − r)(p, q) + r(p′ , q′ ) is also in
T for 0 ≤ r ≤ 1.]
11. Consider the two-player non-constant-sum game represented by the payoff matrix
Player B
B1
B2
(2, −1) (−2, 1)
(−2, 1) (2, −1)
A1
A2
Player A
Show that this game has no Nash equilibrium. Find the payoff function for each
player when A uses the mixed strategy (p, 1 − p) and B uses (q, 1 − q).
Show that for all q, A’s payoff is maximized when p = 12 , and that for all p, B’s payoff
is maximized when q = 21 . Hence write down an equilibrium pair for this game.
12. Consider the two-player non-constant-sum game represented by the payoff matrix
Player A
Verify that the mixed strategies
equilibrium pair.
A1
A2
A3
Player B
B1
B2
(4, 1) (2, 3)
(1, 5) (3, 2)
(2, 4) (1, 5)
7 5 1
, ,
13 13 13
41
B3
(1, 4)
(5, 1)
(6, 3)
for A and
3 1 5
, ,
7 3 21
for B are an
Chapter 4
Input-Output Analysis
4.1
Leontief models
Suppose an economy is divided into sectors, each of which produces an output using a fixed
linear combination of inputs from the sectors. In practice, the parts of a large economy are
aggregated into a smaller number of sectors.
This model was developed by Wassily Leontief, who in 1949 modelled the U.S. economy using 42 linear equations in 42 unknowns. A solution was obtained after 56 hours of computer
time. Leontief won the Nobel prize for Economics in 1973.
Suppose that the production of 1 unit of output in sector j requires aij units of input from
sector i. The input-output coefficients aij form a square technology matrix A.
Each sector must use less of its own product than it creates, so the diagonal entries aii are
all less than 1.
The technology matrix can be used to find the inputs needed for a given level of production.
n
P
If xj units of commodity j are produced, the total input from Sector i is
aij xj units.
j=1
For a given output vector x, the required input vector is Ax.
Example Suppose the technology matrix for a two-sector economy is A =
0.6 0.3
0.2 0.5
,
so every unit of output in Sector (1) requires inputs of 0.6 unit from Sector (1) and 0.2 unit
from Sector (2), etc.
A total output of 200 units from (1) and 150 unitsfrom (2) would
needinputs
of 165
units
0.6 0.3
200
165
from (1) and 115 units from (2). This is given by
=
.
0.2 0.5
150
115
A surplus of 35 units is then generated in each sector.
An output vector (100, 250) needs inputs (135, 145); this is not feasible as 135 > 100.
The matrix can also be used to find the prices of the inputs needed for a given output
value. The unit cost of a commodity as an input is assumed to be the same as its value as
an output. If the value of each unit in Sector i is £yi , the total value of the inputs used to
n
P
produce 1 unit of output from Sector i is
aji yj .
j=1
42
Suppose each unit from Sector (1) has value £150 and from Sector (2), £100. To produce
one unit in Sector (1) costs 0.6 × £150 + 0.2 × £100 = £110. To produce one unit in Sector
(2) costs 0.3 × £150 + 0.5 × £100 = £95.
0.6 0.2
150
110
This is given by
=
.
0.3 0.5
100
95
For a given output price vector y, the vector of input costs is At y.
4.2
The closed Leontief model
In a closed or exchange model, all the output is consumed by the sectors themselves.
Example An economy has three sectors: coal, electricity and steel. 70% of the coal
produced is used in making electricity and the rest in making steel. The electricity sector
consumes 10% of its own output; 40% is used by the coal sector and 50% by steel. Of steel’s
output, 60% is used by the coal sector, 30% by electricity and the rest by steel itself.
Taking 1 unit in each sector to be that sector’s total output, the technology matrix is


0 0.7 0.3
A =  0.4 0.1 0.5 
0.6 0.3 0.1
Suppose the values of the outputs are £y1 for coal, £y2 for electricity and £y3 for steel.
Then the input values are £(0.4y2 + 0.6y3 ) for coal, etc.
To find equilibrium prices which make each sector’s output equal in value to its input,
we solve the system of linear equations
y1 =
0.4y2 + 0.6y3 ,
y2 = 0.7y1 + 0.1y2 + 0.3y3 ,
y3 = 0.3y1 + 0.5y2 + 0.1y3 ,
i.e.
i.e.
i.e.
y1 − 0.4y2 − 0.6y3 = 0
−0.7y1 + 0.9y2 − 0.3y3 = 0
−0.3y1 − 0.5y2 + 0.9y3 = 0
The solution (to 3 significant figures) is (y1 , y2 , y3 ) = a(1.06, 1.16, 1.00). The outputs could
be priced at £106, £116, £100, or any multiples of these.
Letting y = (y1 , y2 , y3 )t , the above equations are y = At y, or (I − At )y = 0. Thus the
vector y is an eigenvector of At , with associated eigenvalue 1.
We wish to know whether there is always a valid (non-negative) vector of equilibrium prices.
The first step in answering this question is to notice that, in the closed model, each row of
the technology matrix A adds up to 1. Let 1 be the vector in Rn whose entries are all 1;
then clearly A1 = 1 so 1 is an eigenvector of A associated with the eigenvalue 1.
A matrix and its transpose have the same eigenvalues, though their eigenvectors are different
in general. Hence At has 1 as an eigenvalue, so the system of equations At y = y has a
solution. However, this solution will not be valid unless its entries are all non-negative.
A matrix A is said to be
• non-negative (A ≥ O) if all its entries are ≥ 0,
• strictly positive (A > O) if all its entries are > 0,
• stochastic if it is square, non-negative and the entries in each column add up to 1.
If A is the technology matrix in a closed Leontief model, At is a stochastic matrix.
43
Proposition 4.1 Suppose A > O, x ≥ 0 and x 6= 0. Then Ax > 0.
Proof
If x ≥ 0 and x 6= 0 then at least one entry of x is strictly positive: say xj > 0.
For i = 1, . . . , n the ith entry of Ax is ai1 x1 + · · · + aij xj + · · · + ain xn . Every term in this
is non-negative and the term aij xj is positive, so every entry of Ax is positive, i.e. Ax > 0.
An eigenvalue µ of a matrix A is called dominant if |µ| ≥ |λ| for all eigenvalues λ of A.
Proposition 4.2 (Frobenius-Perron theorem) Let A be a non-negative, non-zero n×n
real matrix. Then the dominant eigenvalue of A is real, positive and associated with a nonnegative real eigenvector.
Proof
We prove this for the case A > O
Let 1 be the vector in Rn whose entries are all 1.
Let X = {x ∈ Rn+ : 1t x = 1}. X is a compact, convex set.
1
Let f(x) = t
Ax. If x ∈ X then Ax > 0 by Proposition 4.1, so 1t Ax > 0 and thus
1 Ax
f(x) > 0. Also 1t f(x) = 1, so f(x) ∈ X.
Hence f is a continuous function from X to X so by the Brouwer fixed-point theorem f has
1
a fixed point v ∈ X such that f(v) = v, i.e. t
Av = v.
1 Av
Thus Av = µv where µ = 1t Av, which is > 0,
so A has a positive real eigenvalue µ associated with a non-negative, non-zero real eigenvector v.
We now show that µ is dominant.
Av = µv.
Av > 0 by Proposition 4.1, so µv > 0, so v > 0.
Let λ be any eigenvalue of A (real or complex), so Az = λz for some z = (z1 , . . . , zn ) ∈ Cn .
Then for i = 1, . . . , n, ai1 z1 + · · · + ain zn = λzi so |λ||zi | = |ai1 z1 + · · · + ain zn |
≤ ai1 |z1 | + · · · + ain |zn | by the triangle inequality, using |aij | = aij since A > O.


|z1 |


Thus |λ|w ≤ Aw, where w =  ...  ∈ Rn+ . If w = v then |λ|v ≤ Av = µv so |λ| ≤ µ.
|zn |
Now assume w 6= v. Let t be the smallest real number for which v − tw has a zero entry,
vi
so t = min . Clearly t > 0 and v − tw ≥ 0.
wi 6=0 wi
Let j be a position in which this zero entry occurs, so wj 6= 0 and vj − twj = 0.
Suppose (for a contradiction) that |λ| > µ.
Then as Aw ≥ |λ|w we have (Aw)j ≥ |λ|wj > µwj , so −(Aw)j < −µwj .
Thus A(v − tw) = µv − tAw has jth entry µvj − t(Aw)j < µvj − tµwj = µ(vj − twj ) = 0,
so A(v − tw) has a negative entry. But this is impossible, because A is positive and v − tw
is non-negative.
We conclude that |λ| ≤ µ in this case also, so µ is the eigenvalue of greatest modulus, i.e.
it is dominant. The corresponding eigenvector is v > 0.
44
The extension to a general non-negative matrix A is accomplished by considering a sequence
of strictly positive matrices converging to A.
The dominant eigenvalue is called the Frobenius-Perron eigenvalue of A.
Proposition 4.3 Let m and M be the the minimum and maximum column sums respectively of a non-negative real matrix A. Let λ be a real eigenvalue of A associated with a
non-negative eigenvector. Then m ≤ λ ≤ M .
Proof
Let sj be the sum of the entries in column j of A, so m ≤ sj ≤ M for all j.
Let 1 = (1 1 · · · 1)t , so 1t A = (s1 s2 · · · sn ).
Let v = (v1 · · · vn )t be a non-negative eigenvector associated with λ ∈ R, so Av = λv.
v ≥ 0 and v 6= 0 (as 0 cannot be an eigenvector) so 1t v = v1 + · · · + vn > 0.
λ1t v = 1t λv = 1t Av, so λ(v1 + · · · + vn ) = s1 v1 + · · · + sn vn ≥ m(v1 + · · · + vn ), so λ ≥ m.
Similarly λ(v1 + · · · + vn ) ≤ M (v1 + · · · + vn ), so λ ≤ M .
Note that since A and At have the same eigenvalues, Proposition 4.3 could equally well be
stated in terms of minimum and maximum row sums. In particular, the dominant eigenvalue is bounded above by both the maximum row sum and maximum column sum of A.
A matrix A is regular if, for some positive integer k, every entry of Ak is strictly positive.
Example (i)




0.5 0 0.5
0.15 0.3 0.45
Let A =  0.2 0.8 0 . Then A2 =  0.26 0.64 0.1  > O, so A is regular.
0 0.6 0.4
0.12 0.72 0.16
Example (ii)

0.8 0.3 0.1 0.2 0.6
 0.2 0.5 0
0 0.7

Let A = 
0
0
0.9
0.3
0

 0
0 0.2 0.7 0.4
0
0
0 0.3 0.6



.


In any power of A there is always a block of six zeros in the bottom left, so A is not regular.
In a decomposable economy, not every sector buys directly or indirectly from all the
others, i.e. the economy has a proper subset S such that outputs from the sectors in S all
stay inside S.
45
The diagram illustrates an economy with 5 sectors where S consists of sectors 3, 4 and 5.
(The input from each sector to itself is not shown.) The matrix A in Example (ii) above
could be the technology matrix for this economy.
An economy is indecomposable if it is not decomposable, i.e. every sector uses some input
(perhaps indirectly) from every other sector.
Given a technology matrix A, to generate an output vector x needs an input vector Ax.
To generate this needs an input A(Ax) = A2 x, and so on. Thus the (i, j) entry of Ak
represents the input needed from Sector i to yield a unit of output in Sector j after k
stages. In Example (i) above, A2 shows that 0.72 units are needed from Sector 3 to produce
all the inputs needed for 1 unit of output in Sector 2.
A regular technology matrix represents an indecomposable economy.
Proposition 4.4 Let A be a real stochastic square matrix. Then
(i) the only real eigenvalue of A associated with a non-negative eigenvector is 1,
(ii) if A is regular, this eigenvector is positive and is unique up to a scalar multiple.
Proof
(i) By Proposition 4.2 A has a real eigenvalue µ associated with a non-negative eigenvector.
By Proposition 4.3 any such eigenvalue lies between the minimum and maximum column
sums of A, both of which are 1, so it can only be 1.
(ii) Let y be a non-negative eigenvector of A. We have Ay = y, so Ak y = y for all k ∈ N.
If A is regular, Ak > O for some k. Then Ak y > 0 by Proposition 4.1, so y > 0.
Suppose there are linearly independent eigenvectors v > 0 and w > 0 such that Av = v
and Aw = w. Let y = v − tw. y 6= 0 for any t, as v is not a multiple of w.
Ay = Av − tAw = v − tw, so Ay = y.
vi
. For this
i wi
t, all entries of y are non-negative and at least one entry is zero. But we showed above that
y > 0.
Take t to be the smallest real number for which y has a zero entry, so t = min
This contradiction shows that there cannot be a linearly independent set of two eigenvectors
of A associated with the eigenvalue 1.
The significance of this result is that in a closed Leontief model of an indecomposable economy, where the total output is taken as 1 unit, there is always a valid (non-negative) vector
of equilibrium prices with unique proportions.
Example
Consider a three-sector closed economy where all the output is used, with no surplus. Of
Sector 1’s output, 50% goes to each of Sectors 1 and 3. Of Sector 2’s output, 20% goes
to Sector 1 and 80% to Sector 2. Of Sector 3’s output, 60% goes to Sector 2 and 40% to
Sector 3.


0.5 0 0.5
If we regard each sector’s output as 1 unit, the technology matrix is A =  0.2 0.8 0 .
0 0.6 0.4
46
At is regular and stochastic, so we know there is a non-negative vector y such that At y = y.
Solving this, we find that y = (6 15 5)t .
Note that Ax = x yields only the eigenvector (1 1 1)t . This simply means that output
quantity equals input quantity only when all sectors produce the same number of units.
Suppose we want output prices to exceed production costs by a factor 1 + r, i.e. 100r%
profit is made by all sectors. Then y = (1 + r)At y. In other words, y is an eigenvector of
1
At , with eigenvalue
. We have shown that this cannot happen except when r = 0.
1+r
4.3
The open Leontief model
In an open model, as well as the requirements of the sectors from each other there is an
external final demand, e.g. consumption by households. As the sectors try to satisfy this
demand, they create intermediate demands among themselves. In equilibrium supply
equals demand, i.e.
total production = intermediate demand + final demand.
Thus x = Ax + d, where Ax represents the intermediate demand and d is the vector of
final demands.
This production equation can be written as (I − A)x = d, so if (I − A) is non-singular
there is a unique solution x = (I − A)−1 d.
(I−A)−1 is called the Leontief Inverse of A. If it is non-negative, we say A is productive.
The equilibrium costs are determined by the price equation y = At y+c, where the entries
of y are unit prices and c is a vector of external production costs.
Example
An economy consists of two sectors, producing goods G1 , G2 . To produce 1 kg of G1 requires
0.2 kg of G1 and 0.4 tonnes of G2 . To produce 1 tonne of G2 requires 0.6 kg of G1 and 0.3
tonnes of G2 .
In one year, let x1 kg of G1 and x2 t of G2 be produced. This requires inputs
of (0.2x1+0.6x2 )
0.2 0.6
kg of G1 and (0.4x1 + 0.3x2 ) t ogf G2 . The technology matrix is A =
.
0.4 0.3
Suppose consumers require d1 = 5000 kg of G1 and d2 = 2000 tonnes of G2 in a year. Let
d = (d1 d2 )t .
For equilibrium x = Ax + d, or (I − A)x = d.
0.8 −0.6
5000
x1
Thus
=
. Solving, we get
−0.4 0.7
x2
2000
(x1 , x2 ) = (14687.5, 11250), so 14687.5 kg of G1 and 11250 tonnes of G2 must be produced.
Alternatively, we might want to find the prices of the two goods to cover unit production
costs c = (c1 c2 )t .
Let the prices be £y1 per kg of G1 and £y2 per tonne of G2 .
The cost of the inputs for 1 kg of G1 is £(0.2y1 + 0.4y2 ). Similarly the cost for 1 tonne of
G2 is £(0.6y1 + 0.3y2 ).
For equilibrium, y = At y + c, or (I − At )y = c.
47
If c1 = £7 and c2 = £3, then
0.8 −0.4
−0.6 0.7
y1
y2
=
7
3
.
Solving, we get (y1 , y2 ) = (19.0625, 20.625). Thus G1 must be priced at £19.06 per kg and
G2 at £20.63 per tonne.
The equilibrium quantities and prices are related.
The value of the consumer demand is dt y =5000(19.0625) + 2000(20.625) = £136562.50.
Also the cost of producing the equilibrium quantities is
ct x = 7(14687.5) + 3(11250) = £136562.50.
Proposition 4.5 If x = Ax + d and y = At y + c then ct x = dt y.
Proof
yt x = yt Ax + yt d and xt y = xt At y + xt c.
yt x = xt y and yt Ax = xt At y, so yt d = xt c, or equivalently ct x = dt y.
We now establish conditions for an equilibrium to exist in the open model.
Proposition 4.6 Let A be a non-negative real square matrix. If Ak → O as k → ∞ then
(I − A) is non-singular and its inverse is a non-negative matrix. Furthermore,
−1
(I − A)
=
∞
X
Ak .
k=0
Proof
Let Sk = I + A + A2 + · · · + Ak .
(Note that A0 = I.)
Then Sk (I − A) = I + A + A2 + · · · + Ak − A − A2 − · · · − Ak − Ak+1 = I − Ak+1 .
Assuming that Ak → O as k → ∞, we get S∞ (I − A) = I.
Thus S∞ = (I − A)−1 , so (I − A)−1 exists and
−1
(I − A)
2
k
= lim (I + A + A + · · · + A ) =
k→∞
∞
X
Ak .
k=0
As all the terms are non-negative matrices, so is their sum. Hence (I − A)−1 is non-negative.
Example
0.2 0.4
Let A =
. Provided that Ak → O, (I − A)−1 exists and is given by
0.3 0.1
(I − A)−1 = I + A + A2 + A3 + · · ·
1 0
0.2 0.4
0.16 0.12
0.068 0.076
which equals
+
+
+
+ ···
0 1
0.3 0.1
0.09 0.13
0.057 0.049
1.46 0.63
3/2 2/3
−1
Using just these four terms gives (I−A) ≈
; the true value is
.
0.47 1.31
1/2 4/3
We can interpret the series formula for the Leontief Inverse as follows :
48
Given a demand vector d, Ad is the intermediate demand for inputs, i.e. its components
are the quantities of the goods needed to meet the final demand. The amounts required to
produce these quantities are given by A(Ad) = A2 d, and this in turn requires an input of
A(A2 d) = A3 d, and so on.
The overall production level to meet the final demand is thus
x = d + Ad + A2 d + · · · = (I + A + A2 + · · · )d.
If this does not converge then infinite quantities are required and there is no solution. If it
converges, its limit is x = (I − A)−1 d which is then the unique solution to the open model.
Proposition 4.7 Let A be a non-negative real square matrix. Each of the following is a
sufficient condition for I − A to be non-singular and have a non-negative inverse:
(i) all the eigenvalues of A have modulus less than 1,
(ii) the sum of the entries in each row (or each column) of A is strictly less than 1,
(iii) All the leading principal minors of I − A are strictly positive. (The Hawkins-Simon
Condition).
Proof
(i) Assume A is diagonalisable: A = PDP−1 where D is a diagonal matrix whose diagonal
entries are the eigenvalues of A. Then Ak = PDk P−1 for all k ∈ N.
If all eigenvalues are less than 1 in modulus then Dk → O, and hence Ak → O, as
k → ∞. The result then follows by Proposition 4.6.
(If A is not diagonalisable the same reasoning can be used on the Jordan Form of
A, which is an upper triangular matrix similar to A with the eigenvalues on its main
diagonal.)
(ii) The dominant eigenvalue of A is non-negative and less than or equal to the maximum
row (or column) sum by Proposition 4.3, so condition (ii) implies condition (i).
(iii) We prove the Hawkins-Simon Condition for the case A > O, by showing that it implies
condition (i).
Let A be n × n. We proceed by induction on n. When n = 1, A has the single entry
a11 which is thus the only eigenvalue of A. Then I1 − A = (1 − a11 ) whose only l.p.m.
is 1 − a11 . If this is positive then 0 < a11 < 1, so (i) holds.
Thus the result is true for n = 1. Now assume that (iii) ⇒ (i) when n ≤ k, i.e. for
all square positive real matrices up to k × k. Let A have dimensions (k + 1) × (k + 1)
and suppose (iii) holds for Ik+1 − A. We must show that (i) holds for A.
For i = 1, . . . , k let µi be the dominant eigenvalue of the top-left i × i submatrix of A.
Let Mi be the ith leading principal minor of Ik+1 − A, so Mi > 0 for i = 1, . . . , k + 1.
Let A′ be the top-left k × k submatrix of A. For i = 1, . . . , k, Mi is also the ith leading
principal minor of Ik − A′ so (iii) holds for Ik − A′ , hence by the inductive assumption
(i) holds for A′ , i.e. |µi | < 1 for i = 1, . . . , k.
Mk
Let q =
, so q > 0.
Mk+1
49
Let ek+1 be the vector in Rk+1 with (k + 1) entry 1 and all other entries 0. As
det(Ik+1 − A) = Mk+1 > 0, the system of equations (Ik+1 − A)x = ek+1 has a unique
solution y = (Ik+1 − A)−1 ek+1 .
y0
.
By Cramer’s rule the (k + 1)th entry of y is q, so we can write y as
q
′
A b
Partition A as
, so the equation (Ik+1 − A)y = ek+1 is
ct d
y0
0
Ik − A′ −b
=
.
−ct
1−d
q
1
Thus (Ik − A′ )y0 − qb = 0, so (Ik − A′ )y0 = qb > 0. By the inductive assumption
Ik − A′ has a non-negative inverse so y0 = q(Ik − A′ )−1 b > 0 and thus y > 0.
We have now shown that the system (Ik+1 − A)x = ek+1 has a unique positive solution
y ∈ Rk+1 . Ay = y − ek+1 ≤ y, so y − Ay ≥ 0.
As A > O it follows from Proposition 4.1 that Ay − A2 y > 0, so A2 y < Ay ≤ y.
Let µk+1 be the dominant eigenvalue of A, which is non-negative by Proposition
4.2. Then µk+1 2 is the dominant eigenvalue of A2 and hence of (A2 )t so, again by
Proposition 4.2, there is an associated non-negative eigenvector v of (A2 )t such that
(A2 )t v = µk+1 2 v. Transposing both sides gives vt A2 = µk+1 2 vt .
A2 y < y, so vt A2 y < vt y. Thus µk+1 2 vt y < vt y, so 0 ≤ µk+1 2 < 1, so |µk+1 | < 1.
Hence |µi | < 1 for i = 1, . . . , k + 1, i.e. condition (i) holds for A when n = k + 1. By
induction the result follows for all n ∈ N.
Thus if (I − A)x = d, where A is a non-negative square matrix, and any of the three sufficient conditions applies, x is uniquely determined and is non-negative. The open Leontief
model then has a solution.
Example
The technology matrix for a three-sector economy is


0.2 0.6 0.3
A =  0.1 0.1 0.6  ,
0.2 0.4 0.3
Given that det(A) = 0.018 and one of the eigenvalues of A is 0.9, show that A has a nonnegative Leontief Inverse.
Neither the row sums nor the column sums are all less than 1. The eigenvalues of A
have sum tr(A)= 0.6 and product det(A) = 0.9 so the other two are –0.2 and –0.1. All
eigenvalues have modulus < 1 so the eigenvalue condition is satisfied. Thus by Proposition
4.7 (ii) (I − A) has a non-negative inverse, so the economy is productive. We could also
have checked the Hawkins-Simon condition: the leading principal minors of (I − A) are 0.8,
0.72, 0.132, so the condition holds.
Suppose there is an exogenous demand given by d = (100, 250, 300). Then the required
production levels are given by (I − A)−1 d = (2750, 2250, 2500).
Exercises 4
1. An economy has three sectors: Agriculture, Mining and Manufacturing, whose output
is produced using only inputs from these sectors.
50
Agriculture sells 5% of its output to Mining, 30% to Manufacturing, and keeps the
rest. Mining sells 20% of its output to Agriculture, 70% to Manufacturing, and keeps
the rest. Manufacturing sells 20% of its output to Agriculture, 30% to Mining, and
keeps the rest. Regarding the total output from each sector as 1 unit, and assuming
no surplus is generated,
(a) write down the technology matrix A for this economy.
(b) Show that 1 is an eigenvalue of A, and find associated eigenvectors of
(i) A, (ii) At .
(c) Find the equilibrium prices that make each sector’s income equal to its expenditure, if the total value of the outputs of all three sectors is £264 million.
2. An economy can be modelled as consisting of two sectors, producing goods G1 , G2 .
The number of units of Gi required in the production of 1 unit of Gj is aij . There
is an exogenous demand for d1 units of G1 and d2 units of G2 . There are exogenous
production costs £c1 per unit of G1 and £c2 per unit of G2 .
Attempt to solve the input-output model in the following cases, finding both equilibrium quantities and equilibrium prices. Comment on your answers.
(a) a11 = 0.2, a12 = 0.8, a21 = 0.3, a22 = 0.7, c1 = c2 = d1 = d2 = 0.
(b) As in (a), but d1 = 10, d2 = 15, c1 = 4, c2 = 3
(c) a11 = 0.2, a12 = 0.6, a21 = 0.4, a22 = 0.2, c1 = c2 = d1 = d2 = 0.
(d) As in (c), but d1 = 10, d2 = 15, c1 = 4, c2 = 3
(e) a11 = 0.2, a12 = 0.6, a21 = 0.4, a22 = 0.7, c1 = c2 = d1 = d2 = 0.
(f) As in (e), but d1 = 10, d2 = 15, c1 = 4, c2 = 3
3. An economy is divided into three sectors : manufacturing, agriculture and services.
For each unit of output, manufacturing requires 0.1 unit from its own sector, 0.3 unit
from agriculture and 0.3 unit from services.
For each unit of output, agriculture requires 0.2 unit from its own sector, 0.6 unit
from manufacturing and 0.1 unit from services.
For each unit of output, services requires 0.1 unit from its own sector, 0.6 unit from
manufacturing and nothing from agriculture.
(a) Find the technology matrix A. Verify the Hawkins-Simon condition for A.
(b) Find the number of units each sector must produce to satisfy an external demand
for 10 units from each sector.
(c) Show that the ith column of (I − A)−1 gives the extra amounts the sectors must
produce to satisfy an increase of 1 unit in the final demand from sector i.
0.2 0.5
4. Let A =
. Find the exact value of (I − A)−1 . Also find the sum of the
0.5 0.3
first five terms of the series for (I − A)−1 . Compare your answers.


0.5 0.3 0.1
5. Let A =  0.2 0.5 0.4 . Does Ak → O as k → ∞?
0.3 0.1 0.6
6. Let A be a non-negative real matrix with dominant eigenvalue µ. Let B = tI − A
where t > µ. Prove that every real eigenvalue of B is positive.
51