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An experiment can result in any of the outcomes 1, 2, or 3.  If there are two different wagers with r1(1)=4, r1(2)=8, r1(3)=-10 and r2(1)=6, r2(2)=12, r2(3)=-16, is an arbitrage possible?

Question

An experiment can result in any of the outcomes 1, 2, or 3.  If there are two different wagers with r1(1)=4, r1(2)=8, r1(3)=-10 and r2(1)=6, r2(2)=12, r2(3)=-16, is an arbitrage possible?

check_circleAnswer
Step 1

Let's define two matrix as shown on white board.

The first matrix is R matrix which shows three possible outcomes across two wagers.

The second matrix is the probability matrix, P that has probability of three outcomes as its elements.

Please note that p1, p2 and p3 being probabilities can't be negative and they can't be greater than 1.

Hence, 1 ≥ p1, p2, p3 ≥ 0.

[r1(1) r1(2) r1(3)|
8 -10
4
R =
Ir2(1) r2(2) r2(3)]
16 12 -16]
P = p2
[p3]|
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Image Transcriptionclose

[r1(1) r1(2) r1(3)| 8 -10 4 R = Ir2(1) r2(2) r2(3)] 16 12 -16] P = p2 [p3]|

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Step 2

According to Arbitrage Theorem, an arbitrage exists if we can\'t find a solution to this equation:

R. P = 0

The expanded equation is shown on the white board.

p1]
4 8 -10.|p2
0
6
12 -16
[p3]
help_outline

Image Transcriptionclose

p1] 4 8 -10.|p2 0 6 12 -16 [p3]

fullscreen
Step 3

So, we get two equations from above:

Equation (1): 4p1 + 8p2 - 10p3 = 0

Equation (2): 6p1 + 12p2 - 16p3 = 0

and we get one ...

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