# 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?

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.

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.

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 ...

### Want to see the full answer?

See Solution

#### Want to see this answer and more?

Our solutions are written by experts, many with advanced degrees, and available 24/7

See Solution
Tagged in