Position Sizing

How much to bet in each trade

J. Ignacio Ulacia F. (15.6.2002)

file (Betsize)

It is important to know what is the optimum amount of money that one should place on every trade. This is a function of the risk that is intended to carry on every trade and the place that one should leave a trade and accept the loss.

This decision determines the real risk that of every trade. If you risk too little, you will not be rewarded sufficiently to cover your commission and administrative costs. If you risk too much you increase the probability of loosing all. Then what is the adequate amount of risk? From Van Tharp [1] we can see that he has two examples and computes the probability of risk/reward according to previous trades.

In general from game theory, the return form "N" number of games, with probabilities of outcome "Pi", and with payoffs of winning and loosing determined by "Wi" where winning will be positive and loosing will be negative and "R" being the percentage risked form a nominal investment "Io". The resulting investment after playing "N" games is

I = Io * (1 + RW1)(NP1) * (1 + RW2)(NP2) * .............* (1 + RWi)(NPi)

In Van Tharp's book [1] there are some examples that represent true distributions, however the best position size is not mentioned. It can be observed from the previous function that it contains a Maximum that falls in the optimum bet for all games. Solving numerically for the equations represented by both examples we obtain.

Example 1

Payoff
-3
-2
-1
0
5
10
20

Probability
4%
10%
50%
20%
10%
3%
3%

Example 2

Payoff
-3
-2
-1
1
2
3
5
10
20
30

Probability
4%
10%
56%
10%
6%
4%
3%
3%
2%
2%

The maximum for each example will be at R=5.14% for Example 1; and R=6.43% for Example 2 as shown in the following graphs.

 

 

 

From both figures it is readily observed there is a point after which higher risk will only lead to overall losses. This point is at 11.5% for Example 1, and 15% for Example 2.

The payoffs and probabilities tables come form the system used for trading.

Form another example form Ed Seykota, as follows. Having a fair coin, that gives 50% probability of winning and loosing the game is to obtain a twice the bet size if you win and you only loose the amount at risk otherwise. Therefore the equation simplifies to

I = Io * (1+2R)(N/2) * (1-R)(N/2)

that for simplicity, if only two game are played

I = Io * (1 + 2R) * (1-R) = Io * (1 + R - 2R2)

From this equation it is simple to obtain the maximum differentiating and equating to zero.

dI/dR = Io* (1 - 4R) = 0 => R = 0.25 or 25%

It is important to note that the example provided assumes a 50% win/loose ratio. Which in most of the times in the stock market is less than ideal. To complicate a little the problem, If the coin is skewed and gives 40% wins and 60% looses with the same payoffs. Plugging the numbers in the previous equation and solving numerically for R. The result provides a maximum at R=10%, that shifts the amount risked considerably.

Even with this excellent winning payoff 2:1, if the coin is skewed to one third (33%) winning, and two thirds (66%) loosing, the game has no maximum. Solving for this equation, we obtain

 

dI/dR = Io * (6R2 + 6R) = 0 => R1 = 0, R2=1

 

 

References

[1] Van K. Tharp, "Trade your way to financial Freedom", Mc Graw Hill, New York (1999).
[2] Ed Sykota, http://www.seykota.com/tribe/risk/index.htm

 

Copyright 2005© J. Ignacio Ulacia F., All rights reserved