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mathematician/trader sought
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mathematician/trader sought

  #1 (permalink)
Elite Member
boulder, colorado
Futures Experience: Advanced
Platform: TradeStation
Favorite Futures: emini's
Posts: 22 since Jun 2010
Thanks: 2 given, 8 received

mathematician/trader sought

A trade with a target of k ticks and a stop loss of s ticks, with probability p of gaining k ticks before losing s ticks and a strategy of entering the trade and exiting only by stopping out or hitting the target, gives
- E{profit} = p*k (1-p)*s = p*(k+s) s, and
- sd{profit} = (k+s) √ (p*(1-p))
E{profit} is positive iff p>s/(k+s). This is trivial probability theory but it has interesting consequences.
Trading n contracts, each such trade has E{profit} = n*( p*(k+s) s). Suppose you make this bet a number of times, starting with fortune F0 (F measured as a number of ticks) and achieving fortunes F1, F2, . . . by betting the fraction α of your fortune at each trade; that is, ni*si <= α*Fi for the ith bet, with n0 <= α*F0/s0.
The trade is equivalent to a coin toss, betting $1 on heads where P{H} = p: letting w = k/s, if H comes up you win $w and if T comes up you lose $1. After one toss your fortune is F0 + w with probability p and F0 1 with probability 1-p. To maximize the growth rate of your fortune, the optimal strategy is to bet α = p (1-p)/(1+w) of your fortune on each play.
Translating back to trading, the best ni would clearly vary with Fi and also depend upon many other factors, like time of day, etc. some of which may be difficult to estimate without order data that only larger brokerages possess, but in any case the supply of opposing contracts or shares with the initial conditions (p,k,s) will be exhausted by the taking of large enough positions, typically much smaller than αi*Fi, so the limit is determined not by the Kelly Criterion but by the market itself, and the optimal strategy would be to choose ni so that the market can absorb the ni volume without altering the (p,k,s) relationship.
As an example, for a gamble with k=5, s=2 and p =.8, you would bet .8-.2/3.5 = .743, or about 75% of your fortune, and with k=2, s=2, p=.8, the maximal growth rate is achieved by betting .8 - .2/2 = 70% of your fortune, and for k=1, s=4, p=.8, you bet .8 - .2/(5/4) = .64. For p = .8, whatever are k and s, you would always bet at least 60% of your fortune to maximize its growth rate, assuming the initial conditions, and thus, for such trades, the limit is determined by the order flow.
However, some markets, in particular the ES, can absorb a lot of volume without being moved, as you can see by looking at a 10k chart. They thus can handle considerable size, not unlimited but certainly ni large enough for most needs and interests.
Consider a trade with p=1- ε, with k = 8 and s = 1:
- E{profit}=8-9*ε
- sd{profit} = 9 √(ε*(1- ε))
For n trades a day:
E{profit}=(8-9*ε)*n, and
sd{profit} = 9 √(n*ε*(1- ε)).
p = .8 is where it starts to get interesting, as that is when E{profit} > 2*sd{profit} begins to be possible, and even n=1 is quite useful. Zero-ε trades dont exist, of course, as anything can happen at any time, but micro-ε trades exist.
We are a private hedge fund and would be interested in a partner.

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