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Инвестиционные фонды NordFx: профессиональное управление и прозрачность


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Вероятность слива. Чрезвычайно важная статья об мат. расчета


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#1 Fanat

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Опубликовано 27 Ноябрь 2005 - 08:24

Suppose that by setting your stop-loss appropriately, you are risking x% of your account (with a balance of $B, starting at $B1 let's say) in any one trade (and to keep it simple let's say there's only ever one trade running in parallel).

Now obviously if you get N losing trades where (1-(x/100))^N < some arbitrarily small amount like 5% (the exact level will depend on B and, I suppose, the policy of your broker on how much is the minimum allowed in the account), you are wiped out immediately.

If you assume (on the surface it looks pessimistic, but perhaps not) that the probability of being stopped out for any one trade is 0.5, then the chance of being wiped out in this way is obviously very small for x < 10% ..


e.g. if x= 2%, B1 = $10,000, then in this model the chance of that happening is p = 0.5^N where N is given by:
N = ln(0.05)/ln(1-(x/100)) = 150ish (148 nearly)
p = 0.5^150 = 7*10^-46 ... won't happen before the Universe ends or whatever.

However, how could you calculate the probability of being wiped out at all, through any combination of winning and losing trades for a fixed (large) number of trades T?

Suppose you set a profit target = stop loss, therefore target = B + x% (probably this is nowhere near a normal way to set a target, but I'd guess it's OK for order of magnitude purposes).

In this model we have p(win)=p(lose)=0.5, gain=loss=x%, and a 'hard' stop of 0.05B. What's the probability of hitting the stop within T trades?

Do we have to use something nasty like Markov chains in the general case? Or is there a good way to get an estimate, making reasonable estimates like T large, x << 1 or 'hard stop' << B etc.? Or is this question easy because of 'random walk' theory, since I've used p=0.5?

If I read this site correctly:
http://mathworld.wol...imensional.html
... then the probability of being at a distance of d after M steps in a 1D random walk is given by:
P = (2^-M) * (M!/(((M+d)/2)!((M-d)/2)!)).
And the expected distance d after M steps is given by:
d(M) = (M-1)!!/(M-2)!! where !! is the 'double factorial' for M even
d(M) = M!!/(M-1)!! for M odd
leading to an easier-to-calculate asymptotic value for large M:
d ~ sqrt(2*M/pi)

This formula seems to be more or less what I want, correct?

Well, if that asymptotic formula is right, then plugging this into my scenario above, we have T = number of trades = number of steps, M and d = N, the number of monotonic (unidirectional) steps to get down to 5% of B. Then you get these numbers :

so with x = 2% and N = 150, the likely value of T needed to reach wipeout would be (pi/2)*150^2 or about 35000 trades.
With x = 10%, N is about 28.4. So then T is about (pi/2)*(28.4)^2 or 1270 trades. Probably the asymptotic calculation is not valid in this case, so how indicative this number really is I don't know.


started thinking about this when I was getting to grips with backtesting in Metatrader. I tried a few basic strategies and noticed that in some cases there would be very big swings in equity. It starts at $10,000 and it goes up and down all over the place (not very good strategies ). In many cases it hit the default lower bound of $1,000. Of course, other times it would go down a long way and then back up to $20,000 etc.
So obviously if you have a strategy which you think is going to yield a profit long term (not that I have any such beast yet), you also have to take account of the volatility - not just of the market but also of your account balance. No good if, on average, it makes a profit over 1 year but has a 60% chance of hitting the bottom before it gets there.

So say you make, I don't know, 5 trades a day, and you guess about 1200 trades per year. What's the chance that you're going to hit bottom?
After all, isn't this why people say you must limit your risk to 1% or 2% or whatever?
I was trying to put actual figures on it (admittedly I had to make a few big simplifications).

The numbers seem to suggest that if you work with 2% as your risk per trade, then, on average, you'd have to wait 35,000 trades before you hit the bottom. So with those kind of numbers it might take 30+ years on average before you have that kind of appalling run of luck.
But that was with an equal chance of taking a profit or being stopped out. Also with an equal percentage gain in your winners (i.e. 2%) as percentage loss in your losers. So, big simplifications. Might be interesting, might not.


http://www.moneytec....ead.php?t=12429

#2 Fanat

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Опубликовано 27 Ноябрь 2005 - 08:32

Часть 2


Well I finally had a chance to look at this again. I revisited the Seykota website and looked at how he calculated the optimum bet fraction. Have a look at this and tell me if you think I've understood the idea correctly:

What he does is calculate the return as a function of the bet fraction b (i.e. the percentage of your equity you risk) ... if you apply his approach to this Van Tharp simulation (I'm talking about Level 1 and Level 2 here), you get:

Return = (1+B)^11 * (1+10b) * (1-B)^7 * (1-5b)

That would be for 20 "ideal" trades. Of course you could multiply it up by 1 million and then it would be accurate, but this will do.

The optimal return would be at a maximum of that function (return plotted against b on a graph). In order to find that analytically, you'd need to differentiate using the chain rule and then find the root of some hideous polynomial, but the job is done well enough by plotting it in Excel. The maximum is at 0.08 (8%). At that bet fraction, the expected return for 20 trades is 40.5%, roughly, which would give you 358% average return for 75 trades.

However, I'd say it doesn't follow that 8% is the best fraction to use for this game, because of the fact that there are only 75 trades. It's really a very small number when you compare with the 1 in 20 chance of a 5R drawdown. Those two numbers (75 and 20) would need to be much further apart to make the statistics smooth.




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