Express optimal nth (just like optimal f) as a function of probability(win,lose) or some of the above quantities mentioned.
Trading Math - Part I
- Thread starter intradaybill
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Express optimal nth (just like optimal f) as a function of probability(win,lose) or some of the above quantities mentioned.
Ooops indeed; I flipped the inequality sign there, which means what I wrote on that line was a typo - but my conclusion is right: the multiplication of a series of a positive numbers less than 1 and greater than 0 converges to 0 in infinity.
Thanks for pointing out the typo - and by seeing that I made a typo on that line, which means you see that each conditional probability is less than 1 and greater than 0, and that each further step in conditional survival probability is less than previous, I've established my proof, no?
Thanks for pointing out the typo - and by seeing that I made a typo on that line, which means you see that each conditional probability is less than 1 and greater than 0, and that each further step in conditional survival probability is less than previous, I've established my proof, no?
No.
Gambler's Ruin:
Let p be the probability of winning one betting unit.
Let q be the probability of losing one betting unit.
p + q = 1
Let P{T|S} be the probability of reaching target bankroll T with starting bankroll S.
P{T|S} = p*P{T|S+1} + q*P{T|S-1}
P{T|0} = 0
P{T|T} = 1
(p + q)*P{T|S} = p*P{T|S+1} + q*P{T|S-1}
P{T|S+1} - P{T|S} = (q/p)*( P{T|S} - P{T|S-1} )
P{T|2} - P{T|1} = (q/p)*( P{T|1} - P{T|0} )
P{T|2} - P{T|1} = (q/p)*P{T|1}
P{T|3} - P{T|2} = (q/p)*( P{T|2} - P{T|1} )
P{T|3} - P{T|2} = (q/p)²*P{T|1}
P{T|4} - P{T|3} = (q/p)*( P{T|3} - P{T|2} )
P{T|4} - P{T|3} = (q/p)³*P{T|1}
P{T|n+1} - P{T|n} = (q/p)ⁿ*P{T|1}
P{T|S} = SUM[ P{T|n+1} - P{T|n} ]_n=0...S-1
P{T|S} = SUM[ (q/p)ⁿ*P{T|1} ]_n=0...S-1
P{T|S} = P{T|1}*SUM[ (q/p)ⁿ ]_n=0...S-1
P{T|S} = P{T|1}*(1 - (q/p)^S)/(1-(q/p))
P{T|T} = P{T|1}*(1 - (q/p)^T)/(1-(q/p))
1 = P{T|1}*(1 - (q/p)^T)/(1-(q/p))
P{T|S} = (1 - (q/p)^S)/(1 - (q/p)^T)
P{infinity|S} = 1 - (q/p)^S if q < p
P{infinity|S} > 0 if q < p
QED
So you showed at the end of your proof that the probability of gaining infinite amount of money starting with S is greater than 0 if probability of winning is greater than the probability of losing?
(1). Yes, that's true (I actually pointed this out in my first or second post on this thread.
(2). How is this gambler's ruin? Isn't the usual formulation of gambler's ruin have to do with whether a gambler is... 'ruined'? not having more money than god?
(3). You offered a proof of something completely different and not related to either my statement (which you say is wrong) or the proof that I offered in support (which is more puzzling because you caught a careless typo, <- which shows you understand why the proof is correct)
(1). Yes, that's true (I actually pointed this out in my first or second post on this thread.
(2). How is this gambler's ruin? Isn't the usual formulation of gambler's ruin have to do with whether a gambler is... 'ruined'? not having more money than god?
(3). You offered a proof of something completely different and not related to either my statement (which you say is wrong) or the proof that I offered in support (which is more puzzling because you caught a careless typo, <- which shows you understand why the proof is correct)
It is clear he doesn't understand the problem or even what is being discussed here. He copied and pasted an informal proof of the probability of reaching T before reaching S for the trivial case when win amount = +1 and loss amount = -1, given that p>q.
As I said this is an informal proof. In a graduate exam in finance this proof is not accepted. You must get the characteristic polynomial of the recurrence relationship and solve for its roots for an acceptable formal proof. It's been a long time and I won't do it here. We can see from the trivial result though that the probability of reaching T before reaching S=0 is equal to :
P{T/0} = (1 - (q/p)^0)/(1 - (q/p)^T) = 1 /(1 - (q/p)^T)
as T gets very large
P{T/0} ->1 /1 = 1
Again, this is a trivial result and not related to the problem posed, which is very complex and has no unique solution.
Really? Because you completely contradict that here:
So I didn't belabor the point that the gambler is ruined if q is greater than or equal to p. Sorry, I thought that was obvious.
It's certainly related to the OP. The OP asked for a proof or a disproof. I posted a disproof. Case closed.
If what I offered is wrong then simply point to the error(s) and we can proceed from there. Otherwise this is just nitpicking.
We are talking here about a serious nut case. Are you going to stop farting around moron? Do you understand that what you wrote down is a very special, almost trivial, case of risk of ruin? Do you understand you moron that there are trading systems with p < q and avg. win >> avg. loss your equation doesn't cover? Stop farting you dirty ass.
As far as Bill's question, there is a variable t that will make the ROR equal to 1. If the duration of T is signifcant, as we've already pointed out, the ROR is certain.
However, over shorter duration periods the probability of ruin cannot possibly be 1 if you are trading market based derivative products because as was also pointed out depending on whether you have a stop no stop or a target but no stop or a stop and a target the reality is that while the market can decline, there isn't any risk of ruin when there isn't any leverage involved.
But, say there is leverage, like 3x or 2x, we know mathematically that they have a statistical likelihood of approaching zero nearly 100% of the time, but even with leverage if you have some stops that would make it really hard to lose everything, I don't think that even if you could know your p's and q's well enough, I find the page here where the P(win)=0.99 yet ROR=1, and I don't believe this is possible unless there aren't any stops, I'm not talking about profit targets but if there are stops I still don't see where the gambler's ruin has much applicability to high level finance. It is a method to teach opportunity risk, sure; I mean you might know your probabilities but if there aren't enough degrees of freedom or enough trades to determine them then it seems all we're really trying to do here is re-hash a gambling problem, which is not the same as examining trading systems.
I made a speech on technical analysis at the end of college in 06, and that presentation had about 20 slides and about a half hour but it did cover nearly all of these subjects like defining the trading universe, setting parameters, discussing linear regression analysis techniques with R squares greater than 90%.
Anyway that was all laid out in the speech and I don't have it anymore, but the idea was essentially refuting the premise that you cannot ever expect your system to always work.
Now, we go into the always works mode:
Always works to some might be daily, but to each investor that is probably at most quarterly but most likely they pay more attention to monthly and weekly results, so if that's <i>always working</i> we'd have to create some probabilities of winning, run some backtests, optimize, and if you're not winning 95% of the trades I don't think that that's any different from year to year if you have win percentages greater than 97%, so that temporary drop of 95% win percentages might not be that significantly different than the strategy with a P(WIN)=0.97
Apparently we can have an ROR of 1 even if we have a P win prob of 0.99, but in this extreme example the probability approaches 1 because we have an essentially unlimited time horizon, which isn't reasonable but if you have that I still don't think ROR will ever be 1 in the short run but the longer terms in the half century will always have ROR's of 1, but this is more related to the fact that the statistic holds the longer you are an investor or trader, and I think that's really all that needs to be said here because pretending that gambling, which we know has a negative return expectation, is the same as investing, which has a positive return expectation and I'd lump the investing/trading category together even though there's a little more of a gambling factor when trading than when you're investing, is not a similar comparison.
If we wanted to risk 100% of capital, doing so in individual companies is probably suicidal, but if you drop 100% of your portfolio into ETF's or leveraged ETF's the probability of losing everything is zero because it is a safe assumption to presume that the market will never go away. So even though the probability of ruin increases as your betting size increases, absolute ruin is essentially impossible unless the world happens to be in Armaggedon, and, I'm sorry to add the cliche, you'll probably have more important things to worry about than money in this case.
However, over shorter duration periods the probability of ruin cannot possibly be 1 if you are trading market based derivative products because as was also pointed out depending on whether you have a stop no stop or a target but no stop or a stop and a target the reality is that while the market can decline, there isn't any risk of ruin when there isn't any leverage involved.
But, say there is leverage, like 3x or 2x, we know mathematically that they have a statistical likelihood of approaching zero nearly 100% of the time, but even with leverage if you have some stops that would make it really hard to lose everything, I don't think that even if you could know your p's and q's well enough, I find the page here where the P(win)=0.99 yet ROR=1, and I don't believe this is possible unless there aren't any stops, I'm not talking about profit targets but if there are stops I still don't see where the gambler's ruin has much applicability to high level finance. It is a method to teach opportunity risk, sure; I mean you might know your probabilities but if there aren't enough degrees of freedom or enough trades to determine them then it seems all we're really trying to do here is re-hash a gambling problem, which is not the same as examining trading systems.
I made a speech on technical analysis at the end of college in 06, and that presentation had about 20 slides and about a half hour but it did cover nearly all of these subjects like defining the trading universe, setting parameters, discussing linear regression analysis techniques with R squares greater than 90%.
Anyway that was all laid out in the speech and I don't have it anymore, but the idea was essentially refuting the premise that you cannot ever expect your system to always work.
Now, we go into the always works mode:
Always works to some might be daily, but to each investor that is probably at most quarterly but most likely they pay more attention to monthly and weekly results, so if that's <i>always working</i> we'd have to create some probabilities of winning, run some backtests, optimize, and if you're not winning 95% of the trades I don't think that that's any different from year to year if you have win percentages greater than 97%, so that temporary drop of 95% win percentages might not be that significantly different than the strategy with a P(WIN)=0.97
Apparently we can have an ROR of 1 even if we have a P win prob of 0.99, but in this extreme example the probability approaches 1 because we have an essentially unlimited time horizon, which isn't reasonable but if you have that I still don't think ROR will ever be 1 in the short run but the longer terms in the half century will always have ROR's of 1, but this is more related to the fact that the statistic holds the longer you are an investor or trader, and I think that's really all that needs to be said here because pretending that gambling, which we know has a negative return expectation, is the same as investing, which has a positive return expectation and I'd lump the investing/trading category together even though there's a little more of a gambling factor when trading than when you're investing, is not a similar comparison.
If we wanted to risk 100% of capital, doing so in individual companies is probably suicidal, but if you drop 100% of your portfolio into ETF's or leveraged ETF's the probability of losing everything is zero because it is a safe assumption to presume that the market will never go away. So even though the probability of ruin increases as your betting size increases, absolute ruin is essentially impossible unless the world happens to be in Armaggedon, and, I'm sorry to add the cliche, you'll probably have more important things to worry about than money in this case.