Trading Math - Part I

[trade4ever2day]

Quote from bwolinsky:

...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.

Post corrected below:

I think Bill's target was on net P/L and the stop was a terminal time stop. Nevertheless, it is not true that if no leverage is involved there is no risk of ruin. What about if have a $100 account, you trade US bond futures and you trade 1 contract with a 5 point stop. If you get 20 consecutive losers you are ruined. No leverage was involved.

 

[bwolinsky]

Quote from alexandermerwe:

Post corrected below:

I think Bill's target was on net P/L and the stop was a terminal time stop. Nevertheless, it is not true that if no leverage is involved there is no risk of ruin. What about if have a $100 account, you trade US bond futures and you trade 1 contract with a 5 point stop. If you get 20 consecutive losers you are ruined. No leverage was involved.

Actually an extreme amount of leverage is involved in the situation you mentioned.

I guarantee that a 1% move in US Bond futures on this account is a much greater percentage than just 1%. This implies leverage.

Also, it was the terminal time stop that if you were to increase it to infinity then your ROR will be 1 always.

 

[kut2k2]

Quote from ronblack:

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.

You're a slackwitted mouthbreather with delusions of intelligence and reading comprehension. I tried to get you to find out what "disproof" means but once again you proved that a fool can be led to knowledge but it won't make him think.

BTW your presumed knowledge of mathematics is laughable:

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

Actually, asshole, P{T|0} = 0. That was even stated upfront.

It's back to remedial math for you, maybe the eighth time will stick.

 

[intradaybill]

Quote from kut2k2:

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)&#8319;*P{T|1}


P{T|S} = SUM[ P{T|n+1} - P{T|n} ]_n=0...S-1

P{T|S} = SUM[ (q/p)&#8319;*P{T|1} ]_n=0...S-1

P{T|S} = P{T|1}*SUM[ (q/p)&#8319; ]_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

Why does an idiot like you think what you wrote has anything to do with the problem I posed?

Did I speak of unit bets?

Did I speak of q < p?

NO. So why the hell you feel you have to come to this thread and piss all over the place?

Here is how I stated the problem imbecile moron:

Quote from intradaybill:

Define:

1. A trading system that generates entries and exits with some frequency F > 0 per unit time and starts with bankroll B.
2. A Cummulative profit objective P
3. A time to quit for good objective T
4. The probability of ruin R such that bankroll B reaches 0.

The questions (problems) are:

(1) Given P = Pquit and/or T = Tquit are there conditions that may lead to R =1? In other words, even if there is a quit time or quit profit, under which conditions ruin R = 1 is certain? Please exlude the case where expectancy is negative.

(2) Alternative, if there are no P and T constraints is ruin certain (R = 1)?

I think (2) has been answered already. (1) remains.

Look at what you wrote imbecile and ask yourself: what does that have to do with the problem posed? I mean seriously, do you feel compelled to piss all over the place in public?

 

[bwolinsky]

Quote from intradaybill:

Why does an idiot like you think what you wrote has anything to do with the problem I posed?

Did I speak of unit bets?

Did I speak of q < p?

NO. So why the hell you feel you have to come to this thread and piss all over the place?

Here is how I stated the problem imbecile moron:



Look at what you wrote imbecile and ask yourself: what does that have to do with the problem posed? I mean seriously, do you feel compelled to piss all over the place in public?

Bill, we may not have "completely" answered 1, but my answer does the most completely though forgive me for not writing semantics because 1 is completely dependent on how long your time horizon is or how big T gets. If T approaches infinity that is really the most precise and simplistic conclusion you can draw that ROR is nearly perfectly certain when T is large, and even if you don't feel like the p's and q's don't matter, the cases in his answer may have been copied and pasted from some other website.

 

[virtualmoney]

Quote from kut2k2:

Actually, asshole, P{T|0} = 0.

That makes sense since the probability of reaching target T with starting zero bankroll is zero. Can P{0|S} be somehow expressed not only in terms of p & q but other variables such as Nth trade, since we want to find out at what point of trade P{0|S} is max .i.e. include some other conditions/quantities where P(total ruin) is largest?

 

[bwolinsky]

Quote from virtualmoney:

That makes sense since the probability of reaching target T with starting zero bankroll is zero. Can P{0|S} be somehow expressed not only in terms of p & q but other variables such as Nth trade, since we want to find out at what point of trade P{0|S} is max .i.e. include some other conditions/quantities where P(total ruin) is largest?

There isn't enough specific information to truly discover the max, and I don't think you'll be able to prove or disprove that P{0|S}| is maxed whenever certain conditions are met. I don't think there's enough information to find the real max but algebraically I don't feel like doing it but I'll be the first to admit that I did not want to take our 400 level Proofs class.

The question posed by Bill is still too general for any real logical progressions because the question is not that well specified and I don't think even reading to here I quite get why we're trying to find the conditions where ROR=1, or ROR=0, or anywhere in between, what is the question?

And I know Bill posted it but there seems to be a lot of missing assumptions and it doesn't appear to be framed in the classical sense of a proof and simply showing that if ROR=1 because time is large is the same result you'll get if you plug y=x and just watch x increase until you reach infinity. There isn't anything wrong with saying that, but as far as the problem goes I don't think that the question's been correctly specificied and there really is no fundamental basis to try to prove or disprove these things, so if I'm saying we still don't know what you're trying to prove and if I wanted to just ask what you think y will be if you keep increasing the time scale. You'll get an x out to an unknown infinity and not only will it show you by your equations that with your logic there is some assumption of perpetuality and while I think Bill might have changed some of the specs there isn't enough information here to answer these questions.

I can have my Topology PhD from Stanford look at this. I'm sure he'll be able to give you answers but there's no way I'm going to plagiarize like kut2wutkut2 did and claim it is my own work. Not only was the proof nothing but a bunch of cases, but it's most likely copy pasted from some university website.

 

[virtualmoney]

Maybe this will be helpful in understanding path dependence ruin concept better.
http://www.trade2win.com/traderpedia/Risk_of_ruin

ROR = q/(1+p) x current balance

"Conclusion: In agreement with Kaufman the risk of ruin is greatest at the beginning. The risk of ruin also increases the longer your remain a trader because the risk of experiencing a series of losses increases.
The risk of ruin in our example would remain the same as the risk at the beginning if we did not scale back when we started to hit a series of losing trades. By scaling to smaller trade sizes as our portfolio is reduced we lower the risk of ruin and improve our survival rate."

It also mentions about its the no. of string of losses that determines the rate of ROR.
Maybe we can study the characteristics/optimal string of losses for which ROR rate is max.

 

[kut2k2]

Quote from virtualmoney:

That makes sense since the probability of reaching target T with starting zero bankroll is zero. Can P{0|S} be somehow expressed not only in terms of p & q but other variables such as Nth trade, since we want to find out at what point of trade P{0|S} is max .i.e. include some other conditions/quantities where P(total ruin) is largest?

I can't imagine why you'd want to find P{0|S} because the disproof I posted is based on the assumption that T is never smaller than S.

If you're trying to reformulate the problem in terms of an actual probability of ruination, I suggest the following:

R{T|S} = p*R{T|S+1} + q*R{T|S-1}

R{T|0} = 1

R{T|T} = 0

 

[kut2k2]

Quote from bwolinsky:

Quote from intradaybill:

Why does an idiot like you think what you wrote has anything to do with the problem I posed?

Did I speak of unit bets?

Did I speak of q < p?

NO. So why the hell you feel you have to come to this thread and piss all over the place?

Here is how I stated the problem imbecile moron:



Look at what you wrote imbecile and ask yourself: what does that have to do with the problem posed? I mean seriously, do you feel compelled to piss all over the place in public?

Thanks for quoting bill's rant. He knows I have him on ignore so I don't have to directly read his idiocy anymore.

But this does allow the opportunity for a recap:

Quote from intradaybill:

Statement 1:

For ANY trading system, with ANY parameters, if there is no target quit time or target profit after which it quits, eventually there will be total ruin.

I invite all the brains here to either prove or disprove Statement 1.

...and learn something...

Notice the lack of any restrictions of any kind on the trading system, the underlying, or anything else. The statement is essentially that:

ALL trading systems under ALL circumstances will eventually fail unless stopped by a target profit or a time limit.

The challenge is to prove or to disprove the above statement.

So for those of us who actually passed our math courses, we remember that a SINGLE counter-example to a universal statement will DISPROVE the universal statement.

Quote from kut2k2:

Seriously, guys ...

Am I the only one who sees that this is just the old Gambler's Ruin problem crudely dressed up in trading clothes? So long as you have a positive-expectation trading system, your risk of ruin is less than 100%.

Remember: a single counter-example serves as a disproof.

I then posted the counter-example:

Quote from kut2k2:

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{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 now the imbecile is whining and nitpicking over the win-loss ratio above being 1. Seriously? We're supposed to take this bullshit seriously when a blanket challenge about a universal statement was issued? Now suddenly there are restrictions attached?

Like I said before, this isn't really about math, it's about a preening pedant presuming he's bringing some "new insight" to this forum.

However, for those of you wondering whether this is true of all positive-expectation systems, the answer is yes. There is no closed-form proof for the general case, but there have been Monte-Carlo simulations. This can be found with a google search.