Kaufmann Risk of Ruin

[virtualmoney]

Quote from kut2k2:

Interesting...if you have 8 consecutive losses straightaway, your prob(ruin) changes from 0.04 to 0.9999 in the next 100 tosses?
What if you have only 4 consecutive losses that brings you to $6(point of capital path), what is the change in prob(ruin) for the next 100 tosses?

Notice how you never got a response, my friend. That's because math cranks are real good at challenging mathematical claims by others, regardless of evidence, and really bad at defending their own mathematical claims.

Yes, all the brain power to articulate mathematical insults could be compiled to make a bestseller... the mind is a terrible thing to waste.
Seems like the %change in prob(ruin) is greatly affected just by a few consecutive losses...and since unknown consecutive losses can occur at any point in a large sample path... =>prob(ruin) can suddenly change quite a bit.
http://www.quantwolf.com/calculators/recurrencetime.html

 

[intradaybill]

It is known that like animals of certain kinds, cranks also tend to flock.

Nobody owes to you any answers, cranks, for your stupid questions. There is no time to answer stupid questions.

If you don't comply with basic logic there is no point in talking to you.

I am asking you again, this is the 10th time maybe. What sense does it make to say that the failure probability of reaching a target T is one when the target T goes to infinity? Hey, crank, answer this.

That was a trivial problem you cranks, nobody wants to talk about except you, you making a lot of noise pissing and farting all over the place while failing to understand the real issue.

DontMIssTheBu was so right. This is a waste of time.

All of you on ignore cranks.

 

[virtualmoney]

Quote from intradaybill:

The problem is to determine when the probability of failure can be 1 when the target T when T is finite.
This is a practical problem for traders and a solution can be used to filter trading systems. That does not depend just on p,q, C,T,X. It is a very complicated problem.

When the broker goes bust, prob(ruin)->1 immediately even when T is small...
It is not complicated at all. The rest are just trival problems.

 

[trade4ever2day]

Useless disussion with irresponsible math.

When a trading system produces losses - and this is true for all systems other than the Holy Grail System - then the probability of producing a streak of losses just enough to ruin its bankroll B is finite for a number of trials n >= B/b, when b is the bet size and that probability is 1 when n goes to infinity.

There is no need for any math to know this, there is no dependence on p and q, or whether p > q or the other way around.

 

[kut2k2]

Quote from virtualmoney:

When the broker goes bust, prob(ruin)->1 immediately even when T is small...
It is not complicated at all. The rest are just trival problems.

Seriously, dude, you've fallen for this certainty-of-ruin rubbish also?

When the broker goes bust, hopefully you've already pocketed the profit you made from using a positive-expectation system.

 

[ronblack]

Quote from kut2k2:

Seriously, dude, you've fallen for this certainty-of-ruin rubbish also?

When the broker goes bust, hopefully you've already pocketed the profit you made from using a positive-expectation system.

Sure, sure thing, you pocketed 10K of profit due to positive expectation and then lost 100K of capital due to a negative expectation of the broker system existing for longer.

I think you haven't learned any lessons being a member of these fine forums. You can have a positive-expectation system and still lose it all because of capitalization constraints and of other factors that actually turn your expectancy negative. Ruin is certain as times goes by.

 

[kut2k2]

Quote from ronblack:

Sure, sure thing, you pocketed 10K of profit due to positive expectation and then lost 100K of capital due to a negative expectation of the broker system existing for longer.

How often do brokers go out of business? Less than 1% of the time?

I think you haven't learned any lessons being a member of these fine forums. You can have a positive-expectation system and still lose it all because of capitalization constraints and of other factors that actually turn your expectancy negative. Ruin is certain as times goes by.

Somebody forgot to tell that to Warren Buffet.

The contention of the majority of the ruin-is-certain cult here in ET is that positive expectation makes no difference. According to them, a system can be consistently positive-expectation right up to the point where somehow it takes a sharp turn and drives right over the edge of a cliff. Don't ask me to explain it; it's a crank theory that I don't believe in. If you haven't figured out that's where they've been coming from, you're even more clueless than I thought.

 

[Visaria]

Quote from intradaybill:

I think you did not understand what you read. The only difference in the link I gave you between ruin and a high return was the starting capital. The expectancy was the same. When are you going to get these facts straight people?

No, the difference was the inclusion of a $7 commission per trade which apparently wiped out the (positive) edge and made the system as a whole a negative expectation system.

Quote from intradaybill:
Example: fair coin toss. You make $2 with heads and lose $1 with tails.

Expectancy is $0.5

If you start with only $2 your probability of ruin is 0.25 right away and 0.9999 in the next 100 tosses or so.

If you start with $10, your starting probalility of ruin is (0.5)^10, very low and in the next 100 tosses it is only 0.04 (don't ask me how I got that).

Expectancy is the same. Expectancy assumes infinite capital as it is only valid at the limit of large numbers.

Right, so from a practical point of view, trade positive expectancy systems, start with large capital and trade small like 0.5% risk per trade. Note such a staking plan can never lead to ruin theoretically (although if you do have 99% of capital wiped out, it's academic).

 

[ronblack]

R{infinity|X|C} = (q/p)^(C-X) if q < p

OK, I say fine for a moment and let us assume that the win rate is 55% in an unfair coin toss game in which you pay $1 when you lose and receive $1 when you win. According to the formula, if you start with $1 the probability of ruin for growing to an infinite bankroll before dropping to zero holdings is

R{infinity|X|C} = (q/p)^(C-X) = (.45/.55)^1 = 0.8181 or the probability of an infinite bankroll before losing it all is .1818

At the same time, the probability of ruin after just 1 trial is 0.45.

Something is wrong here. We have 18.18% for becoming infinitely rich and 45% for losing it all?

 

[kut2k2]

Quote from ronblack:

R{infinity|X|C} = (q/p)^(C-X) if q < p

OK, I say fine for a moment and let us assume that the win rate is 55% in an unfair coin toss game in which you pay $1 when you lose and receive $1 when you win. According to the formula, if you start with $1 the probability of ruin for growing to an infinite bankroll before dropping to zero holdings is

R{infinity|X|C} = (q/p)^(C-X) = (.45/.55)^1 = 0.8181 or the probability of an infinite bankroll before losing it all is .1818

At the same time, the probability of ruin after just 1 trial is 0.45.

Something is wrong here. We have 18.18% for becoming infinitely rich and 45% for losing it all?

The probability of ruin is based on ALL possible paths to ruin, not just the single path of going directly from C to X. C might grow to 10,000C before dropping back to X. As long as C < T, the many paths to ruin are still ongoing.

R{infinity|X|C} = p*R{infinity|X|C+1} + q*R{infinity|X|C-1}

R{infinity|0|1} = .55*R{infinity|0|2} + .45 = (.45/.55)^1 = .818181818

R{infinity|0|2} = (.8181818 - .45)/.55 = .669421488

R{infinity|0|2} = (.45/.55)^2 = .669421488

QED