Kaufmann Risk of Ruin

virtualmoney · Nov 28, 2011

Quote from alexandermerwe:

the probability of producing a streak of losses just enough to ruin its bankroll B is finite

You got it...The point is to find a mathematical expression of this finite prob.
When is this prob(streak of losses just enough to ruin its bankroll B) > 50%?
i.e. Is it possible to express this prob in terms of current bankroll B, b, N, p, q, etc?

ronblack · Nov 28, 2011

Quote from kut2k2:

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

I guess job well done to this point. But that was a very special case of equal risk and reward at each step of the random-walk. What about if the loss is L(k) and the win is W(k) with avgL < avgW ?

I mean I don't know how the special case relates to the general case where the win and loss amounts are arbitrary. At what point a much larger avgW than avgL makes up for q > p for example so that R # 1?

I think we may be able to draw some general conclusions from this limited case ( W = L = 1) but again I don't know that especially when W and L are not constants.

kut2k2 · Nov 28, 2011

Quote from ronblack:

I guess job well done to this point. But that was a very special case of equal risk and reward at each step of the random-walk. What about if the loss is L(k) and the win is W(k) with avgL < avgW ?

I mean I don't know how the special case relates to the general case where the win and loss amounts are arbitrary. At what point a much larger avgW than avgL makes up for q > p for example so that R # 1?

I think we may be able to draw some general conclusions from this limited case ( W = L = 1) but again I don't know that especially when W and L are not constants.

http://www.elitetrader.com/vb/showthread.php?s=&postid=3355540#post3355540

http://www.elitetrader.com/vb/showthread.php?threadid=231633

Visaria · Nov 28, 2011

Interesting case that i think someone alluded to above was if you have infinite trials, you would have an occurrence where rather than going to ruin, you keep winning and the account goes to infinity.

If the prob of ruin tends to 1 as n tends to infinity, then you can also have prob of reaching infinity in one's acct tending to 1 as n tends to infinity.

virtualmoney · Nov 28, 2011

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

The path to ruin is easier. It takes 100% effort just to bring a 50% loss in capital back to breakeven. Dow index takes the escalator up but elevator down.

Visaria · Nov 28, 2011

Quote from virtualmoney:

The path to ruin is easier. It takes 100% effort just to bring a 50% loss in capital back to breakeven. Dow index takes the escalator up but, maybe not off a cliff, elevator down.

That depends...

virtualmoney · Nov 28, 2011

Quote from Visaria:
If the prob of ruin tends to 1 as n tends to infinity, then you can also have prob of reaching infinity in one's acct tending to 1 as n tends to infinity.

Quote from Visaria:
That depends...

On what? Prob(infinitely rich) overtaking prob(ruin)?

Visaria · Nov 28, 2011

Would depend on the edge of the method i.e. both the probability of a successful trade and the payoff. The bigger the edge, more likely the odds of hitting infinity first rather than hitting zero

virtualmoney · Nov 28, 2011

Quote from Visaria:
Would depend on the edge of the method i.e. both the probability of a successful trade and the payoff. The bigger the edge, more likely the odds of hitting infinity first rather than hitting zero

Ok, express this mathematically for any instantaneous B,bankroll, b,bet size, p-q & risk/reward ratio.
i.e. When is prob(rich) at least = prob(ruin), if it is even possible?

Visaria · Nov 28, 2011

Oh, do i really have to?

Look, suppose you have an edge such that a winning trade has reward /risk of 10/1 and the odds of winning are 90%. Say you did 100 million trades. Repeat this 100 million sequence infinite times. It stands to reason you will have more sequences which end as "rich" as opposed to "ruin".

QED.