Trading Math - Part I

DontMissTheBus · Nov 2, 2011

I hold this is a joke... I can't tell anymore on ET whether something is in jest or in all seriousness anymore

Quote from iplay1515:

It appears that Einstein's theory of evolution is alive and well.

virtualmoney · Nov 2, 2011

Quote from intradaybill:

(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.

http://www.physicsforums.com/showthread.php?t=542040&page=2
Someone mentioned the moment the Markov chain has its first stopped out 'loss', ruin R=1 is certain...even when p(win)=0.99...whatever that means.

iplay1515 · Nov 2, 2011

Quote from DontMissTheBus:

I hold this is a joke... I can't tell anymore on ET whether something is in jest or in all seriousness anymore

My point exactly with a pinch of nonsense added for flavor.

DontMissTheBus · Nov 2, 2011

Ah... you've tried to read a jackie h. thread.... which is, of course, where I got the idea of looking at prices in cauchy-hilbert space in which statically optimized medians of future prices allows me to capture 1e100% of the daily true range.

Quote from iplay1515:

My point exactly with a pinch of nonsense added for flavor.

iplay1515 · Nov 2, 2011

Quote from DontMissTheBus:

Ah... you've tried to read a jackie h. thread.... which is, of course, where I got the idea of looking at prices in cauchy-hilbert space in which statically optimized medians of future prices allows me to capture 1e100% of the daily true range.

Excellent!

I would also include the solutions of the matrix SchrÃ¶dinger equation in the full Cauchy-Lanczos basis.

goodgoing · Nov 3, 2011

Quote from virtualmoney:

http://www.physicsforums.com/showthread.php?t=542040&page=2
Someone mentioned the moment the Markov chain has its first stopped out 'loss', ruin R=1 is certain...even when p(win)=0.99...whatever that means.

IMO that is true only for R:R = 1.

DontMissTheBus · Nov 3, 2011

No. It's not. It's been explained in the previous posts. Given infinite time (intraday bill's specification), risk of ruin is 100%;

Quote from goodgoing:

IMO that is true only for R:R = 1.

virtualmoney · Nov 3, 2011

Quote from goodgoing:
IMO that is true only for R:R = 1.

For this simplified model, we can try to apply that to trading...
If p(win)=0.99 and q(loss)=0.01, and each trade is independent for a fixed bet, then the probability of losing at least once after n trials is 1−0.99^n. It's clear that the probability of losing at least once becomes large for large n, which sets off the initial condition for total ruin.

IMO, For e.g, if system has p(win)=0.8, we should reduce trade size after 6 trials since
1-0.8^6~0.8. In general, reduce size when prob. of losing at least once after n trials = p(win).

This probability value also seems to oscillate back if we view it from the opposite direction .i.e. prob. of winning at least once = 1-0.2^n = q(loss) after a while, when n becomes larger, we can increase trade size again?

DontMissTheBus · Nov 3, 2011

I'm pretty sure this is wrong: the correct solution for even this simple case will require the pay out of the win vs the loss relative to the initial capital level.

Ruin, in finite time and finite capital, is a path-dependent concept; Thus, simply using p(win) and 1-p(win) won't get you all the way there.

Quote from virtualmoney:

For this simplified model, we can try to apply that to trading...
If p(win)=0.99 and q(loss)=0.01, and each trade is independent for a fixed bet, then the probability of losing at least once after n trials is 1−0.99^n. It's clear that the probability of losing at least once becomes large for large n, which sets off the initial condition for total ruin.

IMO, For e.g, if system has p(win)=0.8, we should reduce trade size after 6 trials since
1-0.8^6~0.8. In general, reduce size when prob. of losing at least once after n trials = p(win).

This probability value also seems to oscillate back if we view it from the opposite direction .i.e. prob. of winning at least once = 1-0.2^n = q(loss) after a while, when n becomes larger, we can increase trade size again?

virtualmoney · Nov 3, 2011

Quote from DontMissTheBus:
I'm pretty sure this is wrong: the correct solution for even this simple case will require the pay out of the win vs the loss relative to the initial capital level.
Ruin, in finite time and finite capital, is a path-dependent concept; Thus, simply using p(win) and 1-p(win) won't get you all the way there.

Please elaborate mathematically with examples.