I hold this is a joke... I can't tell anymore on ET whether something is in jest or in all seriousness anymore
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.
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 Schrödinger equation in the full Cauchy-Lanczos basis.
No. It's not. It's been explained in the previous posts. Given infinite time (intraday bill's specification), risk of ruin is 100%;
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?
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.