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% regardless of all the silly "target" bullshit that's been tossed in. What a collossal waste of time and bandwidth. And what a preening pedant to presume he's bringing some "new insight" to this forum. Next
If you actually bothered to read what's been written, the whole point is that the risk of ruin goes to 100% as times goes to infinity (which was how the problem was stated in the first place) - REGARDLESS of whether expectancy is positive or negative.
How is what I said incompatible with gambler's ruin? Gambler's ruin is a specific case of what I said.
No problemo: First, some notations: each trade is a biased coin toss with probability of head being p_win and tail being p_lose; you win +$a if you flip head, and -$b if you flip tail. Let's say at some point, you have NAV of $n and you consider the next $n/$b tosses; The probability of ruin is getting tail for exactly $n/$b tosses - which is of course p_lose^($n/$b). Now let's consider we start with capital $n0. If after $n0/$b tosses, we are still in business and have capital $n1, then the probability of getting to the end of toss $n1/$b is: (p_lose^($n0/$b))*(p_lose^($n1/$b)) Since $n0/$b is always positive, and p_lose > 0, then (p_lose^($n0/$b))*(p_lose^($n1/$b)) > (p_lose^($n0/$b)). That is to say, your probability of survival gets smaller at each step of $x/$b tosses, where $x is whatever your NAV happens to be at the end of the previous step. NOTE: this is key: p_lose^($n/$b) can NEVER be greater than one regardless of of the size of $n and $b because $n >0 (or you are ruined) and $b > 0 (remember, you lose -$b, so $b is the magnitude of the loss). Then as we keep multiplying these together, this series converges of 0 as the number of multiplications converges to infinity because each successive probability is zero. Thus, as the probability of survival goes to zero at infinity, probability of ruin (1 - probability of survival, goes to 1). QED
Recheck your math: Since $n0/$b is always positive, and p_lose > 0, then (p_lose^($n0/$b))*(p_lose^($n1/$b)) > (p_lose^($n0/$b)). Ooops!
This is as useful as saying "as your account balance decays towards zero, your prob of ruin becomes certain." The question is "how large", i.e. Show mathematically, at what nth trade, risk of ruin suddenly becomes significant/asymptotically steep?