I think you do not understand probability theory. Actually, you are one of those who think they understand and this is more dangerous than not understanding and admitting it. You will certainly lose all your money one day if not already. So, in trading you have only two outcomes: win or loss. The resulting random variable takes values from a (generally unknown) distribution F as in P(x = k) = F(x) You confuse random probabilistic outcomes, random variables and values of random variables. You ought to study Probability Theory or shut up because you expose yourself as an idiot although you may not be one.
Did you even bother to click on the links I provided and follow the examples? It's clear as day what's wrong with the two-outcome formula, but not to the lazy or the innumerate. I pretty much agree with The Big D. Anybody who doesn't get it at this point has no business trading. But that's not my concern anymore.
You seem to have too much of a big ego to admit your mistakes. You do not understand basic probabilioty theory, yet you want to give lessons to other people. Are you a moron or what? Your analysis in your links is totally flawed. I have no time to correct you. Read the Wikpedia article http://en.wikipedia.org/wiki/Kelly_criterion Do not reply please. You make no sense to me. P.S. I repeat. You confuse probabilistic outcomes, random variables and values of random variables. Hopeless situation for you...
It's clear the arrogant fool here is you, and unsurprisingly you are completely unaware of just how foolish you've exposed yourself to be. Welcome to my ignore list.
I repeat you egoistic fool. You confuse outcomes of probabilistic setups, random variables and values of random variables. You are a total, complete, cranked up moron. I repeat to you because you seem to lack basic comprehension skills. Trading has two outcomes, win or lose We define a random variable called Trade. Trade takes one of two outcomes, win or loss The value of the random variable is the trade gain or loss. This value comes from a distribution function. The values of random variables are not in general the outcomes you moron.
Actually, they aren't the same expectancy because the winner/loser ratio in the first set would be 1:1 and the winner to loser ratio in the second would be .76:1. Multiplying that by the winning percentage, the expectancy for the first set would be .6 and the second set's would be .46. The reason, of course, is that 1 "negative" black swan in 40 negative outcomes plays a greater weight than 1 "positive" black swan would in the 60 positive outcomes, if the black swans are symmetrical in size. That discrepancy would also flow through to the Kelly outcome. And, since you can always pre-define your maximum loss on a negative black swan by using options and figure that into your Kelly calculation as a maximum "value at risk", I don't see black swans as a barrier to Kelly position-sizing.
Explain, and try to do so without foaming at the mouth like that fool goodgoing. Did you read example 2 in my thread? If so, which part confused you, or seemed to be "disinformation"? If not, why not? This is pure mathematics. You can't do it by your guts, you have to use your head. If you don't get it, just say so. But don't go throwing around words like "disinformation" if you can't back it up, because that makes you look like goodgoing, and that's not a good look.
I just looked at your older thread and I must say I was very surprised of your errors. First you state the equation Gain[f] = (1 + f*R1)*(1 + f*R2)*...*(1 + f*Rn) for the gain. The equation is correct but this is the actual accumulated gain. The R's correspond to a random sequense of the return values R. Now in your example, you equate R1 to the return of win and R2 to the return of loss. This is just one possibility out of four (for two outcomes): R1 = win return R2 = win return R1 = win return R2 = loss return R1 = loss return R3 = win return R1 = loss return R4 = loss return Thus, in simple language, you have four different possibilities for the Gain[f] function and when you average all those you get the expectation E[Gain], which is nothing more than E[Gain] = p * R avg. of win - (1-p) R avg. of loss in the case of the two outcome example, R avg. of win = R win and R avg. of loss is R loss. This E[Gain] is the edge. It is known since Bernouli I think that if you divide this by R avg. win you get the fraction that maximizes the growth of the actual Gain function Gain[f]. This is Kelly. So, I hope you can see your error. It is a trivial error.
Looks like you never got past example 1. More to the point, the gain equation was for the expected gain, not for random gains. Given that trade results are unlikely to duplicate each other, the gain equation was set up to account for all the individual trade returns. No assumptions were made about whether the returns are positive ("win") or negative ("loss"). Now look at example 2. There are ten balls, and they come in three separate colors. That's three separate outcomes. Accounting for all the outcomes, the expected gain over ten draws with replacement becomes Gain[f] = ((1 - f)^7)*((1 - 5*f)^1)*(1 + 10*f)^2 This maximizes at f = k, and k is approximated by the formula I gave. More to the point, I also did the calculations for the exact value of k as well as for the 2-outcome formula, proving that the 2-outcome formula leads to overbetting in cases with more than 2 outcomes (e.g., trading). AFAIC this issue is closed. If you don't like my presentation, read gummy's. BTW he's a math professor, so anybody who says he doesn't know probability is a proven fool.