%% Younger one is== the easier it is to be silly. Sold a homebuilder on the ask today, but it started getting so wild, a market order may have done much better?? LOL Missed a 4 or 5 point buck [deer] 5/+ times @ about 50 yards; easy to miss @ any age LOL
"reality is what your brain tells you it is"! if you think that you need to know all about position sizing..optimal leverage..black scholes..kelly criterion..fibonacci retracements..stochastics..and so on..and on..and on..and on..and on..then..you might want to think about how it is your brain is telling you that you need all this information in order to carry out the simple acts of buying and selling
we are all but fools caught up in the big game where the smart ones gamble with other people's money..the good old Mr & Mrs Gullible the real question to ask oneself..is..do i stay a fool..or do i make a concious decision to do something about it..which..of course..will mean some pain and suffering in my perceived reality it is always far easier to do nothing..and stay on the merry-go-round
Not true, it does NOT lower your risk of ruin. Here is an easy way to calculate a "best guess" estimate of your "Kelly," and that is to use f equals approximately p/2, where p is what you expect the probability of a winning trade (or period) is in the future. If you are trading N different instruments, calculate f for each instrument and divide by N. The reason for these approximations is too lengthy for me to go into in this thread. This too is false. What you call "expectancy" is the probability-weighted mean outcome. But your actual "expectation" is a function of how many trades or periods you will play for, and the derivation of what the actual "expectation" is I will not go into here (for the same reason of it being too lengthy). In short, consider a 1 in 10 chance of winning $10, and a 9 in 10 chance of losing $1. Clearly, to make this situation where you would expect profitablity, you must play for a minimum number of periods (derivation of this also not provided here). Incidentally, though this has a positive expectation in the classical (probability-weighted mean) sense, the growth-optimal fraction to wager is 0 until you have played for a sufficient number of trials. The Kelly answer would be ((B + 1)*P – 1)/B =((10+1)*.1-1)/10 =(11*.1-1)/10 = (1.1-1)/10 =.1/10 =.01 Consider the reverse, where you win $1 90% of the time, and the other 10% of the time you lose $10. Your Kelly formula answer is 0 (because your classical expectancy is negative) yet, the correct, growth-optimal fraction to bet is to bet 100% if you are going to make one play and quit. This gets lower for each "quitting point" up to a certain point (<10 plays, the derivation of the exact number not presented here) and bet nothing thereafter. Correct This is correct but the relative terms "large" and "small" may not be true -- it depends on the parameters of the outcomes (the same with the relative terms of trading a few times, and a large number of times) though, in broad strokes, it is correct.
I believe you'd be way off the mark in many cases with that over-simplified approach to Kelly. Consider a trading system which gives you a 90% probability of making a 1% gain, and a 10% probability of making an 8.8% loss. For this system, full Kelly is 0.227, and half Kelly is 0.1135. According to your formula, you'd use: f = p / 2 = 0.9 / 2 = 0.45 That is, you'd bet almost twice of full Kelly, which is almost certain to lead to the ruin. My own formula would give me f = 0. That is, I would not trade this system at all.
Of course, you can get in trouble with it! The purpose, however, is you ONLY have an estimate of what hte futures percentage of winners is, this is your best guess for sound mathematical reasons.
In a binomially-distributed outcome, such as you describe (two possible outcomes) as the ratio of what you can win to what you can loss gets ever greater, the value for f that maximizes expected geometric growth, asymptotically (i.e. as the number of trades or periods gets ever greater) approaches p, the probability (in the future, over the time you play this game). Similarly, as the ratio of what you can win to what you an lose gets lower, approaching zero, the value for f approaches zero. Thus, f is bound between 0 and p as a function of the ratio of what you can win to what you can lose. Absent knowledge of this ratio, and given that the farther away the f you use is from what the value for f will be the greater the price you will pay is (to the power of N, the number of periods or trades), you minimize the most this difference can be by being at f = p /2. Now if you know what this ratio is going to be, you can be exact, as in your example. I would argue that in trading we can be much more certain with system over a sufficiently long period to know what the percentage of wins will be (this is generally quite consistent over "sufficiently long" periods) far more so than what we stand to win vs what we stand to lose which are far more mercurial amounts into the future.
This is not quite right. The upper bound for f is not 1, but infinity. Here is a simple illustration. Let's say we have a trading system with a 90% probability of a gain of 1%, and a 10% probability of a loss of 1%. The full Kelly is 80. That is, you should borrow 79 times of what you have, and bet that amount on every trade. Now, you may want to be conservative with, say, 1/4 Kelly. That would still prescribe the leverage of 20:1. With your formula, you would bet only 0.9 of your account, which would be way too sub-optimal.
Kelly is what is not bound on the right at 1. The actual growth optimal fraction, the Optimal f calculation, IS bound on the right at 1. (The formula I present in this thread is for an asymptotic best guess when you don't know how the wins and loss amounts will come in. In truth, the Optimal f, the expected growth optimal fraction, is a function of how long you will be in the game).
Thank you for your comments. I have to think about what you said carefully to understand the implications. Regards,