Why do these things always have to become pissing contests or a game of who's bigger? :eek: Here is something worth discussing whether it be right or wrong because it will give others an opportunity to learn. Don't attack the person, instead, discuss the concept, equation etc. Thank you. Joe.
Firstly, sambian, it's a process... I started with a somewhat superficial understanding of what the problem with your calculation was, but, in the end, with the benefit of collective wisdom, I think I can formulate the issue more robustly. Secondly, if you actually consider what I am saying you will realize that I have been saying the same thing, just in a variety of ways. Thus, I claim that your points 1 and 2, where you accuse me of changing my message, are actually me saying the exact same thing. Now, as to your point 3, I tried to expand on what I said initially and illustrate it specifically using the definition of "random walk". In doing so I assumed, erroneously, that when you stated that "eur/usd" follows a "random walk", you meant a random walk, as I described it (SORW). You didn't specify any constraints on the distribution of prices of "eur/usd", like the B&S paper does, which is what led me astray. So I apologize for trying to reject your assumption by using a definition of random walk that was not broad enough. Finally, the discussion we've had clarified things for me and I can now formulate what I see as the "error" in your calculations more accurately and succinctly. Do you want me to do that? Then we can discuss whether my conclusions are incorrect.
there was a trader with the handle electric savant who tried this on here with real money. If I recall he was not successful. I only took a year of stats in college 20 years ago. But isn't his premise just like telling my 5 year old I have 11 fingers. I show both hands and I count 10, 9, 8 , 7 ,6, then I show the other hand and say plus 5 equal 11. That sounds like this type of positive expectancy. if you lose half you money you have to earn 100% to make it back. There is no positive expectancy in such a system. You make 100% in one account than you have to make 100% in the other to be even. You can move you symbols around anyway you like but there is no arguing with the fact that 1 dollar is 100% more than 50 cents. And that if you have two dollars and you lose 50% you are back to 1 dollar.
In my view, the problem is as follows, sambian. You have to be consistent between your distribution assumption and the method of calculating the expected return. So you can use either, but not both, of the two following cases: 1) Assume the price of eur/usd is normally distributed, in which case the two branches of your binomial tree will be eurusd_up = eurusd0 + epsilon and eurusd_down = eurusd0 - epsilon. In this case, your expected return can be calculated the way you do it, i.e. as E1 = P[up] * (eurusd_up - eurusd0) + P[down] * (eurusd_down - eurusd0) 2) Assume the price of eur/usd is log-normally distributed, in which case the two branches of your binomial tree will be eurusd_up = eurusd0 * epsilon and eurusd_down = eurusd0 / epsilon, which is what, effectively, you have used. In this case, expected return should be calculated as E2 = P[up] * log(eurusd_up / eurusd0) + P[down] * log(eurusd_down / eurusd0). Note that in both cases, assuming P[up] = P[down] = 0.5 implies that E1 = E2 = 0. There is no positive expectancy. What you're doing is taking the distribution assumption from case 2, but using the expected return calculation from case 1 with it. That's the fallacy, IMHO.
Risk structures the Markets, without risk you end up with nothing... or even worse given commissions and slippage.
IMHO, your newly created formulas for calculating expected values should be in the schoolbooks. You calculate expected values for trading dollars by multiplying logarithms of prices, instead of the prices themselves. Can you answer why? When you buy eur/usd and usd/eur, what do you expect to get, dollars/euros or logarithms? If you believe in your calculations, I can offer you a deal. We will throw a coin and every time it comes up heads, you pay me $2, while evey time it comes up tails I pay you $0,5. You can multiply the probabilities of heads and tails (0,5) by the logarithms of the dollars which we bet and you will get expected value of zero. If using logarithms of the outcomes is the right way to calculate expected values, prove it by accepting my offer
I must be missing something major here. Let me ask a simple question to any forex traders (I don't trade forex much). 1) Is it possible to purchase eur/usd and usd/eur instruments and simultaneously hold both? 2) Do they track very close to 1/X? Because, ignoring all the math arguments for a moment, I ran sims on both RW and the actual series (from onada). And the results are pretty much always winning. So, assuming you can buy perfectly ratioed 1/x instruments and the underlying instruments move rw in gbm sense (even with different drift/var components on underlyings), then yes, this seems correct (although a neg fat tail on either underlying would smoke you). The only downside on the actual onada data is that the effect of actual % daily moves are so small that it takes a very long time to accumulate the total upside without leveraging. And also, it would be difficult to determine when to optimally exit. I'd love to see an argument, simulated or back-tested example where this fails. I didn't get into the portfolio re-balancing part, but just the case of purchasing equal portions of each instrument (if that's possible). It definitely looks too good to be true, but I'd sure like to see a valid hole (impossibility of 1 and 2 would suffice).