1. If you have dollars in your pocket, you are long usd/eur. If you have euros in the pocket, you are long eur/usd. You can hold them simultaneously. So the answer is yes. 2. 1/(eur/usd) = usd/eur by definition. Therefore forex traders don't need quotes for usd/eur, eur/usd is enough. Central banks print money so fast that it's impossible to profit enough using this system with fiat currencies. But it could be used for gold, silver, stocks etc. The eur/usd is just the best example which I can give to explain the basic idea. If you believe in efficient markets, it's impossible to determine when to optimally exit. This is because according to the theory in every single moment every single price reflects all the available information. Therefore no moment is better than the other to enter or exit.
The answer here is so simple it took me hours to see it. This is a simple Martingale Mean Reversion strategy. In the EURUSD example, note that the exchange rate started at 1:1 and came back to 1:1. What would happen if it fell, then stayed there? Meaning if it did NOT revert to mean? You would be underwater, just like any MR strat. Example: Let's pretend there is a significant shift of value from Europe to USA and the rate only plummetted. start at 1:1 = $500 & â¬500 falls to .5:1; your account still shows $500 and â¬500, but the â¬500 is only worth $250, so your account value is $750. You rebalance to $375 and $375 worth of euro. New account contains $375 and â¬750. falls again to .25:1; your account shows $375 and â¬750, but now your euro are only worth $187.50 for a total account value of $562.50. You rebalance again to $281.25 worth of dollars and $281.25 worth of euro, or â¬1125. falls yet again to .125:1; your account shows $281.25 in USD and â¬1125 in euro, but all those euro are only worth $140.625 in USD. Let's say it stays at this rate. Well, you've lost money. $859.375 to be exact. So the Martingale part is each time a trade goes against you, you increase your bet size. And the Mean Reversion part is that when the underlying moves in one direction, and you make a bet hoping it will come back in the other direction. You make money in the author's gold/silver example because all of the little mean reversions add up to more than the major shift from start to finish. Riskless? Of course not. Your risk is that the underlying moves against you and never comes back, and the wins you find in the jitter of the small moves along the way don't make up for the major move against you.
Black Sholes assumes that the periodic returns are normally distributed, where the return for the i'th period is r_i = log(S_{i+1} / S_i) = log(S_{i+1}) - log(S_i). Thus the differences of the log price for successive periods are normally distributed, which means that the log price undergoes a simple random walk.
if I remember my stochastic calculus classes correctly, the 2 random variables result from the fact that 2 different numeraires are used, something you already pointed out. he used eur/usd and usd/eur rather than the logs.
I have read your article in its entirety. Your assertion that the odds are in your favor (2:1 with equal probability) seem unrealistic. Everybody can make tons of money when you flip a fair coin and you get paid twice as you lose. Besides, if you are so sure your investment is truly risk-free (p = 1), your Kelly ratio should be 1; i.e., you should bet all, and guess what, buy and hold.
After reading your original ideas a bit more, I see that we are on two different wavelengths. You did accidentally stumble on something very useful, but I don't think it's what you intended. Also, I don't believe buying the underlyings in ratios is equivalent to literally buying the two instruments I outlined. Anyone trading forex related instruments have a more concrete answer? Can you purchase one contract/share of eur/usd and simultaneously one contract/share of usd/eur and do the two move exactly 1/x. If that is the case, you were on to something.
I tried solving this and i get the same idea as the above quote. can someone please explain to me how the above does not disprove this whole thing?
You focus on only one possible outcome - eur/usd only falls and stays at some level. You draw the conclusion that the system would lose. But you calculate the losses in dollars. And you have errors in the calculations. This is how they should be: 1. Start at eur/usd =1, we buy 500 euros and 500 dollars. 2. eur/usd = 0.5, we have $750. We buy euros with half of our dollars, and keep the other half, i.e. we buy 750 euros and keep 375 dollars. 3. eur/usd = 0.25, we have $562.5. We buy euros with half of our dollars, and keep the other half, i.e. we buy 1125 euros and keep 281,25 dollars. 4. eur/usd = 0.125, we have $421.875. You came up with $140.625 because of erroneous calculations. But why don't we calculate the balance in euros and see what happens? The account started from 1000 euros and finished with 421.875*8 = 3375. So the system actually made profit Another thing - let's consider what other possibilities did you have at the starting point, when the price was 1. You could have chosen to buy only euros or only dollars. If the markets are efficient and the price follows a random walk, you have no reason to prefer holding dollars instead of holding euros. So let's say you decided to hold euros. Then eur/usd falls to 0,125. You have only 125 dollars, which is significantly worse than if you followed my system. The loss of my system is lower than if you bought and held, this is why I use the term "riskless". As I wrote already, the "risk" of the system as you define it is lower when calculated in both euros and dollars. Of course at some points in time the system might be worse off than just buying and holding something, but it is only when the balance is calculated in one of the currencies. When calculated in the other it will be profitable. If markets are efficient and the price follows a random walk, we can't predict what exactly to buy and hold, so the system is always optimal.
It might seem unrealistic, but it is true. If the price follows a random walk, the probability of an up movement to x is equal to a probability of a down movement to 1/x. Ok, now think about this example: I offer you a game, in which we flip a fair coin. You pay me some amount of money to participate in each flip. If it comes up heads, I give you back twice your bet, and if it comes up tails, I keep half of your bet. How much will you bet each time? For example: You have $10. You believe that the Kelly ratio should be 1 and bet all - you pay $10 for the first flip. It comes up heads and I pay you back $20. Then you bet them all, the coin comes up tails and I keep half of your bet, so you are left again with $10. In this situation the Kelly formula gives f* = 0.25, therefore you should risk $2.5 in the first coin flip. The optimal bet is to bet half of your money each time, and the same logic is used in my article. For example: You have $10. You know that the Kelly ratio is to bet half of your money and pay $5 for the first flip. It comes up tails and I keep $2.5. Then you bet half of your $7.5 = 3.75, the coin comes up tails and I pay you back twice your bet, so you are left with $11.25.
Here should be written "the coin comes up heads" Another remark - the sequence of heads and tails doesn't matter, the end result is always the same. This is a property of the Kelly bet.