[sixhigh]

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This seems counterintuitive - it's like we are beating the perfectly efficient market. Thoughts?

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Jason basically gave the answer. Let me state my thoughts anyway. If I understand you correctly, you mean the following strategy 'beats' an efficient market:

We buy 100€ today and wait until the exchange rate hits a certain level l greater than one and sell the money. For simplicity we assume the interest rate r is zero, the initial exchange rate l_0=1 and we sell at l=2. We can describe the exchange rate as a fair random walk l_n and thus know it will hit l=2 in some period with probability 1. Once this point is reached we will have doubled our money and we know this will be reached with certainty.

Yes, this strategy beats the market. But it needs one assumption: infinite time. The problem here is, we cannot give a period n, where the exchangerate will have hit l=2 with probability 1. For every n (smaller than infinity) there will be some probability that the exchange rate never hit l=2 before.

Thus there will be no period in which we have doubled our money with certainty and we don't beat the market. I.e. this strategy doesn't violate the arbitrage equilibrium of the market.

But this doesn't mean it has no positive expectation. Let's take a simple example: We play the exchange rate game for 4 periods. The initial exchange rate is 1 and every period it halves or doubles with probability 1/2. We buy 100€ before the first period for $100 and whenever the exchange rate hits 2 we sell and quit the game. If it hasn't reached 2 in the 4th period we sell for whatever the exchange rate is.

The probability of the exchange rate hitting 2 before the 4th period is (1/2+1/8). With prob. 1/2 it hits it in the first period (up), with prob. 1/8 in the third (down, up, up). The remaining cases are 1 (p=1/8), 1/4 (p=3/16), 1/16 p=(1/16) in the 4th period.

So the EV for this game is

5/8 * 200$ + 1/8 * 100$ + 3/16 * 25$ + 1/16 * 6$ = 142$.

Weird ... [img]/images/graemlins/smile.gif[/img]