# Last edited on 2014-04-26 22:50:07 by stolfilocal

[quote author=sleger link=topic=178336.msg6391939#msg6391939 date=1398437091]
To come back to yours, I think the error is in the first sentence: "the difference between successive values of Z(i) = log(P(i)) are independent random variables with probability distributions that are symmetric about zero"
Small proof: under log brownian (with no drift) the important basic concept is that the best expectation of price in the future is the current value of the price. 
[/quote]
OK, it seems that I was using a definition of "log Brownian" that is not the standard one used in finance.

Indeed I was assuming that the increments D(i) = Z(i+1)-Z(i)  = log(P(i+1)/P(i)) were normal variables with [b]zero[/b] mean, so that Z(i) would be a Brownian variable as it is usually defined in other areas - with no trend.  (Note that I am a prof of [i]computer science[/i], not economics!)

However, as you point out, by that definition the expected price E(P(i0+n)) would grow exponentially with n; which does not make sense in the trading context, where the "efficient market hypothesis" demands E(P(i0+n)) = P(i0).  (Or does it? See below.)

We agree at least that in a log-Brownian model the increments D(i) = Z(i+1)-Z(i)  = log(P(i+1)/P(i)) should be assumed to be independent random variables, yes? 

The standard way to achieve E(P(i0+n)) = P(i0), in finance, seems to be: assume that the increments D(i) are Gaussian variables with slightly negative mean, mu = -sigma^2/2.  That is, one assumes a slight negative trend in the log-price Z(i) so that the broadening of the log-normal distribution of P(i0+n) as n increases preserves the mean P(i0).   Is that correct?

That assumption satisfies the "efficient market hypothesis", but implies (as you pointed out) that the price is slightly more likely to go down than to go up at each step.  Then Prob(P(i0+n) < P(i0)) increases with with the stride n.  Which seems weird too.

We can get rid of this weidness by assuming a probability distribution for D(i) such that E(D(i)) = 0, E(exp(D(i))) = 1.  It seems that these two conditions cannot be obtained with a Gaussian distribution, except in the limit when sigma → 0.  However, they can be achieved with other distributions that are symmetric about zero, especially if they have fatter tails than the Gaussian.  And, indeed, the most obvious deficiency of the log-Brownian model seems to be that, in real data, the distribution of the increments is not Gaussian.

If my math is correct, the distribution of the n-step increments too would satisfy both conditions: E(Z(i0+n)) = Z(i0), and E(P(i0+n)) = P(i0).  Moreover the distribution of Z(i0+n), being the convolution of n symmetric distributions, would be symmetric about Z(i0), implying that Prob(Z(i0+n) < Z(i0)) = 1/2, and hence Prob(P(i0+n) < P(i0)) = 1/2. 

With these assumptions, even though the distribution of P(i0+n)/P(i0) approaches a log-normal distribution as n increases (by the central limit theorem), it remains sufficiently "log-abnormal" to satisfy those conditions (which a true log-normal distribution cannot achieve, it seems). 

Perhaps you can tell me what would be a convenient "fat-tailed" [i]symmetric[/i] distribution to assume for the increments D(i) that would satisfy both conditions.  (Perhaps a mixture of Gaussians with zero mean whose variance has log-normal distribution?  I would have a justification for that choice...) 

Finally, about the "efficient market hypothesis": shouldn't it say that E(P(i0 + n)) = P(i0)*Q^n, where Q > 1 is the typical ROI factor of a generic investment per time step?  That is, if E(P(i0+n)) = X, then P(i0) should be [i]less[i] than X, otherwise other investments would be more profitable. 

With that modification to the "efficient market hypothesis", the distribution of D(i) must satisfy E(exp(D(i)) = Q, not 1; and one can achieve that even with a [i]zero-mean[/i] Gaussian if desired.  In that case one would have a legitimate log-Browninan model (with Gaussian increments) such that that E(Z(i0+n)) = Z(i0), E(P(i0+n)) = P(i0)*Q^n, and Prob(P(i0+n) < P(i0)) = 1/2.  Does this make sense?