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Ernest Chan (blog) writes in his book Quantitative Trading - How to Build Your Own Algorithmic Trading Business:
Here is a little puzzle that may stymie many a professional trader. Suppose a certain stock exhibits a true (geometric) random walk, by which I mean there is a 50-50 chance that the stock is going up 1 percent or down 1 percent every [day]. If you buy this stock, are you most likely--in the long run and ignoring financing costs--to make money, lose money, or be flat?
Most traders will blur out the answer "Flat!," and that is wrong. The correct answer is that you will lose money, at the rate of 0.005 percent (or 0.5 basis points) every [day]! This is because for a geometric random walk, the average compounded rate of return is not the return μ, but is g = μ - σ^2/2.
For this very reason, geometric Brownian Motion is often written as
bmwhere "μ - σ^2/2" is the expected return and "μ" is the return of the expected prices, i.e. ln(E[St]/E[St-1]).

If you lose 50 percent of your portfolio, you have to make 100 percent to get back to even... that's what everybody knows. But it's also interesting to see how mild volatiltiy hurts over time. Here are ten (random - no cherry picking) realizations of a geometric Brownian Motion with a daily volatility of 1% (i.e. a yearly volatility of 16% when having 252 trading days) over the period of 100 years:
bm02
bm01
bm03
bm04
bm05
bm06
bm07
bm08
bm09
bm10
R Development Core Team (2008). R: A language and environment for statistical computing.

PS: Hey, this looks promising:
veryprom
OneEyedMan (guest) meinte am 26. Dec, 16:17:
What you described isn't Brownian motion
Because increments of all scales are not normally distributed. The distribution you described has the exact geometric mean of 0.
ln (1+g) = 1/2 ln (.99)+1/2 ln (1.01)
->2 ln (1+g) = ln (.99 * 1.01) = ln(1) = 0 -> g =0 
Mahalanobis antwortete am 26. Dec, 19:28:
Re:
First, we are talking here about geometric Brownian motion. If a stockprice follows geometric Brownian motion, its returns are normally distributed but the increments (the absolute changes in the stock price) are log-normally distributed. Chan used a binomial model but it can be shown that the distribution of a stock that follows a binomial model converges to the distribution of a stock that follows a geometric Brownian motion. For my simulation, I drew from a normal distribution.

Second, you made a computational error: 0.99*1.01 does not equal one, it's (1-0.01^2).

S(t) = S(0)*Z(1)*Z(2)*Z(3)...Z(n)
log(S(t)/S(0)) = log(Z(1))+log(Z(2))+...+log(Z(n))

E(log(Z(s))) = 0.5*log(0.99) + 0.5*log(1.01)
E(log(Z(s))) = 0.5*(log(0.99*1.01))
E(log(Z(s))) = 0.5*(log(0.9999))

Since log(0.9999) < 0, S(t) goes to minus infinity. QED

NB: log() = ln() 
OneEyedMan (guest) antwortete am 27. Dec, 23:19:
Woops
Point taken on the arithmetic error. My mistake and since my point turned on it, it was a big mistake.

Is Chan's point just that you shouldn't confuse the arithmetic and geometric mean return? That's a critical fact that all asset managers should know.

Is he making a point about the second order aproximation to the geometric mean using mean and variance? If so, I don't get it. Because why would you use the approximation when the exact answer is so readily available. 
Mahalanobis antwortete am 28. Dec, 10:44:
Well,
it's not only about ra ≥ rg, but also about the extent of the difference.

I think these days most asset managers look at VAMI charts and compounded rates of return since that is what their software (e.g. PerTrac) gives them anyway. I worked as a Hedge Fund Analyst until recently (Woops) and the track record of most funds was somewhere between 2 and 5 years (monthly data). And especially when having such short time series, one shouldn't assume μ and σ fixed. It's more about how much those parameters can vary given certain environments. (And in case the funds do mark-to-model, you can forget most of the statistical analysis anyway ;-D).

Chan's book is more about "How to build your own algorithmic trading business" than on "quantitative trading". Take a look at the table of contents (Amazon). For somebody who comes straight out of University and has no practical experience, it is probably a good read and the price is really ok too (< $40). 
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nickgogerty (guest) meinte am 28. Jan, 21:45:
FYI it is called Siegels paradaox
http://mathworld.wolfram.com/SiegelsParadox.html

nice demonstration. 
Mahalanobis antwortete am 29. Jan, 09:24:
Yup,
I've actually blogged about it already. 
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