I do not know what the price will be but the log brownian model does NOT say that (I am just repeating myself here) ...
Look at a call closed formula price, N(d2) represents the probability of the call being in the money, if you put S=K (our example here) you are left with N(something negative) which means that the probability that the price ends up above current value is ... less than 50% !
And below current value is ... more than 50%.
I don't understand your notation,
here is mineBasically, in the log-Brownian model the difference between successive values of Z(i) = log(P(i)) are independent random variables with probability distributions that are symmetric about zero. Therefore after any number n of steps the probability distribution of Z(i+n) will be symmstric about the starting value Z(i). That means Z(i+n) wil be less than Z(i) with 50% probability. Since log is monotonic, it preserves cumulative probabilities, therefore P(i+n) will be less than P(i) with 50% probability. What is wrong with this argument?
(Strictly speaking, "Brownian" requires a normal distribution of increments with zero mean and fixed variance. In practice the variance varies slowly and the distributions have fatter tails than the norma; but by the law of large numbers they become near-normal for large n.)
My notation is the standard financial notation as found under wikipedia black-scholes article (S=spot price, K=strike ...). So now you have my notation please take a look at my argument again.
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. Your hypothesis does not respect that.
For instance let's say Pi=exp(1) so Z(i)=1, now we simulate a 1 step tree with 50% chance of going up 0.1 and 50% chance of going down 0.1 to simplify (doesnt change the result), and we compute Expectation[P(i+1)] which should be equal to exp(1). Well, Expectation[P(i+1)]=0.5*exp(1.1)+0.5*exp(0.9) which is not exp(1).
Also you can ask any bank quant, under simple log brownian mode, an at the money binary option (pays 1 if stock is above, 0 below) its theoretical price is strictly less than 0.5, which is exactly the probability of ending above current value, which is also exactly the value of the N(d2) I mentioned in my post before.
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