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Suppose an asset is described by geometric Brownian motion with a drift, i.e.

$$dS_t = S_t\mu dt + S_t \sigma dW_t$$ for a Wiener process $W_t$ and $S_0=1$. By some consequence of Girsanov's theorem (which I don't understand completely yet) we can conclude that there is a unique risk free measure under which

$$dS_t = S_t\sigma d\tilde{W}_t$$ for some Wiener process $\tilde{W}_t$ is the risk free measure. In particular, the $\sigma$ in both equations is the same.

I tried to get an intuition for this theorem by using a binomial approximation. Choose some large $N \in \mathbb{N}$ and $a,b \in \mathbb{R}$ and let $(A_n)_n$ be a sequence of i.i.d. random variables with $\mathbb{P}(A_n=a)=p$ and $\mathbb{P}(A_n=b)=1-p$. Then the process defined by $$S_{n/N} = A_nA_{n-1}...A_1$$ approximates our original process in some sense (I don't understand yet how to make this statement precise). Since

$$S_{(n+1)/N}-S_{n/N}=(A_{n+1}-1)S_{n/N}$$

it approximates geometric Brownian motion with $$\mu/N = \mathbb{E}[A_{n+1}-1] = p(a-1)+(1-p)(b-1)$$ and $$\sigma^2/N = \mathrm{Var}[A_{n+1}-1]=pa(a-1)+(1-p)b(b-1)$$

Now I found the risk free measure $\mathbb{Q}$ of this process. It is easy to see that in this risk free measure the $A_n$ are also i.i.d. with $\mathbb{Q}(A_n=a)=q$ and $\mathbb{Q}(A_n=b)=1-q$ and

$$\mathbb{E}_\mathbb{Q}[A_n-1]=q(a-1)+(1-q)(b-1)=0$$

But the volatility in the risk neutral measure is given by $$\tilde{\sigma}^2/N = qa(1-a)+(1-q)b(1-b)$$

which is not necessarily the same as $\sigma^2/N$.

This seems to contradict the statement for true unapproximated geometric Brownian motion. So where is my mistake? Is maybe the approximation of Brownian motion with a binomial model more subtle than I think?

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