# Risk neutral measure in the binomial approximation of geometric Brownian motion

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?