# Geometric Brownian Motion as the limit of a Binomial Tree?

Consider the price of a stock whose drift and volatility parameters are $$\mu, \sigma$$ respectively, over the time interval $$[0, t]$$. Suppose we use an $$n$$-stage binomial tree to model the price dynamics with up and down factors $$u=e^{\mu\Delta t+\sigma\sqrt{\Delta t}}, \qquad d=e^{\mu\Delta t-\sigma\sqrt{\Delta t}}$$ where $$\Delta t=t/n.$$ Following the Ross, we can argue that the probability of an up move is $$p=\dfrac12\left(1+\dfrac{\mu}{\sigma}\sqrt{\Delta t}\right).$$

Price at the end is given by $$S_t=S_0u^jd^{n-j}=S_0 e^{\mu t+2\left(\frac{j-n/2}{\sqrt{n}}\right)\sigma\sqrt{ t}},$$ where $$j$$ is the number of up moves during the process. Since $$j$$ has a binomial distribution, for a very large (but fixed) $$n$$, by the CLT we can argue that $$j=np+\sqrt{np(1-p)}z$$ where $$z\sim N(0, 1).$$ Then $$\frac{2j-n}{\sqrt{n}}=(2p-1)\sqrt{n}+2\sqrt{p(1-p)}\sigma z$$ and it implies that $$S_t=S_0 e^{2\mu t+2\sqrt{p(1-p)t}\sigma z}\to S_0 e^{2\mu t+\sigma\sqrt{t}z}.$$ However, I was hoping that this would converge to the classical model $$S_0 e^{(\mu-\sigma^2/2) t+\sigma\sqrt{t}z}.$$ What is the mistake, if any, I made?
How can I derive this last model using the initial setup?

• @JanStuller: For $n$ values $\Delta t$ approach to zero. Commented Jun 11 at 3:09
• @JanStuller: since the term inside the square root is positive it should not be a problem. As long as negative numbers are not involved, by algebra (mathematics) of limits, limit of a square root is the square root of the limit. Commented Jun 11 at 7:29
• @JanStuller: Also, because of the limit $\Delta t\to 0,$ the choice $p=1/2$ actually makes sense. Commented Jun 11 at 12:00
• I think what you are looking for is a version of Donsker's Theorem for GBM. Donsker's theorem gives conditions under which a discreet Random walk converges in Distribution to the Wiener process. However, a similar theorem does not exist for GBM. So what you are looking for does not exist. I think the answer as to why it doesn't exist is Ito's Lemma: Ito's formula produces the term "$-0.5\sigma^2 t$" as a result of $dW_t^2$ converging to $dt$: a similar result does not exist for a discrete process so you cannot obtain it from a binomial tree. Commented Jun 18 at 9:30

We can show that the moments of the Binomial tree agree with the moments of the continuous model for the case where we pick symmetrical probability value $$p=0.5$$.

I will change the notation slightly (you will see later on why: I deliberately use $$\eta$$ instead of $$\mu$$):

$$u=e^{\eta\Delta t+\sigma\sqrt{\Delta t}}, \qquad d=e^{\eta\Delta t-\sigma\sqrt{\Delta t}}$$

Also, let's introduce the following notation: the maturity we are interested in will be denoted $$T$$ (instead of $$t$$), and $$\Delta t$$ can then be expressed as $$\frac{T}{n}$$, where $$n$$ is the number of steps on the tree.

So the dynamics for $$S_t$$ is given as:

$$S_T=S_0u^jd^{n-j}$$

Taking the log makes things a lot easier:

$$\ln{\left(\frac{S_T}{S_0}\right)}=j\ln(u)+(n-j)\ln(d)=\\=j\left(\ln(u)-\ln(d)\right)+n\ln(d)=\\=2j\sigma\sqrt{\Delta t}+n\eta\Delta t-n\sigma\sqrt{\Delta t}=\\=j\left(2\sigma\sqrt{\frac{T}{n}}\right)+n\left(\eta\frac{T}{n}-\sqrt{\frac{T}{n}}\sigma\right)$$

We know that $$j\sim Bin(n,p)$$, so the moments are:

\begin{align*} \tag{1} \mathbb{E}\left[\ln{\left(\frac{S_T}{S_0}\right)}\right]=\left(2\sigma\sqrt{\frac{T}{n}}\right)\mathbb{E}\left[j\right]+n\left(\eta\frac{T}{n}-\sqrt{\frac{T}{n}}\sigma\right)=\\=\left(2\sigma\sqrt{\frac{T}{n}}\right)(np)+n\left(\eta\frac{T}{n}-\sqrt{\frac{T}{n}}\sigma\right)=\\=2p(\sigma\sqrt{T}\sqrt{n})-\sigma\sqrt{T}\sqrt{n}+\eta T=\\=(2p-1)(\sigma\sqrt{T}\sqrt{n})+\eta T \end{align*}

The variance is: \begin{align*} \tag{2} V\left(\ln{\left(\frac{S_T}{S_0}\right)}\right)=V\left(j2\sigma\sqrt{\frac{T}{n}}\right)=\\=V(j)\left(4\sigma^2\frac{T}{n}\right)\\=np(1-p)\left(4\sigma^2\frac{T}{n}\right)=\\=p(1-p)4T\sigma^2 \end{align*}

If $$p=0.5$$, then the moments above are:

$$\mathbb{E}\left[\ln{\left(\frac{S_T}{S_0}\right)}\right]=\eta T \qquad V\left(\ln{\left(\frac{S_T}{S_0}\right)}\right)=T\sigma^2$$

So if we pick $$\eta:=\mu-0.5\sigma^2$$, then the moments of the binomial tree model agree with the continuous model as long as $$p=0.5$$.

(the beauty is that if we pick $$p=0.5$$ the moments agree with the continuous model for any $$n$$).

So with the set-up above, for a (smallish) finite number of steps "$$n$$" in the Binomial tree model, the random variable $$\ln{\left(\frac{S_T}{S_0}\right)}$$ will follow a Binomial distribution, but its mean and variance will agree with the continuous GBM model.

What about convergence of the Binomial model to the GBM model for large $$n$$? Here we can show convergence in distribution:

Using CLT, we can say that for large $$n$$:

$$j\xrightarrow{d}N(np,\sqrt{np(1-p)})$$

So (using the results (1) and (2) ) we can say that for large $$n$$:

$$$$\boxed{\left( \frac{S_T}{S_0} \right)\xrightarrow{d}N\left(\sqrt{Tn}\sigma(2p-1)+\eta T,p(1-p)4T\sigma^2\right)}$$$$

Again, this will agree with the continuous model as long as $$p=0.5$$ and $$\eta=\mu-0.5\sigma^2$$.

# Summary:

• For any finite ("small") number of steps $$n$$ in the binomial tree model , the moments will agree with the continuous model as long as $$p=0.5$$ and $$\eta = \mu-0.5\sigma^2$$ (and under this set-up the variable $$\ln\left(\frac{S_T}{S_0}\right)$$ will be distributed Binomially with these moments)
• If, in addition, $$n$$ becomes large, the distribution of the variable $$\ln\left(\frac{S_T}{S_0}\right)$$ will converge from Binomial to Normal (with the same moments)

Final note: above we looked at the case where $$\mu$$ is the historical drift. This gives us the freedom to chose $$p=0.5$$. If instead we considered the risk-neutral model with $$\eta=r-0.5\sigma^2$$, then the parameter $$p$$ is given by $$p:=\frac{e^{r\frac{T}{n}-d}}{u-d}$$ and it's a lot more difficult to show the convergence.

Working line by line I get:

$$S_t=S_0e^{j\left(2\sigma\sqrt{\Delta t}\right)+n\left(\mu\Delta t-\sqrt{\Delta t}\sigma\right)}=\\=S_0\exp{\left(\mu t+2\left(j\sigma \sqrt{\Delta t} \right) - n\sigma\sqrt{\Delta t}\right)}=\\=S_0\exp{\left(\mu t+\sigma \sqrt{\Delta t}\left(2j - n \right) \right)}$$

Substituting for $$j=np+\sqrt{np(1-p)}z$$, we get:

$$S_0\exp{\left(\mu t+\sigma \sqrt{\Delta t}\left(2np+2\sqrt{np(1-p)}z - n \right) \right)}$$

Now evaluating the terms, we get:

$$np=n(0.5+0.5\frac{\mu}{\sigma}\sqrt{\Delta t})=0.5n+0.5\frac{\mu}{\sigma}\frac{t}{\sqrt{\Delta t}}$$

Now the other term:

$$2\sqrt{np(1-p)}z=2\left(np-np^2\right)^{\frac{1}{2}}z=\\=2\left(0.5n+0.5\frac{\mu}{\sigma}\frac{t}{\sqrt{\Delta t}}-n(0.5 -0.5\frac{\mu}{\sigma}\sqrt{\Delta t})^2\right)^{\frac{1}{2}}z=\\=2\left(0.5n+0.5\frac{\mu}{\sigma}\frac{t}{\sqrt{\Delta t}}-n(0.25-0.5\frac{\mu}{\sigma}\sqrt{\Delta t}+0.25\frac{\mu^2}{\sigma^2}\Delta t)\right)^{\frac{1}{2}}z=\\=2\left(0.5n+0.5\frac{\mu}{\sigma}\frac{t}{\sqrt{\Delta t}}-0.25n-0.5\frac{\mu}{\sigma}\frac{t}{\sqrt{\Delta t}}+0.25\frac{\mu^2}{\sigma^2}t\right)^{\frac{1}{2}}z=\\=2\left(0.25n+0.25\frac{\mu^2}{\sigma^2}t\right)^{\frac{1}{2}}z$$

Plugging it all back (using $$\sigma \sqrt{\Delta t}*2np=\mu t)$$:

$$S_t=S_0\exp{\left(2\mu t+2\sigma \sqrt{\Delta t}\left(0.25n+0.25\frac{\mu^2}{\sigma^2}t\right)^{\frac{1}{2}}z\right)}=\\=S_0\exp{\left(2\mu t+2\sigma \sqrt{\Delta t} \left(0.25n+0.25\frac{\mu^2}{\sigma^2}n \Delta t\right)^{\frac{1}{2}}z\right)}=\\=S_0\exp{\left(2\mu t+\sigma \sqrt{t} \left(1+\frac{\mu^2}{\sigma^2} \Delta t\right)^{\frac{1}{2}}z\right)}$$

To replicate your result fully, we must show that:

$$\lim_{n\to\infty}\left(1+\frac{\mu^2}{\sigma^2} \Delta t\right)^{\frac{1}{2}}=0$$

• Thank you for your detailed response. I was under the impression that we can derive the continuous model from the binomial model as the limit in this way. Suppose we do not know the parameters of the continuous model. Then how can we justify the reason for $\eta=\mu-0.5\sigma^2$? Commented Jun 10 at 14:11
• I guess that's the advantage of Stochastic Calculus. It gives the term $-0.5 \sigma^2$ via Ito's formula. Starting from the Binomial tree, it is clearly difficult to re-create the Stochastic calculus result just via convergence arguments, unless we make the substitution for $\eta$ (which - as you point out - defeats the purpose, because we make that substitution to match the continuous result, as opposed to the continuous result being a natural outcome of the convergence argument). Commented Jun 10 at 16:31
• The substitute $\eta$ is very interesting. It doesn't feels like just a mathematical gymnastic. It should have some meaning. Commented Jun 11 at 7:35
• @Bumblebee I am in the procrss of writing a paper on the meaning of that term. When I finish it, I will link it here :) Commented Jun 11 at 8:34

First, the factors $$u$$ and $$d$$ in the binomial tree are given by $$u=e^{\sigma\sqrt{\Delta t}},d=e^{-\sigma\sqrt{\Delta t}}.$$ Using the Taylor expansion till terms of order $$\Delta t$$ gives for example for $$u$$ $$u \approx 1 - \sigma\sqrt{\Delta t} + \frac{1}{2}\sigma^2 \Delta t.$$ Second, the risk-neutral probability is $$p = \frac{e^{\mu\Delta t} - d}{u-d} \approx \frac{(1 + \mu\Delta t) - (1 - \sigma\sqrt{\Delta t} + \frac{1}{2}\sigma^2 \Delta t)} {(1 + \sigma\sqrt{\Delta t} + \frac{1}{2}\sigma^2 \Delta t) - (1 - \sigma\sqrt{\Delta t} + \frac{1}{2}\sigma^2 \Delta t)} = \frac{\mu\Delta t + \sigma\sqrt{\Delta t} - \frac{1}{2}\sigma^2 \Delta t} {2\sigma\sqrt{\Delta t}}\\ p \approx \frac{1}{2}\left(1 + (\mu - \frac{1}{2}\sigma^2)\frac{\sqrt{\Delta t}}{\sigma}\right).$$ Further, we can follow your steps $$S_t=S_0u^jd^{n−j}=S_0e^{j\sigma\sqrt{\Delta t}}e^{-(n-j)\sigma\sqrt{\Delta t}} = S_0 e^{\frac{2j-n}{\sqrt{n}}\sigma\sqrt{t}}.$$ Variable $$j\sim B(n,p)$$ has the binomial distribution with mean $$np$$ and variance $$np(1-p)$$. We approximate it by a normal variable $$Z\sim N(np, np(1-p))$$. Or using a standard normal variable $$z\sim N(0,1)$$ it can be put as $$j = np + \sqrt{np(1-p)}z.$$ The term in the exponent for $$S_t$$ becomes $$\frac{2j-n}{\sqrt{n}}\sigma\sqrt{t} = \frac{(2p-1)n + 2\sqrt{np(1-p)}z}{\sqrt{n}}\sigma\sqrt{t} = \frac{(\mu - \frac{1}{2}\sigma^2)\frac{\sqrt{\Delta t}}{\sigma}n + 2\sqrt{np(1-p)}z}{\sqrt{n}}\sigma\sqrt{t} = (\mu - \frac{1}{2}\sigma^2)t + \sigma\sqrt{t}z$$ and $$S_t = S_0e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma\sqrt{t}z}.$$

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• I have seen this derivation with risk-neutral probability before. What I'm asking is how/why we cannot get the same geometric Brownian motion with the natural probability $p=\dfrac12\left(1+\dfrac{\mu}{\sigma}\sqrt{\Delta t}\right).$ Commented yesterday
• @Bumblebee Why do you call this expression for p a natural probability? This is just a simplification of a more correct formula. Maybe you can provide a reference where the equations for p, u, and d are coming from, as they are quite non-standard. They lead to clear errors, like 2mu in the exponent. With the given formulas, it is not possible to match either the first or the second moment of spot at time t+1. If you look in Luenberger "Investment science", you will see p like you gave, but then it is corrected to a more proper value. Commented yesterday
• In my mind, these formulas for up/down factors take the time value of money into account, hence more accurate than the ones given in the CRR model. Secondly, this natural probability is not a simplification of risk-neutral probability. It is a different probability that does not come from "arbitrage-free" assumption. Commented yesterday