Geometric Brownian Motion as the limit of a Binomial Tree?

Question

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?

Answer 1

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 $n$:

\begin{equation} \boxed{\ln{\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)} \end{equation}

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.

Replicating your results:

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:

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

Answer 2

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 $$ S_t = S_0e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma\sqrt{t}z}. $$

Real measure (additional considerations) The discussion above is related to the risk-neutral measure. Similar, results can be obtained for the real measure.

In case of the real probabilities it is important to clearly define the starting point, whether we consider log-returns or relative returns.

Log returns If we start from the definition that we want to equate log returns with the historically observed returns both in the expectation and the variance, i.e. $$ \mathbb{E}[\ln S(\Delta t)/S(0)] = \nu \Delta t\\ Var[\ln S(\Delta t)/S(0)] = \sigma^2 \Delta t $$ and we want to find $u = d^{-1}$ and the probability $p$, we get the expressions as in the question: $$ u=e^{\sigma\sqrt{\Delta t}},d=e^{-\sigma\sqrt{\Delta t}}\\ p = \frac{1}{2}\left( 1+ \frac{\nu}{\sigma} \sqrt{\Delta t}\right) $$ This will inevitably lead to the continues case without the Ito term $-\frac{1}{2}\sigma^2 t$ in the exponent. This is because in the Black-Scholes model $\nu$ already includes this term. Indeed, if we compute the expectations above not in the binomial tree model but in the Black-Scholes model we get $$ \mathbb{E}[\ln S(t)/S(0)] =\mathbb{E}[\mu t - \frac{1}{2}\sigma^2 t + \sigma Z_t] = \left(\mu - \frac{1}{2}\sigma^2 \right)t = \nu t\\ Var[\ln S(t)/S(0)] = \sigma^2 t $$ If the Ito term is not in the Black-Scholes case, it will not appear in the binomial tree case.

Relative returns In order to get the desired form with the Ito term in the exponent, we need to start from the relative returns to find the proper $u$, $d$, and $p$. This is because in the Black-Scholes case we have for the relative returns $$ \mathbb{E}[S(\Delta t)/S(0) - 1] = e^{\mu \Delta t} -1 \approx \mu \Delta t\\ Var[S(\Delta t)/S(0)] = e^{2\mu \Delta t} \left(e^{\sigma^2 \Delta t} -1 \right) \approx \sigma^2 \Delta t. $$ If we start from the definition $$ \mathbb{E}[S(\Delta t)/S(0) - 1] = \mu \Delta t\\ Var[S(\Delta t)/S(0)] = \sigma^2 \Delta t. $$ for the binomial tree to find $u$, $d$, and $p$, we come inevitably to the familiar form for $S(t)$ with the Ito term. But those parameters will be different then in the topic question.