6 Brownian Motion

Brownian motion arising as the limit of a Binomial model. Definition. Geometric Brownian motion. Transformations of Brownian motion. Brownian motion as a martingale. Limit of squared increments.

Suppose we fix a terminal time \(T>0\), and we want to think about the behaviour of a Binomial model with a large number of intermediate steps. For example, we could consider \(T=\) 1 year, and think of each step as being a day, an hour, a minute, a second,…. Can we say anything about the limit?

Suppose we divide \([0,T]\) into \(n\) steps, of length \(T/n\). We can first think of what we should do with the bank account. Perhaps we initially get paid an interest rate \(r\) over 1 year, so at time \(1\) we have \((1+r)\). An alternative would be to receive the interest rate \(r\) compounded biannually, which means that we receive the interest \(\frac{r}{2}\) paid every half-a-year. So at the end of a year, \(1\) is worth \((1+\frac{r}{2})^2\).

This extends to compounding \(n\) times a year, so after one year, we receive \((1+\frac{r}{n})^n\), and if we let \(n \to \infty\), we get continuous compounding: if we have 1 at time \(0\), at time 1 we have \[\text{} \lim_{n \to \infty}\left( 1+\frac{r}{n}\right)^n \to \text{} \mathrm{e}^{r}.\] More generally, if we keep \(1\) in the bank for \(t\) units of time, then at the end of the period we have \(\mathrm{e}^{rt}\). (Can you see why?)

So the question now is, what happens to the asset under the risk-neutral measure as we let \(n \to \infty\)? Recall that, under \(\widetilde{\mathbb{P}}\), we know the value of the asset at time 1 is: \(S_n = S_0 d^{n} \left(\frac{u}{d}\right)^{H_n}\) where \(H_n\) is the number of heads we get on the first \(n\) flips, so \(H_n \sim \text{Bin}(n,q)\). Let’s fix \(q=\frac{1}{2}\), and think about how to choose \(u\) and \(d\). Since we want a sensible limit as \(n\to\infty\), we should choose the parameters of the model, \(u_n, r_n\) and \(d_n\) to depend on \(n\), and the argument above suggests we should take \(r_n = \frac{r}{n}\). In addition, if \(q=\frac{1}{2}\), then there must exist \(s_n\) such that \(u_n = 1+r_n + s_n\) and \(d_n = 1+r_n-s_n\).

We now want to choose \(s_n\) so that in the limit as \(n\to \infty\), we get something that is random, but not ‘too random’ that is, in the limit, we should have a well-defined random variable.

It turns out to be easier to consider \(\log(S_n)\), which is: \[\begin{align} \log(S_n) & = & \log\left(S_0 d_n^n \left( \frac{u_n}{d_n} \right)^{H_n}\right)\nonumber\\ & = & \log(S_0) + n \log(d_n) + H_n(\log(u_n)-\log(d_n)). \tag{6.1}\end{align}\] We note that \(\log(1+x) \approx x-\frac{x^2}{2}\) for small \(x\). Suppose that \(s_n = \frac{\sigma}{\sqrt{n}}\), so that \(u_n = 1+\frac{r}{n} + \frac{\sigma}{\sqrt{n}}\), and \(d_n = 1+ \frac{r}{n} - \frac{\sigma}{\sqrt{n}}\).

Then we have \[\begin{align} \log(u_n) & = & \log\left(1+\frac{r}{n} + \frac{\sigma}{\sqrt{n}}\right) \\ & \approx & \frac{r}{n} + \frac{\sigma}{\sqrt{n}} - \frac{1}{2}\left(\frac{r}{n} + \frac{\sigma}{\sqrt{n}}\right)^2\\ & \approx & \frac{r}{n} + \frac{\sigma}{\sqrt{n}} - \frac{\sigma^2}{2n}\end{align}\] where we have ignored terms of order \(n^{-\frac{3}{2}}\) and smaller. Similarly, we get: \[\log(d_n) \approx \frac{r}{n} - \frac{\sigma}{\sqrt{n}} - \frac{\sigma^2}{2n}.\]

If we put these into (6.1) we get: \[\begin{align} \log(S_n) & \approx & \log(S_0) + \left(\frac{r}{n} - \frac{\sigma}{\sqrt{n}} - \frac{\sigma^2}{2n}\right)n + 2\frac{\sigma}{\sqrt{n}}H_n\\ & \approx & \log(S_0) + 2 \sigma \frac{H_n - \frac{1}{2}n}{\sqrt{n}} + r - \frac{\sigma^2}{2}\\ & \approx & \log(S_0) + \sigma \frac{H_n - \frac{1}{2}n}{\sqrt{\frac{1}{4}n}} + r - \frac{\sigma^2}{2}.\end{align}\]

Now, recall that \(H_n\) is the number of heads, so we can write \(H_n = Y_1 + Y_2 + \cdots + Y_n\), where \(Y_i\) are i.i.d. with mean \(\frac{1}{2}\) and variance \(\frac{1}{2}\times \left(1-\frac{1}{2}\right)=\frac{1}{4}\). The Central Limit Theorem tells us that we get the convergence in distribution: \[\frac{H_n - \frac{1}{2}n}{\sqrt{\frac{1}{4}n}} \to Z \sim N(0,1)\] that is, for large \(n\) this looks like a Normal random variable with mean 0 and variance 1. If we call this \(B\), we see that: \[S = S_0 \mathrm{e}^{\sigma B + (r-\frac{1}{2}\sigma^2)}.\]

Now, this was all assuming that we are looking for the price of the asset at time 1. We can also do a similar analysis for the asset at a general time \(t\) (we do a similar analysis, but look at the asset after \(m\) steps, where \(m \approx tn\)), and we get an expression for the value of the asset at time \(t\), \(S_t\) as: \[S_t = S_0 \mathrm{e}^{\sigma B_t + (r-\frac{1}{2}\sigma^2)t},\] where \(B_t\) is now \(N(0,t)\) for each \(t\). What can we say about \(B_t\)?

Definition 43. A process \((B_t)_{t \ge 0}\) is a (standard) Brownian motion if \(B_0 = 0\) and:

  1. \(t\mapsto B_t\) is continuous

  2. \(B_t\) has independent increments: i.e. \[(B_{t_1} - B_{0}), (B_{t_2}-B_{t_1}), \dots, (B_{t_n}-B_{t_{n-1}}) \mbox{ are independent}\] if \(0 \le t_1 \le t_2 \le \dots \le t_{n-1} \le t_n\)

  3. \(B_t - B_s \sim N(0,(t-s))\) for \(0 \le s \le t\).

Note that the third condition implies \(B_t \sim N(0,t)\), by taking \(s=0\).

‘Standard’ here refers to the fact that \(\mathbb{E}B_1 = 0 = B_0\), and \(\mathbb{V}\mbox{ar}(B_1) = 1\). More generally, for \(\nu \in \mathbb{R}\) and \(\xi >0\) we can have: \[\widetilde{B}_t = \nu t + \xi B_t\] where \(B_t\) is a standard Brownian motion. Then we say that \(\widetilde{B}_t\) is a Brownian motion with drift \(\nu\) and diffusion coefficient \(\xi\).

In fact, a rather surprising (and hard) theorem says that the standard Brownian motion actually exists!

Definition 44. Suppose that \(B_t\) is a standard Brownian motion, and \(x_0,\xi > 0\). Then the process \[X_t = x_0 \exp\left\{\left(\nu - \frac{1}{2}\xi^2\right) t + \xi B_t\right\}\] is a geometric Brownian motion (GBM) with infinitesimal drift \(\nu\) and diffusion coefficient \(\xi\).

  • \(x_0\) is the initial value, so \(X_0 = x_0\)

  • \(\exp(\cdot) >0\), so \(X_t > 0\) for all \(t \ge 0\)

  • From the connection to the binomial model, we see that \(\nu\) is the average proportionate change (per unit of time) in \(X_t\) over a small time-step, and \(\xi^2\) is the variance of this change.

Since Brownian motion will form a key part of what follows, we will look at a few properties of Brownian motion.

Theorem 45. If \(B_t\) is a standard Brownian motion, then:

  1. \(X_t^1 = -B_t\),

  2. \(X_t^2 = B_{t+s}-B_s\), for \(s>0\),

  3. \(X_t^3 = a B_{a^{-2}t}\), for \(a\neq 0\)

are all (standard) Brownian motion.

The final property corresponds to ‘zooming’ into the origin. If we rescale correctly, the process when we zoom in is still a Brownian motion.

Solution:

  • Note \(X_t^1=X_t^3\) for \(a=-1\), so 1. is a special case of 3. and we don’t need to prove it explicitly.

    • \(t\mapsto X_t^2\) is continuous since \(t\mapsto B_t\) is and thus \(t\mapsto B_{t+s}\) is too.

    • Independence of increments follows from independence of increments for \(B\) and \[\begin{align} X_{t_k}^2-X_{t_{k-1}}^2=\left(B_{t_k+s}-B_s\right)-\left(B_{t_{k-1}+s}-B_s\right)=B_{t_k+s}-B_{t_{k-1}+s}\end{align}\]

    • \(X_t^2-X_r^2\sim N(0,t-r)\) and \(X_0^2=0\) follows from \[\begin{align} X_t^2-X_r^2=B_{t+s}-B_{r+s}\sim N(0,(t+s)-(r+s))=N(0,t-r)\end{align}\] and \(X_0^2=B_{0+s}-B_s=0\).

    • \(t\mapsto X_t^3\) is continuous since \(t\mapsto B_t\) is

    • Independence of increments follows from independence of increments for \(B\) and \[\begin{align} X_{t_k}^3-X_{t_{k-1}}^3=a(B_{a^{-2}t_k}-B_{a^{-2}t_{k-1}})\end{align}\]

    • \(X_t^3-X_r^3\sim N(0,t-r)\) and \(X_0^3=0\) follows from \[\begin{align} X_t^3-X_r^3=a(B_{a^{-2}t}-B_{a^{-2}r})\sim aN(0,a^{-2}t-a^{-2}r)=N(0,t-r)\end{align}\] and \(X_0^3=aB_{0}=0\) using \[\begin{align} Z\sim N(\mu,\sigma)\ \implies \ aZ\sim N(a\mu,a^2\sigma)\end{align}\]

\(\square\)

Think of \(B, X^1, X^2, X^3\) etc. as being “statistically” identical. It is not true that \(B_t = X_t^1\) always, even though both are Brownian motions. However, it is still true that Brownian motion is unique as a random variable, in the sense that the probability that \(B\) does something is the same as the probability that \(X^1\) or \(X^2\) do the same thing. This is a common thing in probability: the Normal distribution with mean 0 and variance 1 is unique in the sense that e.g. the distribution function is unique, however if \(X \sim N(0,1)\), then \(-X \sim N(0,1)\) as well, but they are not equal, in fact \(\mathbb{P}(X = -X) = 0\).

Theorem 46. Brownian Motion is a martingale.

Note that we will informally assume that many of the properties we had in discrete time, such as being able to take conditional expectations, also hold for continuous time. To prove these results rigorously is rather harder, but for the rest of this course, we will be a bit more informal regarding such details.

Solution: Need to show:

  • \(\mathbb{E}|B_t|<\infty\) for all \(t\ge 0\)

    • Note \(|x|\le 1+x^2\) and so \[\begin{align} \mathbb{E}|B_t|\le\mathbb{E}\left[ 1+B_t^2 \right]=1+\mathbb{E}\left[ B_t^2 \right]-\mathbb{E}\left[ B_t \right]^2=1+\mathbb{V}\mbox{ar}(B_t)=1+t<\infty\end{align}\] since \(B_t\sim N(0,t)\).

  • \(\mathbb{E}_s[B_t]=B_s\) for all \(t\ge s\ge 0\)

    • We have \[\begin{align} \mathbb{E}_s[B_t]=&\mathbb{E}_s[B_s]+\mathbb{E}_s[B_t-B_s]=B_s\end{align}\] since \(B_s\) adapted and \(B_t-B_s\sim N(0,t-s)\) independent of \(B_r=B_r-B_0\) for all \(r\le s\).

  • \(B_t\) adapted

    • \(B_t\) is adapted because \(B_t\) is known at time \(t\).

\(\square\)

One of the features of Brownian motion is that the paths of the process are not very smooth at all. It can be shown (we won’t!) that the path \(B_t\) is not differentiable anywhere. We will be interested in a related property: suppose \(f(t)\) is a differentiable function, with a bounded derivative. Then there exists some constant \(K\) such that \(|f(t)-f(s)| \le K|t-s|\). Suppose now we consider the function evaluated at smaller and smaller time steps. Specifically, suppose that we set \(t_i^n = \frac{i}{n}\), so \(t_{i+1}^n - t_i^n = \frac{1}{n}\), and \(T_n = \lfloor tn \rfloor\), so \(t_{T_n}^n \approx t\).

Then we get: \[\begin{align} \sum_{i=0}^{T_n-1} (f(t_{i+1}^n) - f(t_i^n))^2 & \le & K^2 \sum_{i=0}^{T_n-1} |t_{i+1}^n-t_i^n|^2 = K^2 \sum_{i=0}^{T_n-1} \frac{1}{n^2} = \frac{K^2}{n^2} \times \lfloor tn \rfloor \to 0,\end{align}\] as \(n \to \infty\).

However, if we do the same for Brownian motion: \[\sum_{i=0}^{T_n-1} (B_{t_{i+1}^n} - B_{t_i^n})^2\] is a sum of \(T_n \approx tn\) i.i.d. r.v.’s with mean \((t_{i+1}^n-t_i^n)=1/n\), and finite variance. This has a non-zero average, so perhaps the limit is not zero.

We make this precise by considering the limit in the following sense:

Lemma 47. We have: \[\lim_{n \to \infty}\mathbb{E}\left[ \left| \sum_{i=0}^{T_n-1} (B_{t_{i+1}^n} - B_{t_i^n})^2 - t\right|^2 \right] = 0.\] That is, \(\sum_{i=0}^{T_n-1} (B_{t_{i+1}^n} - B_{t_i^n})^2 \to t\) in mean-square, or (equivalently) in \(\mathcal{L}^2\).8

Proof. To simplify the mathematics, we will assume that \(T_n = \lfloor tn \rfloor = tn\). The result remains true in general, but we need to be a bit more careful in the proof.

Note that if we write \(\Delta_i^n = (B_{t_{i+1}^n}-B_{t_i^n})\), then we know that \(\Delta_i^n \sim N(0,\frac{1}{n})\) and so, using standard properties9 of the Normal distribution: \[\mathbb{E}\left[ \left(\Delta_i^n\right)^2 \right] = \frac{1}{n}, \quad \mathbb{E}\left[ \left(\Delta_i^n\right)^4 \right] = \frac{3}{n^2}.\] In addition, if \(i \neq j\), \(\Delta_i^n\) and \(\Delta_j^n\) are independent, since they are increments of the Brownian motion.

So: \[\begin{align} \mathbb{E}\left[ \left| \sum_{i=0}^{T_n-1} (B_{t_{i+1}^n} - B_{t_i^n})^2 - t\right|^2 \right] & = \mathbb{E}\left[ \left( \sum_{i=0}^{T_n-1} \left((\Delta_i^n)^2 - \frac{1}{n}\right)\right)^2 \right]\\ & = \mathbb{E}\left[ \sum_{i,j=0}^{T_n-1} \left((\Delta_i^n)^2-\frac{1}{n}\right) \left((\Delta_j^n)^2 - \frac{1}{n}\right) \right] \\ & = \sum_{i,j=0}^{T_n-1}\mathbb{E}\left[ \left((\Delta_i^n)^2-\frac{1}{n}\right) \left((\Delta_j^n)^2 - \frac{1}{n}\right) \right] \end{align}\] But by the independence of Brownian increments, and since \(\mathbb{E}\left[ (\Delta_i^n)^2-\frac{1}{n} \right] = 0\), the terms where \(i\neq j\) are all zero.

We therefore have: \[\begin{align} \mathbb{E}\left[ \left| \sum_{i=0}^{T_n-1} (B_{t_{i+1}^n} - B_{t_i^n})^2 - t\right|^2 \right] & = \sum_{i=0}^{T_n-1}\mathbb{E}\left[ \left((\Delta_i^n)^2-\frac{1}{n}\right)^2 \right] \\ & = \sum_{i=0}^{T_n-1} \mathbb{E}\left[ (\Delta_i^n)^4 - \frac{2}{n}(\Delta_i^n)^2 + \frac{1}{n^2} \right] \\ & = \sum_{i=0}^{T_n-1}\left( \frac{3}{n^2}-\frac{2}{n^2} + \frac{1}{n^2}\right) \\ & = t \, n \frac{2}{n^2} = \frac{2t}{n} \to 0,\end{align}\] where we have used the values for \(\mathbb{E}\left[ (\Delta_i^n)^4 \right]\) and \(\mathbb{E}\left[ (\Delta_i^n)^2 \right]\) we gave earlier and \(T_n = tn\).

  1. Recall (e.g. from MA20224) that \(\mathcal{L}^2\) is the set of r.v.’s which have \(\mathbb{E}\left[ X^2 \right]< \infty\) and we say that a sequence of r.v.’s \(X_n\) in \(\mathcal{L}^2\) converge to \(X \in \mathcal{L}^2\) if and only if \(\mathbb{E}\left[ |X_n-X|^2 \right] \to 0\) as \(n\to \infty\).↩︎

  2. For example, one can recall that if \(X \sim N(0,\sigma^2)\), then the m.g.f. of \(X\) is \(M_X(\alpha) = \mathbb{E}\left[ \mathrm{e}^{\alpha X} \right] = \mathrm{e}^{\alpha^2\sigma^2/2}\), and \(\mathbb{E}\left[ X^n \right] = M_X^{(n)}(0)\), the \(n\)th derivative, evaluated at \(0\).↩︎