7 Stochastic Integration
Stochastic integral against Brownian motion. Properties of the Stochastic Integral.
The material in this section is somewhat harder than the previous material. The goal is to give you a brief introduction to how the stochastic integral is defined. I would not anticipate the material in this section featuring heavily in an exam question (except perhaps Example53 and the derivation up to (7.1) ). In the next section of the notes, we will see how we can practically manipulate the integral, and this will be more important.
Why is the roughness of Brownian motion a problem? Suppose we consider the path of a financial asset, and suppose we choose to invest in the asset in a way that depends on the behaviour of the asset, so perhaps we always choose to hold a number of units of the asset which is equal to the current value of the asset. How much profit/loss will we make?
If the asset price at time \(t\) is \(S_t\), and we approximate by changing the number of units of the asset we hold at times \(t_i^n\), then our profit/loss from this strategy \(V_t^n\) is: \[V_t^n = \sum_{i=0}^{T_n-1} S_{t_i^n}(S_{t_{i+1}^n}-S_{t_i^n}).\] If we now let \(n\to \infty\), then we expect to get the value of following this strategy ‘in continuous time’.
First we observe that, for \(a,b \in \mathbb{R}\): \[a(b-a) = \frac{1}{2}(b^2-a^2) - \frac{1}{2}(b-a)^2.\]
Then if we take \(a = S_{t_i^n}\) and \(b = S_{t_{i+1}^n}\) \[\begin{align} V_t^n & = & \sum_{i=0}^{T_n-1} S_{t_i^n}(S_{t_{i+1}^n}-S_{t_i^n})\\ & = & \sum_{i=0}^{T_n-1}\left[ \frac{1}{2}\left(S_{t_{i+1}^n}^2-S_{t_i^n}^2\right) - \frac{1}{2}\left(S_{t_{i+1}^n}-S_{t_i^n}\right)^2\right]\\ & = & \frac{1}{2}\left(S_{\lfloor tn\rfloor /n}^2-S_{0}^2\right) - \frac{1}{2} \sum_{i=0}^{T_n-1}\left(S_{t_{i+1}^n}-S_{t_i^n}\right)^2\end{align}\] where we have used the fact that \(t_{T_n}^n = \lfloor tn \rfloor/n\), which is approximately \(t\).
But we have just observed above: if \(S_t = f(t)\), a continuous, differentiable function, then the second term on the right disappears as \(n\to\infty\). However, if \(S_t=B_t\) is Brownian motion, then this term will not disappear. In fact, it will converge to \(t\) in the limit. At least, in the sense of convergence in \(\mathcal{L}^2\).
Since we are thinking about small time steps, we will write (informally) \(\mathrm{d}t\) for the time-step (in this case, \(\frac{1}{n}\)), \(\mathrm{d}S_t = S_{t+\mathrm{d}t} - S_t\) for the change in \(S_t\), and similarly, \(\mathrm{d}f(t) = f({t+\mathrm{d}t}) - f(t)\), and \(\mathrm{d}B_t = B_{t+\mathrm{d}t} - B_t\).
If we are thinking about letting the time step go to 0, then the sum will become an integral, and we can rephrase these results as: \[\sum_{i=0}^{T_n-1} f(t_i^n)(f(t_{i+1}^n)-f(t_i^n)) \to \int_0^t f(s) \, \mathrm{d}f(s) = \frac{1}{2} (f(t)^2 - f(0)^2),\] but for Brownian motion: \[\begin{align} \tag{7.1} \sum_{i=0}^{T_n-1} B_{t_i^n}(B_{t_{i+1}^n}-B_{t_i^n}) \to \int_0^t B_s \, \mathrm{d}B_s = \frac{1}{2}(B_t^2 - B_0^2) -\frac{1}{2}t.\end{align}\]
For differentiable \(f\), this is nothing more than the product rule: since \(\mathrm{d}f(t) = f(t+\mathrm{d}t) - f(t) \approx f'(t) \mathrm{d}t\), we have: \[\begin{align} \int_0^t f(s) \mathrm{d}f(s) & = & \int_0^t f(s) f'(s) \, \mathrm{d}s\, = \, \int_0^t \left(\frac{1}{2}f(s)^2\right)' \mathrm{d}s\\ & = & \left[\frac{1}{2} f(s)^2\right]_0^t \, = \, \frac{1}{2}(f(t)^2 - f(0)^2).\end{align}\] But for Brownian motion, which is not differentiable, the roughness of the paths gives us the additional terms involving \(t\).
Recall that when we considered processes in discrete time, the processes of the form \(\sum_{i=0}^{n-1} \phi_i (S_{i+1}-S_i)\) played an important role to understanding our possible trading portfolios. The same will be true in continuous time, but we need to make sense of the limit as the gap between times where we trade get smaller.
In particular, (using the \(\mathrm{d}S_t\) notation above), we are interested in the limit \[\sum_{i=0}^{T_n } \phi_{t_i^n} (S_{t_{i+1}^n} - S_{t_i^n}) \to \int_0^t \phi_s \mathrm{d}S_s\] and we will first consider the case where \(S_t = B_t\) is Brownian motion10 so we want to make sense of: \[\sum_{i=0}^{T_n } \phi_{t_i^n} (B_{t_{i+1}^n} - B_{t_i^n}) \to \int_0^t \phi_s \mathrm{d}B_s,\] and later see how this relates to the financial interpretation.
So we begin by considering what class of processes \((\phi_s)_{s \ge 0}\) which we are allowed to take as the integrand in the integral. In general, we will want \(\phi_s\), the value of the integrand at time \(s\), to only depend on the information available up to time \(s\).
An easy case is when we allow \(\phi\) only to change a finite number of times. These integrands will play a special role, so we define:
Definition 48. A process \((\phi_t)_{t \ge 0}\) is called a simple process if it can be written as \[\begin{align} \tag{7.2} \phi_t = \sum_{i = 0}^{n-1} X_i \boldsymbol{1}\{t \in [t_i,t_{i+1})\},\end{align}\] where \(0 = t_0 < t_1 < \dots < t_n\), and each \(X_i\) is a random variable, known at time \(t_i\), such that \(\mathbb{E}\left[ X_i^2 \right]<\infty\).
This definition is nice, since we are essentially in the discrete case! We hold \(\phi_{t_i}\) units of the asset between time \(t_i\) and time \(t_{i+1}\), and we can compute the ‘gains’ from this as the sum of the terms \(\phi_{t_i} (B_{t_{i+1}}-B_{t_i})\).
This leads us to the definition:
Definition 49. We define the stochastic integral \((\mathcal{I}_t)_{t \ge 0}\) (with respect to a Brownian motion \(B_t\)) of a simple process \((\phi_t)_{t \ge 0}\) given by (7.2) as: for \(t \in [t_k,t_{k+1})\) \[\mathcal{I}_t = \sum_{i = 0}^{k-1} X_i (B_{t_{i+1}} - B_{t_i}) + X_{k} (B_{t}-B_{t_k}).\]
Solution: Define the integrand \(\phi\) by: \[\phi_t = \begin{cases} X_0 = 1, \quad & t \in [t_0,t_1)\\ X_1 = 2, \quad & t \in [t_1,t_2)\\ X_2 = \frac{1}{2}, \quad & t \in [t_2,t_3)\\ X_3 = -B_{t_3}, \quad & t \in [t_3,t_4) \end{cases}\]
image
Here \(\mathcal{I}_t = \sum_{i = 0}^{k-1} X_i (B_{t_{i+1}} - B_{t_i}) + X_{k} (B_{t}-B_{t_k}) = \int_0^t\phi_s \, \mathrm{d}B_s\). \(\square\)
Then we have some important properties of the stochastic integral for a simple process:
Theorem 50. The stochastic integral of a simple process is a martingale.
And the Stochastic integral has another important property:
Theorem 51 (It isometry). The stochastic integral of a simple process satisfies: \[\mathbb{E}\left[ \mathcal{I}_t^2 \right] = \mathbb{E}\left[ \int_0^t \phi_s^2 \, \mathrm{d}s \right]\]
Solution: The stochastic integral of a simple process is a martingale: We show \(\mathbb{E}_s[\mathcal{I}_t]=\mathcal{I}_s\).
Consider \(t\in[t_k,t_{k+1})\) and \(s<t_k\) (other cases similar). If \(s\le t_j<t_k\): \[\begin{align} \mathbb{E}_s[X_j(B_{t_{j+1}}-B_{t_j})]=\mathbb{E}_s[\mathbb{E}_{t_j}[X_j(B_{t_{j+1}}-B_{t_j})]]=\mathbb{E}_s[X_j\underbrace{\mathbb{E}_{t_j}[(B_{t_{j+1}}-B_{t_j})]}_{=0\text{ (BM is martingale)}}]=0\end{align}\]
If \(s\ge t_{j+1}\): \(\mathbb{E}_s[X_j(B_{t_{j+1}}-B_{t_j})]=X_j(B_{t_{j+1}}-B_{t_j})\).
If \(t_j\le s\le t_{j+1}\): \(\mathbb{E}_s[X_j(B_{t_{j+1}}-B_{t_j})]=X_j\mathbb{E}_s[(B_{t_{j+1}}-B_{t_j})] = X_j(B_s-B_{t_j})\). Since \(s< t_k\): \(\mathbb{E}_s[X_k(B_{t}-B_{t_k})]=\mathbb{E}_s[\mathbb{E}_{t_k}[X_k(B_{t}-B_{t_k})]]=\mathbb{E}_s[X_k\mathbb{E}_{t_k}[(B_{t}-B_{t_k})]]=0\), and adding up non-zero terms for \(t_m\le s<t_{m+1}\): \[\begin{align} \mathbb{E}_s\mathcal{I}_t=\sum_{j=0}^{m-1}X_j(B_{t_{j+1}}-B_{t_j})+X_m(B_s-B_{t_m})=\mathcal{I}_s.\end{align}\] \(\square\)
Solution:
It Isometry for simple processes: We show the stochastic integral of a simple process satisfies: \[\begin{align} \mathbb{E}\left[ \mathcal{I}_t^2 \right] = \mathbb{E}\left[ \int_0^t \phi_s^2ds \right]\end{align}\] Suppose \(t\in[t_k,t_{k+1})\), and set \(\Delta_i=\begin{cases}B_{t_{i+1}}-B_{t_i}&,\ i<k\\B_t-B_{t_k}&,\ i=k\end{cases}\). Then \(\mathcal{I}_t=\sum_{i=0}^k X_i\Delta_i\).
We have \[\begin{align} \mathbb{E}\left[ \mathcal{I}_t^2 \right] & =\mathbb{E}\left[ \sum_{i,j=0}^k X_iX_j\Delta_i\Delta_j \right]\\ & =\sum_{i=0}^k\mathbb{E}\left[ X_i^2\Delta_i^2 \right]+2\sum_{0\le i<j\le k}\mathbb{E}\left[ X_iX_j\Delta_i\Delta_j \right]\end{align}\]
Consider \(i<j\) (note \(t_{j+1}=t\) if \(j=k\)) \[\begin{align} \mathbb{E}\left[ X_iX_j\Delta_i\Delta_j \right]= &\mathbb{E}\left[ X_iX_j(B_{t_{i+1}}-B_{t_i})(B_{t_{j+1}}-B_{t_j}) \right]\\ =&\mathbb{E}\left[ X_iX_j(B_{t_{i+1}}-B_{t_i})\mathbb{E}_{t_j}[(B_{t_{j+1}}-B_{t_j})] \right]=0\end{align}\] since \(\mathbb{E}_{t_j}[(B_{t_{j+1}}-B_{t_j})] = 0\).
Consider \(i=j\) (note \(t_{i+1}=t\) if \(i=k\)) \[\begin{align} \mathbb{E}\left[ X_i^2\Delta_i^2 \right]= &\mathbb{E}\left[ X_i^2(B_{t_{i+1}}-B_{t_i})^2 \right]\\ =& \mathbb{E}\left[ X_i^2 \right]\mathbb{E}\left[ (B_{t_{i+1}}-B_{t_i})^2 \right]\\ =&\mathbb{E}\left[ X_i^2 \right] \begin{cases} t_{i+1}-t_i&,\ i\ne k\\ t-t_k&,\ i=k \end{cases}.\end{align}\]
Then \(\mathbb{E}\left[ \mathcal{I}_t^2 \right]=\sum_{i=0}^{k-1}\mathbb{E}\left[ X_i^2 \right](t_{i+1}-t_i)+\mathbb{E}\left[ X_k^2 \right](t-t_k)\). To show: \(\mathbb{E}\left[ \mathcal{I}_t^2 \right]=\sum_{i=0}^{k-1}\mathbb{E}\left[ X_i^2 \right](t_{i+1}-t_i)+\mathbb{E}\left[ X_k^2 \right](t-t_k)=\mathbb{E}\left[ \int_0^t \phi_s^2ds \right]\), note \(\phi_t=X_i\) on \([t_i,t_{i+1})\), so \(\mathbb{E}\left[ \int_{t_i}^{t_{i+1}} \phi_s^2ds \right]=\mathbb{E}\left[ X_i^2(t_{i+1}-t_i) \right]\). Summing all values of \(i\): \[\begin{align} \mathbb{E}\left[ \int_0^t \phi_s^2ds \right]= &\sum_{i=0}^{k-1}\mathbb{E}\left[ \int_{t_i}^{t_{i+1}} \phi_s^2ds \right]+\mathbb{E}\left[ \int_{t_k}^{t} \phi_s^2ds \right]\\ =&\sum_{i=0}^{k-1}\mathbb{E}\left[ X_i^2 \right](t_{i+1}-t_i)+\mathbb{E}\left[ X_k^2 \right](t-t_k)=\mathbb{E}\left[ \mathcal{I}_t^2 \right].\end{align}\]
\(\square\)
Finally, we define our class of potential integrands. Let \(\mathcal{V}\) be the set of stochastic processes, \((\phi_t)_{t \ge 0}\) such that \(\phi_t\) is known at time \(t\), and: \[\mathbb{E}\left[ \int_0^t \phi_s^2 \, \mathrm{d}s \right] < \infty.\] The key idea is that we can now define the stochastic integral for a general integrand in \(\mathcal{V}\) by considering the limit of integrals of simple processes.
Definition 52. Suppose \(\phi_t \in \mathcal{V}\), and suppose \(\phi^n_t\) is a sequence of simple processes such that: \[\mathbb{E}\int_0^t |\phi_s - \phi_s^n|^2 \,\mathrm{d}s\to 0 \quad \text{ as } n \to \infty.\] Then we define the stochastic integral of \(\phi_t\) with respect to a Brownian motion \(B_t\) by: \[\begin{align} \tag{7.3} \mathcal{I}_t = \int_0^t \phi_s \, \mathrm{d}B_s = \lim_{n \to \infty} \int_0^t \phi_s^n \, \mathrm{d} B_s = \lim_{n\to \infty} \mathcal{I}^n_t,\end{align}\] where the limit is in the sense of \(\mathbb{E}\left[ (\mathcal{I}^n_t-\mathcal{I}_t)^2 \right] \to 0\) as \(n \to \infty\).
Of course, there are a lot of mathematical questions which, if we were being rigorous, we would try to answer — for example does there always exist an approximating sequence of simple functions, and does the limit in (7.3) exist, and is it unique? The answer to these questions are yes but we will not go into the details.
(Highly non-examinable aside: Roughly, the argument runs like this: if we have convergence of a sequence \(\phi_n\) of simple processes to the process \(\phi \in \mathcal{V}\), then the sequence \(\phi_n\) must be a Cauchy sequence in \(\mathcal{V}\) in the sense that: \[\mathbb{E}\left[ \int_0^t |\phi_s^m - \phi_s^n|^2 \,\mathrm{d}s \right] \to 0\] as \(n,m \to \infty\). Now, for each \(n\), the integral of the simple process, \(\mathcal{I}^n_t = \int_0^t \phi_s^n \, \mathrm{d}B_s\) exists and is an element of the space \(\mathcal{L}^2\) by the It isometry. Moreover, it can be shown that the space \(\mathcal{L}^2\) is a Complete metric space with the metric \(d(X,Y) = \mathbb{E}\left[ (X-Y)^2 \right]\) — that is, every Cauchy sequence in \(\mathcal{L}^2\) converges to a (unique) limit. So all we need to show is that \(\mathcal{I}^n_t\) is a Cauchy sequence. Now, with a bit of work, it is possible to show that the difference of two simple processes, \(\phi_t^n - \phi_t^m\) is also a simple process, and the stochastic integral of \(\phi_t^n-\phi_t^m\) is given by: \[\int_0^t (\phi_s^n-\phi_s^m) \, \mathrm{d}B_s = \int_0^t \phi_s^n \, \mathrm{d} B_s - \int_0^t \phi_s^m \, \mathrm{d}B_s = \mathcal{I}_t^n-\mathcal{I}_t^m.\] So we can use the It isometry, Theorem51, to see that: \[\mathbb{E}\left[ (\mathcal{I}_t^n - \mathcal{I}_t^m)^2 \right] = \mathbb{E}\left[ \left(\int_0^t (\phi_s^n-\phi_s^m) \, \mathrm{d}B_s\right)^2 \right] = \mathbb{E}\left[ \int_0^t |\phi_s^m - \phi_s^n|^2 \,\mathrm{d}s \right].\] We assumed that this final term converged to zero as \(n,m \to \infty\), and so \(\mathcal{I}^n_t\) is also a Cauchy sequence in \(\mathcal{L}^2\). This means that it converges to some limit in \(\mathcal{L}^2\) as \(n,m \to \infty\), and we set this limit to be the integral of the process \(\phi_t\). We can deduce that the limit is unique by interleaving Cauchy sequences of simple integrands, and seeing that these sequences still converge to a limit, which must therefore be unique.
End of highly non-examinable aside.)
Example 53. This is all very well, but can we actually find the stochastic integral for any processes? Well, recall the calculations at the beginning of this chapter. There, we showed (replacing \(S\) with \(B\)) that if we take \(T_n = \lfloor tn\rfloor\) and \(t_i^n = i/n\), we can write: \[\sum_{i=0}^{T_n-1} B_{{t_i}^n} (B_{t_{i+1}^n} - B_{{t_i}^n}) = \frac{1}{2}\left(B_{\lfloor tn \rfloor /n}^2 - B_0^2\right) - \frac{1}{2}\sum_{i=0}^{T_n-1} \left( B_{t_{i+1}^n}-B_{t_i^n}\right)^2.\]
Now, the left-hand side is really just the stochastic integral of the simple process: \[\phi^n_t = \sum_{i=0}^{T_n-1} B_{t_i^n} \boldsymbol{1}\{t \in [t_i^n,t_{i+1}^n)\}\] and as \(n \to \infty\), this looks more and more like Brownian motion, and in particular, it can be shown that \(\int_0^t |\phi_s^n - B_s|^2\, \mathrm{d}s\to 0\).
So we have convergence as required in the definition of the stochastic integral. This means that the right-hand side should converge to the stochastic integral as \(n \to \infty\). But by the continuity of Brownian motion, \(B_{\lfloor tn \rfloor /n} \to B_{t}\) since \(\lfloor tn \rfloor /n \to t\), and the sum of the squared Brownian increments converges to \(t\) by Lemma47. Hence \[\begin{align} \tag{7.4} \int_0^t B_s \, \mathrm{d}B_s = \frac{1}{2}\left(B_t^2 - B_0^2\right) - \frac{1}{2}t.\end{align}\]
We can also summarise some key properties of the stochastic integral:
Theorem 54 (Properties of the Stochastic Integral). Suppose \((\phi_t)_{t \ge 0}, (\psi_t)_{t \ge 0} \in \mathcal{V}\), and \((B_t)_{t \ge 0}\) is a Brownian motion. Then the stochastic integrals11: \(\mathcal{I}_t = \int_0^t \phi_s \, \mathrm{d}B_s, \mathcal{J}_t = \int_0^t \psi_s \, \mathrm{d}B_s\) satisfy:
Continuity: \(t \mapsto \mathcal{I}_t\) is continuous;
Known: \(\mathcal{I}_t\) is known at time \(t\);
Linearity: if \(\alpha, \beta \in \mathbb{R}\), then \(\alpha \phi_t + \beta \psi_t \in \mathcal{V}\) and: \[\int_0^t (\alpha \phi_s + \beta \psi_s) \, \mathrm{d}B_s = \alpha \mathcal{I}_t + \beta \mathcal{J}_t;\]
(iv) Martingale: \(\mathcal{I}_t\) is a martingale;
It Isometry: \(\mathbb{E}\left[ \mathcal{I}_t^2 \right] = \mathbb{E}\left[ \int_0^t \phi_s^2 \, \mathrm{d}s \right]\).
We will not prove these results, although we observe that we have already shown \((iv)\) and \((v)\) for simple processes, and \((iii)\) is also not too difficult to show for simple processes.
This is called the Bachelier model when we model an asset price by a Brownian motion. It is not a good model for asset prices, since it can go negative with positive probability.↩︎
we define \(\mathcal{J}_t\) here only for the linearity property; of course, all the other properties hold for \(\mathcal{J}_t\) as well as for \(\mathcal{I}_t\).↩︎