last updated: October 16, 2024

Stochastic Processes and Financial Mathematics
(part two)

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13.4 Stochastic exponentials and martingale representation

In this section we look at a close relative of the SDE (13.8) for geometric Brownian motion. In particular, we look at

\begin{equation} \label {eq:stoch_exp} dX_t=\sigma _tX_t\,dB_t \end{equation}

with the initial condition \(X_0=1\). Here, \(\sigma _t\) is a stochastic process. By comparison to (13.8), we have set \(\alpha =0\) (which makes our life easier) but \(\sigma \) is no longer a deterministic constant (which makes our life harder).

The key idea is the same: we assume that a strictly positive solution exists, take logarithms \(Z_t=\log X_t\), then look for an explicit formula for \(Z\), and in turn an explicit formula for \(X\), which we can then go back and check is really a solution.

From Ito’s formula we have

\begin{align*} dZ_t&=\l (0+0+\frac 12(\sigma _t X_t)^2\frac {-1}{X_t^2}\r )\,dt + \sigma _t X_t \frac {1}{X_t}\,dB_t\\ &=-\frac 12\sigma _t^2\,dt+\sigma _t\,dB_t. \end{align*} This gives us

\[Z_t=Z_0+\int _0^t \sigma _u\,dB_u-\frac 12\int _0^t\sigma _u^2\,du\]

and hence

\begin{equation} \label {eq:stoch_exp_pre} X_t=X_0\exp \l (\int _0^t\sigma _u\,dB_u -\frac {1}{2}\int _0^t\sigma _u^2\,du\r ). \end{equation}

It can be checked (again, left for you, see 13.13) that this formula really does solve (13.13).

  • Remark 13.4.1 If \(\sigma \) is a deterministic constant, then (13.14) becomes precisely (13.9) with \(\alpha =0\); as we would expect since this is also how (13.13) is connected too (13.8).

In view of (13.14) we have:

  • Definition 13.4.2 The stochastic exponential of the process \(\sigma _t\) is

    \[\mc {E}_\sigma (t)=\exp \l (\int _0^t\sigma _u\,dB_u-\frac {1}{2}\int _0^t\sigma _u^2\,du\r ).\]

Of course, we have shown that \(\mc {E}_\sigma (t)\) solves (13.13), and noting that \(\mc {E}_\sigma (0)=1\) we thus have

\begin{equation} \label {eq:stoch_exp_int} \mc {E}_\sigma (t)=1+\int _0^t\sigma _u\mc {E}_\sigma (u)\,dB_u. \end{equation}

We record this equation here because we’ll need it in the next section.

The martingale representation theorem \(\offsyl \)

Note that this section is off-syllabus, since it is marked with a \(\offsyl \). However, since it covers a result that we will need in our analysis of the Black-Scholes model, it will still be covered in lectures.

Recall from Theorem 12.2.1 that Ito integrals \(\int _0^t F_u\,dB_u\) are martingales. This might make us wonder if, given a martingale \(M_t\in \mc {H}^2\), whether it is possible to write \(M\) as

\[M_t=M_0+\int _0^t h_u\,dB_u\]

for some stochastic process \(h\). Here, we follow common convention in denoting \(h_t\) with a lower case letter. The answer is strongly positive:

  • Theorem 13.4.3 (Martingale Representation Theorem) Let \(M_t\in \mc {H}^2\) be a continuous martingale. Fix \(T\in (0,\infty )\). Then there exists a stochastic process \(h_t\in \mc {H}^2\) such that

    \[M_t=M_0+\int _0^t h_u\,dB_u\]

    for all \(t\in [0,T]\).

Sketch of Proof: Thanks to (13.15), we already know that this theorem holds if \(M_0=1\) and \(M_t\) is the stochastic exponential of some stochastic process \(\sigma _t\) – in this case we take \(h_t=\sigma _t\mc {E}_\sigma (t)\). The proof of the martingale representation theorem, which we don’t include in this course, works by showing that any continuous martingale \(M_t\in \mc {H}^2\) can be approximated by a sequence \(M^{(n)}_t\) of continuous martingales that are themselves stochastic exponentials. As a consequence, the martingale representation theorem tells us that the process \(h_t\) exists, but does not provide us with a formula for \(h_t\).   ∎

The martingale representation theorem illustrates the importance of Ito integrals, and suggests that they are likely to be helpful in situations involving continuous time martingales. In fact, Theorem 13.4.3 will sit right at the heart of the argument that we will use (in Section 15.2) to show that hedging strategies exist in continuous time.