Greeks and Martingale Representation

In complete diffusion models, martingale representation yields a conceptual foundation for delta hedging: delta is the integrand in the stochastic integral representation of the discounted price.

---

Discounted asset and wealth

In Black–Scholes under \(\mathbb{Q}\),

\[ \mathrm{d}S_t = rS_t\,\mathrm{d}t + \sigma S_t\,\mathrm{d}W_t, \qquad B_t = e^{rt} \]

Define the discounted asset \(\widetilde{S}_t := B_t^{-1}S_t\). Then

\[ \mathrm{d}\widetilde{S}_t = \sigma \widetilde{S}_t\,\mathrm{d}W_t \]

so \(\widetilde{S}\) is a \(\mathbb{Q}\)-martingale.

---

Discounted option price is a martingale

Let \(V(t,S_t)\) be the price process and define \(\widetilde{V}_t := B_t^{-1}V(t,S_t)\). Under \(\mathbb{Q}\),

\[ \widetilde{V}_t = \mathbb{E}^{\mathbb{Q}}[\widetilde{V}_T\mid \mathcal{F}_t] \]

so \(\widetilde{V}\) is a martingale.

---

Martingale representation

In a Brownian filtration, any square-integrable martingale can be represented as

\[ \boxed{ \widetilde{V}_t = \widetilde{V}_0 + \int_0^t Z_s\,\mathrm{d}W_s } \]

for some predictable \(Z\) with \(\mathbb{E}\int_0^T Z_s^2\,\mathrm{d}s<\infty\).

---

Identification of delta

By Itô’s formula and the PDE cancellation of drift,

\[ \mathrm{d}\widetilde{V}_t = B_t^{-1}\sigma S_t V_S(t,S_t)\,\mathrm{d}W_t \]

Thus

\[ \boxed{ Z_t = B_t^{-1}\sigma S_t\,\Delta(t,S_t), \qquad \Delta=V_S } \]

---

What to remember

• Discounted option prices are martingales under \(\mathbb{Q}\).
• Martingale representation gives the stochastic integrand.
• Delta is the hedge ratio in complete diffusion models.

---

Exercises

Exercise 1. Starting from \(\mathrm{d}\widetilde{S}_t = \sigma \widetilde{S}_t \, \mathrm{d}W_t\), verify that \(\widetilde{S}_t\) is a \(\mathbb{Q}\)-martingale by checking that \(\mathbb{E}^{\mathbb{Q}}[\widetilde{S}_t \mid \mathcal{F}_s] = \widetilde{S}_s\) for \(s < t\).

Solution to Exercise 1

We need to show \(\mathbb{E}^{\mathbb{Q}}[\widetilde{S}_t \mid \mathcal{F}_s] = \widetilde{S}_s\) for \(s < t\).

Since \(\mathrm{d}\widetilde{S}_u = \sigma \widetilde{S}_u \,\mathrm{d}W_u\), this SDE has the solution:

\[ \widetilde{S}_t = \widetilde{S}_s \exp\!\left(-\frac{1}{2}\sigma^2(t-s) + \sigma(W_t - W_s)\right) \]

Taking the conditional expectation given \(\mathcal{F}_s\):

\[ \mathbb{E}^{\mathbb{Q}}[\widetilde{S}_t \mid \mathcal{F}_s] = \widetilde{S}_s \cdot \mathbb{E}^{\mathbb{Q}}\!\left[\exp\!\left(-\frac{1}{2}\sigma^2(t-s) + \sigma(W_t - W_s)\right) \,\Big|\, \mathcal{F}_s\right] \]

Since \(W_t - W_s\) is independent of \(\mathcal{F}_s\) and distributed as \(N(0, t-s)\):

\[ \mathbb{E}\!\left[\exp\!\left(-\frac{1}{2}\sigma^2(t-s) + \sigma(W_t - W_s)\right)\right] = \exp\!\left(-\frac{1}{2}\sigma^2(t-s)\right) \cdot \exp\!\left(\frac{1}{2}\sigma^2(t-s)\right) = 1 \]

where we used the moment generating function \(\mathbb{E}[e^{aZ}] = e^{a^2/2}\) for \(Z \sim N(0,1)\) with \(a = \sigma\sqrt{t-s}\). Therefore:

\[ \mathbb{E}^{\mathbb{Q}}[\widetilde{S}_t \mid \mathcal{F}_s] = \widetilde{S}_s \]

confirming that \(\widetilde{S}\) is a \(\mathbb{Q}\)-martingale.

---

Exercise 2. In the martingale representation \(\widetilde{V}_t = \widetilde{V}_0 + \int_0^t Z_s \, \mathrm{d}W_s\), the integrand \(Z_t\) satisfies \(Z_t = B_t^{-1}\sigma S_t \Delta(t, S_t)\). For a European call in the Black--Scholes model, express \(Z_t\) explicitly in terms of \(S_t\), \(\sigma\), \(r\), \(\tau\), and the normal CDF \(N(\cdot)\).

Solution to Exercise 2

In the Black-Scholes model, the European call delta is \(\Delta(t, S_t) = N(d_1)\) where:

\[ d_1 = \frac{\ln(S_t / K) + (r + \frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}, \qquad \tau = T - t \]

The integrand in the martingale representation is:

\[ Z_t = B_t^{-1}\sigma S_t \Delta(t, S_t) = e^{-rt}\sigma S_t N(d_1) \]

Equivalently, using \(\widetilde{S}_t = e^{-rt}S_t\):

\[ Z_t = \sigma \widetilde{S}_t N(d_1) \]

This gives the explicit martingale representation:

\[ \widetilde{V}_t = \widetilde{V}_0 + \int_0^t \sigma \widetilde{S}_s N(d_1(s, S_s)) \,\mathrm{d}W_s \]

where \(d_1(s, S_s)\) is evaluated at time \(s\) with spot \(S_s\) and time-to-maturity \(T - s\).

---

Exercise 3. Show that if \(V(t,S)\) satisfies the Black--Scholes PDE, applying Ito's formula to \(\widetilde{V}_t = e^{-rt}V(t,S_t)\) produces a process with zero \(\mathrm{d}t\) drift, confirming that \(\widetilde{V}_t\) is a martingale.

Solution to Exercise 3

Apply Ito's formula to \(\widetilde{V}_t = e^{-rt}V(t, S_t)\):

\[ \mathrm{d}\widetilde{V}_t = e^{-rt}\!\left[-rV + V_t + rS_t V_S + \frac{1}{2}\sigma^2 S_t^2 V_{SS}\right]\mathrm{d}t + e^{-rt}\sigma S_t V_S \,\mathrm{d}W_t \]

The \(\mathrm{d}t\) coefficient can be rewritten as:

\[ e^{-rt}\!\left[\left(V_t + rS_t V_S + \frac{1}{2}\sigma^2 S_t^2 V_{SS}\right) - rV\right] \]

The Black-Scholes PDE states that \(V_t + rSV_S + \frac{1}{2}\sigma^2 S^2 V_{SS} - rV = 0\). Therefore the entire \(\mathrm{d}t\) drift vanishes, and we are left with:

\[ \mathrm{d}\widetilde{V}_t = e^{-rt}\sigma S_t V_S(t, S_t)\,\mathrm{d}W_t \]

Since the drift is zero, \(\widetilde{V}_t\) is a local martingale. Under standard integrability conditions (which hold for the Black-Scholes call and put), it is a true martingale.

---

Exercise 4. In a complete market, the martingale representation theorem guarantees uniqueness of the integrand \(Z_t\). Explain why this uniqueness is essential for the claim that "delta is the hedge ratio." What happens to this argument in an incomplete market?

Solution to Exercise 4

Uniqueness and hedging. The martingale representation theorem in a Brownian filtration states that for any square-integrable \(\mathcal{F}_T\)-measurable random variable \(\widetilde{V}_T\), there exists a unique predictable process \(Z_t\) such that:

\[ \widetilde{V}_t = \widetilde{V}_0 + \int_0^t Z_s \,\mathrm{d}W_s \]

The uniqueness of \(Z_t\) is essential because it implies there is exactly one hedge ratio \(\Delta_t\) (determined by \(Z_t = B_t^{-1}\sigma S_t \Delta_t\)) that replicates the claim. If the integrand were not unique, multiple hedging strategies could replicate the same claim, and "the" delta hedge would be ambiguous.

Incomplete markets. In an incomplete market (e.g., with jumps or stochastic volatility driven by an additional Brownian motion), the martingale representation theorem with respect to a single Brownian motion fails. Not every square-integrable martingale can be written as a stochastic integral against \(W\) alone. Consequently:

• Perfect replication is generally impossible.
• There is no unique hedge ratio; instead one must choose a hedging criterion (e.g., minimum variance, superhedging, utility-based).
• The risk-neutral measure \(\mathbb{Q}\) is not unique, and different choices of \(\mathbb{Q}\) lead to different prices and different "deltas."

---

Exercise 5. Consider a portfolio that replicates a European claim \(H\) using \(\Delta_t\) shares of the underlying and a position in the money market. Write the self-financing condition for the discounted portfolio and show that it is equivalent to the martingale representation of \(\widetilde{V}_t\).

Solution to Exercise 5

Consider a self-financing portfolio consisting of \(\Delta_t\) shares of \(S\) and a money market position. The portfolio value is \(\Pi_t = \Delta_t S_t + \psi_t B_t\). The self-financing condition is:

\[ \mathrm{d}\Pi_t = \Delta_t \,\mathrm{d}S_t + \psi_t \,\mathrm{d}B_t \]

Define the discounted portfolio \(\widetilde{\Pi}_t = B_t^{-1}\Pi_t\). Then:

\[ \mathrm{d}\widetilde{\Pi}_t = \Delta_t \,\mathrm{d}\widetilde{S}_t = \Delta_t \sigma \widetilde{S}_t \,\mathrm{d}W_t \]

where we used the self-financing condition and the fact that \(\mathrm{d}\widetilde{S}_t = \sigma\widetilde{S}_t\,\mathrm{d}W_t\). Integrating:

\[ \widetilde{\Pi}_t = \widetilde{\Pi}_0 + \int_0^t \Delta_s \sigma \widetilde{S}_s \,\mathrm{d}W_s \]

For this portfolio to replicate the European claim \(H\), we need \(\widetilde{\Pi}_t = \widetilde{V}_t\) for all \(t\). Comparing with the martingale representation:

\[ \widetilde{V}_t = \widetilde{V}_0 + \int_0^t Z_s \,\mathrm{d}W_s \]

we see that replication requires \(\widetilde{\Pi}_0 = \widetilde{V}_0\) (correct initial investment) and:

\[ \Delta_t \sigma \widetilde{S}_t = Z_t = B_t^{-1}\sigma S_t \Delta(t, S_t) \]

Since \(\widetilde{S}_t = B_t^{-1}S_t\), this simplifies to \(\Delta_t = \Delta(t, S_t) = V_S(t, S_t)\), confirming that the self-financing replicating strategy uses the PDE delta as the hedge ratio.

---

Exercise 6. Suppose \(\sigma\) is replaced by a deterministic time-varying function \(\sigma(t)\) in the Black--Scholes model. Does the martingale representation still hold? Does the identification \(Z_t = B_t^{-1} \sigma(t) S_t \Delta(t, S_t)\) remain valid? Justify your answers.

Solution to Exercise 6

Does the martingale representation still hold? Yes. The martingale representation theorem applies in any filtration generated by a Brownian motion \(W\). Whether \(\sigma\) is constant or a deterministic function \(\sigma(t)\) does not affect this: the filtration is still \(\mathcal{F}_t = \sigma(W_s : s \le t)\), and any square-integrable \(\mathcal{F}_T\)-measurable martingale can be written as a stochastic integral against \(W\).

Does the identification remain valid? Yes. The dynamics become \(\mathrm{d}S_t = rS_t\,\mathrm{d}t + \sigma(t)S_t\,\mathrm{d}W_t\), and the discounted asset satisfies:

\[ \mathrm{d}\widetilde{S}_t = \sigma(t)\widetilde{S}_t\,\mathrm{d}W_t \]

Applying Ito's formula to \(\widetilde{V}_t = e^{-rt}V(t, S_t)\) with the time-dependent volatility PDE:

\[ V_t + rSV_S + \frac{1}{2}\sigma(t)^2 S^2 V_{SS} - rV = 0 \]

the drift cancels exactly as before, yielding:

\[ \mathrm{d}\widetilde{V}_t = e^{-rt}\sigma(t)S_t V_S(t, S_t)\,\mathrm{d}W_t \]

Therefore:

\[ Z_t = B_t^{-1}\sigma(t)S_t\,\Delta(t, S_t) \]

The only change is that \(\sigma\) is replaced by \(\sigma(t)\) everywhere. The structural relationship between the martingale integrand and delta is preserved because the model remains complete (one source of randomness, one tradable asset).