9 Continuous-time Finance

Black-Scholes model, Portfolios in continuous time, Girsanov’s Theorem, Change of Measure, Black-Scholes formula for Call options, Fundamental Theorem of Asset Pricing

We begin by choosing a model for our asset price:

Definition 59. We say an asset price \(S_t\) obeys the Black-Scholes model if: \[\begin{align} \tag{9.1} S_t = s_0 \exp\left\{ \sigma B_t + \left(\mu-\frac{1}{2} \sigma^2\right) t\right\}\end{align}\] where \(\sigma\) is the volatility parameter, \(\mu\) is the mean rate of return of the asset, and \(s_0\) the value of the asset at time \(0\).

This means that \(S_t\) is the solution to the Stochastic Differential Equation: \[\mathrm{d}S_t = S_t \sigma \mathrm{d}B_t + S_t \mu \mathrm{d}t, \quad S_0 = s_0.\]

In discrete time, we found it helpful to consider the discounted asset price. In continuous time, we discount from time \(t\) to time \(0\) by multiplying by \(\mathrm{e}^{-rt}\) where \(r\) was the (continuously compounded) interest rate. So the discounted asset price will be: \[S_t^* = \mathrm{e}^{-rt} S_t = s_0 \exp\left\{ \sigma B_t + \left(\mu-r-\frac{1}{2} \sigma^2\right) t\right\}\] and so also \[\begin{align} \mathrm{d}S_t^* & = & S_t^* \sigma \, \mathrm{d}B_t + S_t^* (\mu - r) \, \mathrm{d}t \tag{9.2}\\ & = & \mathrm{e}^{-rt} ( S_t \sigma \, \mathrm{d}B_t + S_t (\mu - r) \, \mathrm{d}t) \nonumber\\ & = & \mathrm{e}^{-rt} ( \mathrm{d}S_t - r S_t \, \mathrm{d}t) \tag{9.3}\end{align}\]

What about the value of our portfolio? Suppose we have a portfolio worth \(V_t\) at time \(t\), which consists of \(\phi_t\) units of the asset (worth \(\phi_t S_t\)), and the remaining cash (\(V_t - \phi_t S_t\)) invested in the bank account. Then at time \(t+\mathrm{d}t\), the portfolio has value: \[V_{t + \mathrm{d}t} = \phi_t S_{t + \mathrm{d}t} + \mathrm{e}^{r \mathrm{d}t} (V_t - \phi_t S_t)\] since \(1\) in the bank at time \(t\) is worth \(\mathrm{e}^{r (t+\mathrm{d}t)-rt} = \text{}\mathrm{e}^{r\mathrm{d}t}\) at time \(t+\mathrm{d}t\). Therefore: \[\begin{align} \mathrm{d}V_t & = & V_{t + \mathrm{d}t} - V_t \\ & = & \left(\phi_t S_{t + \mathrm{d}t} + \mathrm{e}^{r \mathrm{d}t} (V_t - \phi_t S_t)\right) - \left(\phi_t S_t + V_t - \phi_t S_t\right)\\ & = & \phi_t (S_{t + \mathrm{d}t} - S_t) + (\mathrm{e}^{r \mathrm{d}t} - 1)(V_t - \phi_t S_t)\\ & = & \phi_t \, \mathrm{d}S_t + r (V_t - \phi_t S_t) \, \mathrm{d}t\end{align}\] since \(\mathrm{e}^{r \mathrm{d}t} \approx 1 + r \mathrm{d}t\).

By itself, this isn’t very useful, but what happens if we consider the discounted portfolio value, \(V_t^* = \mathrm{e}^{-rt} V_t\)?

For example, by applying It’s Lemma to the function \(f(x,t) = \mathrm{e}^{-rt} x\), we get: \[\begin{align} \mathrm{d}(V_t^*) & = & \mathrm{d}(\mathrm{e}^{-rt} V_t) \nonumber\\ & = & \mathrm{e}^{-rt} \mathrm{d}V_t - r \mathrm{e}^{-rt} V_t \mathrm{d}t\nonumber\\ & = & \mathrm{e}^{-rt}(\phi_t \, \mathrm{d}S_t + r (V_t - \phi_t S_t) \, \mathrm{d}t) - r \mathrm{e}^{-rt} V_t \mathrm{d}t\nonumber\\ & = & \mathrm{e}^{-rt} \phi_t \, \mathrm{d}S_t - r \mathrm{e}^{-rt}\phi_t S_t \, \mathrm{d}t \nonumber\\ & = & \phi_t \, \mathrm{d}S_t^* \tag{9.4}\end{align}\] where we have used (9.3) in the last line.

This result should not be unexpected! Compare this last result with Lemma37 — the result says that the change in the discounted value of the portfolio is equal to the number of units of the asset held, times the change in the (discounted) asset value. In discrete time, we then found a new probability measure, under which the discounted asset price was a martingale. Can we do the same thing here?

Theorem 60 (Girsanov’s Theorem). Fix \(T >0\) and let \((B_t)_{t \in [0,T]}\) be a standard Brownian motion. Define \[Z_t = \exp\left\{\alpha B_t - \frac{1}{2}\alpha^2 t \right\}, \quad t \in [0,T],\] for \(\alpha \in \mathbb{R}\). Let \(\widetilde{\mathbb{E}} \left[ \cdot \right]\) be the change of measure of \(\mathbb{E}\left[ \cdot \right]\) induced by the Radon-Nikodym derivative \(Z_T\), so: \[\widetilde{\mathbb{E}} \left[ Y \right] = \mathbb{E}\left[ Z_T \cdot Y \right]\] for any r.v. \(Y\).

Then \(\widetilde{B}_t = B_t - \alpha t\) is a standard Brownian motion under \(\widetilde{\mathbb{E}}\).

Girsanov’s Theorem says that we can change the drift of the Brownian motion through a suitable change of measure! How might we use this? Suppose we can ‘change’ the drift of the Brownian motion so that \(\mu = r\) by switching from \(B_t\) to \(\widetilde{B}_t\), then (9.2) simplifies to: \[\mathrm{d}S_t^* = \mathrm{e}^{-rt} S_t \sigma \, \mathrm{d}\widetilde{B}_t\] so that \(S_t^*\) is a martingale, which is exactly what we wanted.

(Non-examinable) proof of Theorem60. We prove this result by showing that, under \(\widetilde{\mathbb{E}}\), then \(\widetilde{B}_t = B_t - \alpha t\) is a standard Brownian motion.

Observe first that we have \(\widetilde{B}_0 = B_0 - \alpha \cdot 0 = 0\). In addition, \(B_t\) is continuous, and so is \(\alpha t\), so \(\widetilde{B}_t\) is also continuous.

Now recall that the moment generating function of a random variable, \(g_Y(\gamma) = \mathbb{E}\left[ \mathrm{e}^{\gamma Y} \right]\) determines the distribution of \(Y\) uniquely (that is, if \(X\) and \(Y\) have the same moment generating function, then their distribution is the same), and the moment generating function of \(Y \sim N(\mu, \sigma^2)\) is: \(g_Y(\gamma) = \mathbb{E}\left[ \mathrm{e}^{\gamma Y} \right] = \mathrm{e}^{\mu \gamma + \frac{1}{2} \gamma^2 \sigma^2}\).

It follows that \(\widetilde{B}_t - \widetilde{B}_s \sim N(0,(t-s))\) for \(0 \le s \le t \le T\) under \(\widetilde{\mathbb{E}}\) if we can show: \[\widetilde{\mathbb{E}} \left[ \mathrm{e}^{\gamma (\widetilde{B}_t - \widetilde{B}_s)} \right] = \mathrm{e}^{\frac{1}{2}\gamma^2 (t-s)}.\] Now, by the definition of \(\widetilde{\mathbb{E}}\) and \(\widetilde{B}_t\), we have: \[\begin{align} \widetilde{\mathbb{E}} \left[ \mathrm{e}^{\gamma (\widetilde{B}_t - \widetilde{B}_s)} \right] & = & \mathbb{E}\left[ \mathrm{e}^{\gamma\left( B_t - B_s - \alpha(t-s)\right)} \mathrm{e}^{\alpha B_T - \frac{1}{2}\alpha^2 T} \right]\\ & = & \mathrm{e}^{-\gamma \alpha (t-s)} \mathbb{E}\left[ \mathrm{e}^{ \gamma\left( B_t - B_s\right) + \alpha B_T - \frac{1}{2}\alpha^2 T} \right]\end{align}\] Now, observe that we can write: \[\begin{align} \alpha B_T - \frac{1}{2}\alpha^2 T & = \left(\alpha B_s - \frac{1}{2} \alpha^2 s\right) + \left(\alpha (B_t-B_s) - \frac{1}{2}\alpha^2 (t-s)\right) \\ & \quad {} + \left(\alpha (B_T-B_t) - \frac{1}{2}\alpha^2 (T-t)\right).\end{align}\] Also, by the independence of Brownian increments, we know \(B_s\), \(B_t-B_s\) and \(B_T-B_t\) are all independent (under \(\mathbb{P}\)). Hence: \[\begin{align} \widetilde{\mathbb{E}} \left[ \mathrm{e}^{\gamma (\widetilde{B}_t - \widetilde{B}_s)} \right] & = & \mathrm{e}^{-\gamma \alpha (t-s)} \times \\ && \qquad \mathbb{E}\left[ \mathrm{e}^{ \gamma\left( B_t - B_s\right) +\left(\alpha B_s - \frac{1}{2}\alpha^2 s\right) + \left(\alpha (B_t-B_s) - \frac{1}{2}\alpha^2 (t-s)\right) + \left(\alpha (B_T-B_t) - \frac{1}{2}\alpha^2 (T-t)\right)} \right]\\ & = & \mathrm{e}^{-\gamma \alpha (t-s) - \frac{1}{2}\alpha^2 (t-s)} \mathbb{E}\left[ \mathrm{e}^{ \gamma\left( B_t - B_s\right) + \alpha (B_t-B_s)} \right] \mathbb{E}\left[ \mathrm{e}^{\alpha B_s - \frac{1}{2}\alpha^2 s} \right] \times \\ && \qquad \mathbb{E}\left[ \mathrm{e}^{\alpha (B_T-B_t) - \frac{1}{2}\alpha^2 (T-t)} \right]\\ & = & \mathrm{e}^{-\frac{1}{2}(\alpha^2 + 2 \gamma \alpha) (t-s)} \mathbb{E}\left[ \mathrm{e}^{ (\gamma+ \alpha)\left( B_t - B_s\right)} \right]\\ & = & \mathrm{e}^{-\frac{1}{2}(\alpha^2 + 2 \gamma \alpha) (t-s)} \mathrm{e}^{\frac{1}{2} (\gamma + \alpha)^2 (t-s)}\\ & = & \mathrm{e}^{\frac{1}{2}\gamma^2 (t-s)}\end{align}\] where in the second last line, we have used the fact that \(B_t-B_s \sim N(0,(t-s))\), and we know the moment generating function for a normal random variable. So it follows that \(\widetilde{B}_t-\widetilde{B}_s \sim N(0,t-s)\) under \(\widetilde{\mathbb{E}}\).

Finally, we need to show that if \(0 \le t_1 \le t_2 \le t_3 \le t_4 \le T\) then \(\widetilde{B}_{t_4}-\widetilde{B}_{t_3} \perp \widetilde{B}_{t_2} - \widetilde{B}_{t_1}\). To show this, it is enough to show that \[\widetilde{\mathbb{E}} \left[ f(\widetilde{B}_{t_4}-\widetilde{B}_{t_3}) g(\widetilde{B}_{t_2} - \widetilde{B}_{t_1}) \right] = \widetilde{\mathbb{E}} \left[ f(\widetilde{B}_{t_4}-\widetilde{B}_{t_3}) \right] \widetilde{\mathbb{E}} \left[ g(\widetilde{B}_{t_2} - \widetilde{B}_{t_1}) \right]\] for arbitrary functions \(f\) and \(g\).

Using similar arguments as above, it is not too hard to show: \[\begin{align} && \widetilde{\mathbb{E}} \left[ f(\widetilde{B}_{t_4}-\widetilde{B}_{t_3}) g(\widetilde{B}_{t_2} - \widetilde{B}_{t_1}) \right]\\ & = & \mathbb{E}\Big[f(B_{t_4}-B_{t_3}-\alpha(t_4-t_3)) g(B_{t_2} - B_{t_1}-\alpha(t_2-t_1)) \times \\ && \qquad\mathrm{e}^{\alpha(B_{t_4}-B_{t_3})-\frac{1}{2} \alpha^2(t_4-t_3)} \mathrm{e}^{\alpha(B_{t_2}-B_{t_1})-\frac{1}{2} \alpha^2(t_2-t_1)}\Big] \\ & = & \mathbb{E}\left[ \widetilde{f}(B_{t_4}-B_{t_3}) \mathrm{e}^{\alpha(B_{t_4}-B_{t_3})-\frac{1}{2} \alpha^2(t_4-t_3)} \right] \mathbb{E}\left[ \widetilde{g}(B_{t_2} - B_{t_1}) \mathrm{e}^{\alpha(B_{t_2}-B_{t_1})-\frac{1}{2} \alpha^2(t_2-t_1)} \right]\\ & = & \widetilde{\mathbb{E}} \left[ \widetilde{f}(B_{t_4}-B_{t_3}) \right] \widetilde{\mathbb{E}} \left[ \widetilde{g}(B_{t_2} - B_{t_1}) \right] \\ & = & \widetilde{\mathbb{E}} \left[ f(\widetilde{B}_{t_4}-\widetilde{B}_{t_3}) \right]\widetilde{\mathbb{E}} \left[ g(\widetilde{B}_{t_2} - \widetilde{B}_{t_1}) \right]\end{align}\] where we have written \(\widetilde{f}(x) = f(x-\alpha(t_4-t_3)), \widetilde{g}(x) = g(x-\alpha(t_2-t_1))\). This shows that the increments are independent.

By Girsanov’s Theorem, under the probability measure \(\widetilde{\mathbb{P}}\) (corresponding to \(\widetilde{\mathbb{E}}\)), \(\widetilde{B}_t = B_t - \alpha t\) is a standard Brownian motion. If we substitute \(B_t = \widetilde{B}_t + \alpha t\) into (9.1) , we get: \[\begin{align} S_t & = & s_0 \exp\left\{ \sigma (\widetilde{B}_t + \alpha t) + \left(\mu-\frac{1}{2} \sigma^2\right) t\right\}\\ & = & s_0 \exp\left\{ \sigma \widetilde{B}_t + \left(\mu + \sigma \alpha - \frac{1}{2}\sigma^2 \right) t \right\}.\end{align}\]

We want the discounted asset price to be a martingale under \(\widetilde{\mathbb{P}}\), and we said this would happen exactly when we had \(r\) instead of \(\mu\) in the exponential, so we want: \[\mu + \sigma \alpha = r \quad \implies \quad \alpha = \frac{r-\mu}{\sigma}.\] By analogy with the discrete case, we call \(\widetilde{\mathbb{P}}\) the risk-neutral probability measure, where \(\widetilde{\mathbb{P}}\) is the change of measure of \(\mathbb{P}\) given by the Radon-Nikodym derivative: \[Z_T = \exp\left\{ \frac{r-\mu}{\sigma} B_T- \frac{1}{2} \left(\frac{r-\mu}{\sigma}\right)^2 T \right\}.\]

Then we have the continuous-time analogue of Theorem38:

Lemma 61. In the Black-Scholes model, under the probability measure \(\widetilde{\mathbb{P}}\), the discounted asset price \(S_t^*\) is a martingale.

Proof. Since \(B_t = \widetilde{B}_t + \alpha t\), then \(\mathrm{d}B_t = \mathrm{d} \widetilde{B}_t + \alpha \, \mathrm{d}t\), so (9.2) can be rewritten as: \[\begin{align} \mathrm{d}S_t^* & = & S_t^* \sigma \, \mathrm{d}B_t + S_t^* (\mu - r) \, \mathrm{d}t \\ & = & S_t^* \sigma \, \mathrm{d}\widetilde{B}_t + S_t^* \sigma \left( \frac{r - \mu}{\sigma}\right) \, \mathrm{d}t+ S_t^* (\mu - r) \, \mathrm{d}t\\ & = & S_t^* \sigma \, \mathrm{d}\widetilde{B}_t.\end{align}\] And hence \(S_t^*\) is a martingale under \(\widetilde{\mathbb{P}}\) (since then \(\widetilde{B}_t\) is a Brownian motion).

We now want to prove the Fundamental Theorem of Asset Pricing in continuous time (c.f. Theorem41). This result had two main ingredients: we needed our portfolios of traded assets to be martingales, and we needed the Martingale Representation Theorem (Theorem40). The corresponding version in continuous time says:

Theorem 62 (Martingale Representation Theorem). Let \(\widetilde{\mathbb{P}}\) be a probability space on which \((\widetilde{B}_t)_{t \in [0,T]}\) is a Brownian motion. Let \(Y\) be a random variable with \(\widetilde{\mathbb{E}} \left[ Y^2 \right] < \infty\) which depends only on \((\widetilde{B}_t)_{t \in [0,T]}\). Then there exists a process \(\phi_t\), which depends only on the Brownian motion up to time \(t\) such that \[Y = \widetilde{\mathbb{E}} \left[ Y \right] + \int_0^T \phi_s \, \mathrm{d}\widetilde{B}_s.\]

We will not prove this result, but note simply that this is the continuous-time equivalent of Theorem40.

Then we can use this to show the continuous-time version of the Fundamental Theorem of Asset Pricing (FTAP):

Theorem 63 (Fundamental Theorem of Asset Pricing in Continuous Time). Suppose the asset \((S_t)_{t \in [0,T]}\) behaves according to the Black-Scholes model, and suppose \(\widetilde{\mathbb{P}}\) is the corresponding risk-neutral measure. If \(C_T\) is the payoff at time \(T\) of a derivative, whose value depends only on the path of \((S_t)_{t \in [0,T]}\), then there is no arbitrage¹⁶ if and only if the price of the derivative at time \(t\) is: \[\begin{align} \tag{9.5} C_t = \mathrm{e}^{-r(T-t)} \widetilde{\mathbb{E}}_{t} \left[ C_T \right].\end{align}\]

Non-examinable (sketch) proof. We first observe that, in the usual notation, we have the discounted price of the option: \(C_t^* = \mathrm{e}^{-rt} C_t\). In this notation, (9.5) becomes: \[\begin{align} \tag{9.6} C_t^* = \widetilde{\mathbb{E}}_{t} \left[ C_T^* \right].\end{align}\]

Suppose (9.5) holds for all \(t \in [0,T]\), and there exists an arbitrage. That is (extending (9.4) to include a second asset), we can find \(\phi_t^1, \phi_t^2\) such that \[V_t^* = \int_0^t \phi_s^1 \, \mathrm{d}S_s^* + \int_0^t \phi_s^2 \mathrm{d}C_s^*\] is an arbitrage (i.e. \(\mathbb{P}(V_T \ge 0)\) and \(\mathbb{P}(V_T >0) >0\)). Moreover, by (9.6) , \(C_t^*\) is a martingale under \(\widetilde{\mathbb{P}}\), and so is \(S_t^*\), so \[\int_0^t \phi_s^1 \, \mathrm{d}S_s^* = \int_0^t \phi_s^1 \sigma S_s^* \, \mathrm{d}\widetilde{B}_s\] is also a martingale. It is not so obvious that this is true for the second term in the expression of \(V_t^*\) which involves \(C_t^*\), and indeed this is harder. However, a continuous-time analogue of Theorem26 holds, and it follows that \(\int_0^t \phi_s^2 \mathrm{d}C\) is indeed a martingale. Hence \(V_t^*\) is a martingale with \(V_0^* = 0\) and so: \[0 = V_0^* = \widetilde{\mathbb{E}} \left[ V_T^* \right] = \widetilde{\mathbb{E}} \left[ \mathrm{e}^{-rT}V_T \right] = \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ V_T \right].\] However, if \(V_T\) is an arbitrage, then \(\widetilde{\mathbb{E}} \left[ V_T \right] >0\). So if (9.5) holds, there can be no arbitrage.

Suppose on the other hand that there exists no arbitrage, and consider \(C_T^*\). We show (9.5) holds when \(t=0\) (the more general case being similar in spirit). By Theorem62 we can find \((\phi_t)_{t \in [0,T]}\) such that \[C_T^* = \widetilde{\mathbb{E}} \left[ C_T^* \right] + \int_0^T \phi_t \, \mathrm{d}\widetilde{B}_t.\] Suppose \(C_0^* < \widetilde{\mathbb{E}} \left[ C_T^* \right]\). Then we buy the derivative at time \(0\), and hold \(- \frac{\phi_t}{S_t^* \sigma}\) units of the asset at time \(t\), investing/borrowing any remaining cash from the bank. Then \(V_0^* = 0\) and: \[\begin{align} \mathrm{d}V_t^* & = & \mathrm{d}C_t^* - \frac{\phi_t}{S_t^* \sigma} \,\mathrm{d} S_t^*\\ & = & \mathrm{d}C_t^* - \frac{\phi_t}{S_t^* \sigma} S_t^* \sigma \,\mathrm{d} \widetilde{B}_t \\ & = & \mathrm{d}C_t^* - \phi_t \,\mathrm{d}\widetilde{B}_t\end{align}\] Integrating from \(0\) to \(T\), we get: \[V_T^* - V_0^* = C_T^* - C_0^* - \int_0^T \phi_t \, \mathrm{d}\widetilde{B}_t = C_T^* - C_0^* - C_T^* + \widetilde{\mathbb{E}} \left[ C_T^* \right] = \widetilde{\mathbb{E}} \left[ C_T^* \right]-C_0^*.\] It follows that \(\widetilde{\mathbb{P}}(V_T > 0) = 1\), and so this is certainly an arbitrage.

If \(C_0^* > \widetilde{\mathbb{E}} \left[ C_T^* \right]\). Then we sell the derivative at time \(0\), and hold \(\frac{\phi_t}{S_t^* \sigma}\) units of the asset at time \(t\). An almost identical argument to that above tells us that this is an arbitrage. Hence in the absence of arbitrage, (9.5) holds.

We spend the rest of the course exploring some consequences of this result. The first case we consider is to use the FTAP to find the price of a Call option.

Theorem 64 (Black-Scholes Formula). In the Black-Scholes model, the arbitrage-free price of a European Call option with strike \(K\) and maturity \(T\) is: \[C(K,T) = s_0 \Phi(d_1) - K \mathrm{e}^{-rT} \Phi(d_2),\] where \[\begin{align} d_1 & = & \frac{\log\left(\frac{s_0}{K}\right) + \left(r + \frac{1}{2} \sigma^2\right) T}{\sigma \sqrt{T}}\\ d_2 & = & \frac{\log\left(\frac{s_0}{K}\right) + \left(r - \frac{1}{2} \sigma^2\right) T}{\sigma \sqrt{T}} = d_1 - \sigma \sqrt{T}\end{align}\]

In the theorem, \(\Phi(\cdot)\) is the cumulative (standard) normal distribution function, \[\Phi(x) = \int_{-\infty}^x \frac{\mathrm{e}^{-y^2/2}}{\sqrt{2\pi}} \, \mathrm{d}y.\]
Note that there is no \(\mu\) appearing in the formula! This should not be too surprising: under the risk-neutral measure, \(\mu\) is ‘replaced’ by \(r\), so \(\mu\) should not appear in the option price.
To understand how the Black-Scholes formula behaves, we look at the prices for different parameter values. See the plots in Figure4.

Figure 4: Plots of the Black-Scholes price of a European Call, plotted as a function of the strike, \(K\). The different plots show the effect on the price of varying the respective parameters by \(\pm 25\%\). The ‘straight’ line is the value of the option when \(\sigma = 0\) — which is \(\mathrm{e}^{-rT}(\mathrm{e}^{rT}s_0 - K)_+\). The solid curve in each graph corresponds to: \(s_0 = 100, r = 0.05, \sigma = 0.75, T = 0.5.\)

(a) Volatility \(\sigma\): upper curve is \(\sigma = 0.94\), lower curve is \(\sigma = 0.56\).

(b) Asset price \(s_0\): upper curve is \(s_0 = 125\), lower curve is \(s_0 = 75\).

(d) Interest Rate \(r\): upper curve is \(r = 0.063\), lower curve is \(r = 0.038\).

Proof of Theorem64. We use the fact that the price of the call option is the expectation of the payoff under the risk-neutral probability.

So the price is given by: \[\begin{align} C(K,T)& = & \widetilde{\mathbb{E}} \left[ \mathrm{e}^{-rT} (S_T - K)_+ \right] \\ & = & \widetilde{\mathbb{E}} \left[ \mathrm{e}^{-rT} \left( s_0 \exp\left\{ \sigma \widetilde{B}_T + (r-\frac{1}{2} \sigma^2)T\right\} - K\right)_+ \right] \end{align}\] Since the \((\dots)_+\) means we only need count when this is positive, we can compute the expectation over the set where \[s_0 \exp\left\{ \sigma \widetilde{B}_T + (r-\frac{1}{2} \sigma^2)T\right\} \ge K\]

In particular, the payoff is positive if: \[\widetilde{B}_T \ge \frac{1}{\sigma} \left[ \log\left( \frac{K}{s_0} \right) - (r-\frac{1}{2}\sigma^2) T\right] =: \lambda\] where we write \(\lambda\) for the term on the right.

If we write \(\widetilde{\mathbb{E}} \left[ X;A \right]\) to mean the expectation of \(X\) on the set \(A\) (i.e. the integral of \(X\) times its density on the set \(A\)), we get \[\begin{align} C(K,T) & = & \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ (S_T - K)_+ \right] \\ & = & \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ S_T - K; S_T \ge K \right] \\ & = & \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ S_T;S_T \ge K \right] - \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ K;S_T \ge K \right].\end{align}\]

Since \(K\) is just a constant, the second term here, \(\widetilde{\mathbb{E}} \left[ K;S_T \ge K \right]\), must be \(K\) times the probability (under the risk-neutral probability) that the process ends above \(K\).

For the first term, we get: \[\mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ S_T;S_T \ge K \right] = \]

Solution: \[\begin{align} e^{-rT}\widetilde{\mathbb{E}} \left[ S_T;S_T\ge K \right] & = \mathrm{e}^{-rT}\widetilde{\mathbb{E}} \left[ s_0\exp\left(\sigma \widetilde{B}_T+(r-\sigma^2/2)T\right);S_T\ge K \right]\\ & = s_0e^{-\frac{\sigma^2}{2}T}\widetilde{\mathbb{E}} \left[ \exp\left(\sigma \widetilde{B}_T\right)|;S_T\ge K \right]\\ & = s_0e^{-\frac{\sigma^2}{2}T}\widetilde{\mathbb{E}} \left[ \exp\left(\sigma \widetilde{B}_T\right);\widetilde{B}_T\ge \lambda \right]\end{align}\]

\(\square\)

Since \(\widetilde{B}_T\) is a Brownian motion (under \(\widetilde{\mathbb{P}}\)), \(\widetilde{B}_T \sim N(0,T)\), and we can calculate:

\[\widetilde{\mathbb{E}} \left[ \mathrm{e}^{\sigma \widetilde{B}_T} ; \widetilde{B}_T \ge \lambda \right] = \]

Solution: Using the fact that \(\widetilde{B}_T\) has \(N(0,T)\) distribution, we get \[\begin{align} \widetilde{\mathbb{E}} \left[ \mathrm{e}^{\sigma \widetilde{B}_T} ; \widetilde{B}_T \ge \lambda \right] & = \int_\lambda^\infty \mathrm{e}^{\sigma x} \frac{1}{\sqrt{2\pi T}} \mathrm{e}^{-x^2/2T} \, \mathrm{d}x\\ & = \int_\lambda^\infty \frac{1}{\sqrt{2\pi T}} \exp\left\{-\frac{x^2}{2T} + \frac{2T\sigma x}{2T} \right\} \, \mathrm{d}x\\ & = \int_\lambda^\infty \frac{1}{\sqrt{2\pi T}} \exp\left\{-\frac{(x-T\sigma)^2 - T^2 \sigma^2}{2T} \right\} \, \mathrm{d}x\\ & = \mathrm{e}^{T \sigma^2/2}\int_\lambda^\infty \frac{1}{\sqrt{2\pi T}} \exp\left\{-\frac{(x-T\sigma)^2}{2T} \right\} \, \mathrm{d}x\\ & = \mathrm{e}^{T \sigma^2/2}\int_{\lambda-T \sigma}^\infty \frac{1}{\sqrt{2\pi T}} \exp\left\{-\frac{y^2}{2T} \right\} \, \mathrm{d}y\\\end{align}\] where we used the substitution \(y=x-T\sigma\).

Making a further substitution \(u = y/\sqrt{T}\), we get \[\begin{align} \widetilde{\mathbb{E}} \left[ \mathrm{e}^{\sigma \widetilde{B}_T} ; \widetilde{B}_T \ge \lambda \right] & = \mathrm{e}^{T \sigma^2/2}\int_{\frac{\lambda-T \sigma}{\sqrt{T}}}^\infty \frac{1}{\sqrt{2\pi}} \exp\left\{-\frac{u^2}{2} \right\} \, \mathrm{d}u% \\ % & = \me^{T \sigma^2/2} \left( 1 - \Phi\left( \frac{\lambda-\sigmaT}{\sqrt{T}\right)\right) .\end{align}\]

\(\square\)

This final integrand is the density of an \(N(0,T)\), and this is integrated between \(\lambda - T\sigma\) and \(\infty\), so that this is just: \[1 - \Phi\left(\frac{\lambda - T\sigma}{\sqrt{T}}\right) = \Phi\left(\frac{ T\sigma- \lambda}{\sqrt{T}}\right)\] but note also that \[\begin{align} \frac{ T\sigma - \lambda}{\sqrt{T}} & = & \frac{T \sigma - \frac{1}{\sigma} \left[ \log\left( \frac{K}{s_0} \right) - (r-\frac{1}{2}\sigma^2) T\right]}{\sqrt{T}} \\ & = & \frac{ T \sigma^2 + \log\left( \frac{s_0}{K} \right) + (r-\frac{1}{2}\sigma^2) T}{\sigma \sqrt{T}}\\ & = & \frac{ \log\left( \frac{s_0}{K} \right) + (r+\frac{1}{2}\sigma^2) T}{\sigma \sqrt{T}} = d_1\end{align}\]

This gives us: \[\begin{align} \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ S_T;S_T \ge K \right] & = & s_0 \mathrm{e}^{-\frac{1}{2}\sigma^2 T} \widetilde{\mathbb{E}} \left[ \exp\left\{ \sigma \widetilde{B}_T\right\} ; \widetilde{B}_T \ge \lambda \right] \\ & = & s_0 \mathrm{e}^{-\frac{1}{2}\sigma^2 T} \mathrm{e}^{T \sigma^2/2} \Phi(d_1) \\ & = & s_0 \Phi(d_1)\end{align}\]

We now consider the term: \[\mathrm{e}^{-rT}\widetilde{\mathbb{E}} \left[ K ; S_T \ge K \right]\] But since \(K\) is fixed, and \(\{S_T \ge K\} = \{ \widetilde{B}_T \ge \lambda\}\), this term is just \(K\mathrm{e}^{-rT}\) times the probability a Brownian motion at time \(T\) is above \(\lambda\).

Which is the probability that a \(N(0,T)\) is greater than \(\lambda\): \[\begin{align} 1 - \Phi\left(\frac{\lambda}{\sqrt{T}}\right) & = & \Phi\left(-\frac{\lambda}{\sqrt{T}}\right) \\ & = & \Phi\left(-\frac{\frac{1}{\sigma} \left[ \log\left( \frac{K}{s_0} \right) - (r-\frac{1}{2}\sigma^2) T\right]}{\sqrt{T}} \right) \\ & = & \Phi\left(\frac{ \left[ \log\left( \frac{s_0}{K} \right) + (r-\frac{1}{2}\sigma^2) T\right]}{\sigma\sqrt{ T}}\right) \\ & = & \Phi\left( d_2\right)\end{align}\] Putting this all together, we get the price of the option to be: \[C(K,T) = s_0 \Phi(d_1) - \mathrm{e}^{-rT} K \Phi(d_2).\]

We can also use the Black-Scholes formula to give us the price of a Call option with maturity time \(T\) at any time \(t<T\). If the price of the asset a time \(t\) is \(S_t\), then this is equivalent to having a call option with maturity date \(T-t\) and with initial asset price \(S_t\). So we can find the price of the call option at time \(t\) by replacing \(T\) by \(T-t\) and \(s_0\) by \(S_t\). If we write the price as \(f(S_t,t)\) then \(f\) is given by: \[\begin{align} f(x,t) & = x \Phi\left(\frac{\log\left(\frac{x}{K}\right) + \left(r +\frac{1}{2} \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right) \nonumber \\ &\quad {}- \mathrm{e}^{-r(T-t)} K \Phi\left(\frac{\log\left(\frac{x}{K}\right) + \left(r -\frac{1}{2} \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right). \tag{9.7}\end{align}\]

Once we have a formula for the Call price, we can derive the price of a Put option with the same strike and maturity using Put-Call Parity.

The Payoff of a Call option is: \((S_T - K)_+\), and the payoff of a Put option is: \[\begin{align} (K-S_T)_+ & = & \max\{0,K-S_T\} \\ & = & \max\{0,-(S_T-K)\}\\ & = & - \min\{0,S_T-K\}\end{align}\]

So: \[\begin{align} (S_T - K)_+ - (K - S_T)_+ & = \max\{0,S_T-K\} + \min\{0,S_T-K\} \\ & = S_T-K.\end{align}\]

Solution:

image

\(\square\)

In particular, \[(K-S_T)_+ = (S_T - K)_+ -S_T + K.\] If we write \(P(K,T)\) for the price of the Put option with strike \(K\) and maturity \(T\), and \(C(K,T)\) for the price of the corresponding call option, from the FTAP we get: \[\begin{align} P(K,T) & = & \mathrm{e}^{-rT}\widetilde{\mathbb{E}} \left[ (K-S_T)_+ \right] \\ & = & \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ (S_T - K)_+ -S_T + K \right] \\ & = & \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ (S_T - K)_+ \right] - \widetilde{\mathbb{E}} \left[ \mathrm{e}^{-rT} S_T \right] + \mathrm{e}^{-rT} \widetilde{\mathbb{E}} \left[ K \right] \\ & = & C(K,T) - s_0 + \mathrm{e}^{-rT}K,\end{align}\] where we have used the fact that \(S_t^*\) is a martingale (under \(\widetilde{\mathbb{P}}\)), so \(\widetilde{\mathbb{E}} \left[ \mathrm{e}^{-rT} S_T \right] = S_0^* = s_0\) and \(K\) is a constant, so we can ‘take out what is known’. This formula is known as the Put-Call Parity formula.

Finally, recall that the Black-Scholes formula for the Call price was: \[C(K,T) = s_0 \Phi(d_1) - K \mathrm{e}^{-rT} \Phi(d_2),\] We can rearrange to get a formula for the price of the Put option: \[\begin{align} P(K,T) & = & s_0 \Phi(d_1) - K \mathrm{e}^{-rT} \Phi(d_2) - s_0 + \mathrm{e}^{-rT} K\\ & = & s_0(\Phi(d_1)-1) + K\mathrm{e}^{-rT}(1-\Phi(d_2))\\ & = & K \mathrm{e}^{-rT} \Phi(-d_2) - s_0 \Phi(-d_1).\end{align}\] (Recall that \(1-\Phi(x) = \Phi(-x)\) by properties of the standard Normal distribution).

Finally, we consider the problem of hedging an option in continuous time. Can we find the hedging portfolio? Suppose I sell an option with payoff \(g(S_T)\) at time \(T\), and I wish to trade in the underlying stock in a way that I have \(g(S_T)\) at time \(T\). Suppose the option is worth¹⁷ \(f(S_t,t)\) at time \(t\), and we are short the option, long \(\phi_t\) units of the asset, and hold \(f(S_t,t)-\phi_t S_t\) in the bank, so our portfolio has value \(V_t = (f(S_t,t) - \phi_t S_t) + \phi_t S_t - f(S_t,t)\) (the value of the cash, asset and option terms respectively).

Now consider the change in the value of our portfolio from time \(t\) to time \(t+\mathrm{d}t\). For each pound in the bank, the bank account grows by \(r \, \mathrm{d}t\), while the change in our holding in the asset is \(\phi_t\) times the change in \(S_t\), and the change in the value of the option is \(\mathrm{d}(f(S_t,t))\). Then: \[\mathrm{d}V_t = (f(S_t,t)-\phi_t S_t) r \, \mathrm{d}t+ \phi_t \, \mathrm{d}S_t - \mathrm{d} (f(S_t,t)).\]

By It’s Lemma: \[\mathrm{d}(f(S_t,t)) = \frac{\partial f}{\partial x}(S_t,t) \, \mathrm{d}S_t + \frac{1}{2} \frac{\partial ^2f}{\partial x^2}(S_t,t) \, (\mathrm{d}S_t)^2 + \frac{\partial f}{\partial t}(S_t,t) \, \mathrm{d}t.\] Also \[\begin{align} (\mathrm{d}S_t)^2 & = & (S_t \sigma \, \mathrm{d}\widetilde{B}_t + r S_t \, \mathrm{d}t)^2\\ & = & S_t^2 \sigma^2 \, (\mathrm{d}\widetilde{B}_t)^2 + 2 S_t^2 \sigma r \, \mathrm{d} \widetilde{B}_t \, \mathrm{d}t+r^2 S_t^2 \, (\mathrm{d}t)^2 \\ & = & S_t^2 \sigma^2 \, \mathrm{d}t\end{align}\]

So \[\begin{align} \mathrm{d}V_t = (f(S_t,t)-\phi_t S_t) r \, \mathrm{d}t+ \left(\phi_t - \frac{\partial f}{\partial x}(S_t,t)\right) \, \mathrm{d}S_t - \left(\frac{1}{2} S_t^2 \sigma^2 \frac{\partial ^2f}{\partial x^2}(S_t,t) + \frac{\partial f}{\partial t}(S_t,t)\right) \, \mathrm{d}t.\end{align}\] Now suppose we choose \(\phi_t = \frac{\partial f}{\partial x}(S_t,t)\). Then the \(\mathrm{d} S_t\) term disappears and we have: \[\begin{align} \mathrm{d}V_t = \Big(r f(S_t,t)-r S_t \frac{\partial f}{\partial x}(S_t,t) - \frac{1}{2} S_t^2 \sigma^2 \frac{\partial ^2f}{\partial x^2}(S_t,t) - \frac{\partial f}{\partial t}(S_t,t)\Big) \, \mathrm{d}t.\end{align}\]

Remember that we chose our portfolio such that \(V_t = 0\). If the \(\mathrm{d}t\) term is greater than \(0\), we can create a portfolio worth \(0\) at time \(t\) which will have a strictly positive value at time \(t+\mathrm{d}t\). This means that an arbitrage exists! A similar argument holds if the \(\mathrm{d}t\) term is less than zero if we buy the option, go short \(\frac{\partial f}{\partial x}(S_t,t)\) units of the asset, and borrow/invest the rest in the bank.

So:

If the price of the option at time \(t\) is \(f(S_t,t)\) and we wish to hedge a short position in the option, we should hold \(\frac{\partial f}{\partial x}(S_t,t)\) units of the asset. The amount \(\frac{\partial f}{\partial x}(S_t,t)\) is often called the delta of the option.
If the price of the option at time \(t\) can be written as a function \(f(S_t,t)\), then the function \(f(x,t)\) should satisfy the Black-Scholes PDE: \[r x \frac{\partial f}{\partial x}(x,t) + \frac{1}{2} x^2 \sigma^2 \frac{\partial ^2f}{\partial x^2}(x,t) + \frac{\partial f}{\partial t}(x,t) = r f(x,t).\] The Black-Scholes PDE gives us another way to try and compute option prices: try to solve the PDE together with a boundary condition \(f(x,T) = g(x)\), where \(g(S_T)\) is the payoff of the option at the maturity date.

Example 65. Compute the Delta of a Call option under the Black-Scholes model.

Recall the value of a Call option at time \(t\) with asset price \(S_t\), strike \(K\) and maturity \(T\) was given by (9.7) . To find the delta of the option, we differentiate \(f(x,t)\) by \(x\) (that is, the \(S_t\) variable).

Solution: Write \[\begin{align} d_1(x,t) & = \frac{\log(x/K) + \left( r + \frac{1}{2}\sigma^2 \right) (T-t)}{\sigma \sqrt{T-t}}\\ d_2(x,t) & = d_1(x,t) - \sigma \sqrt{T-t}.\end{align}\] Then \(\frac{\partial d_1}{\partial x}(x,t)=\frac{\partial d_2}{\partial x}(x,t)=\frac{\partial }{\partial x}\left\{\frac{\log x}{\sigma \sqrt{T-t}}\right\} =\frac{1}{x\sigma \sqrt{T-t}}\). Then \(f(x,t) = x \Phi\left(d_1(x,t)\right)- e^{-r(T-t)} K \Phi\left(d_2(x,t)\right)\) implies \[\begin{align} \frac{\partial f}{\partial x}(x,t)= &\Phi(d_1(x,t))+\frac{\partial d_1}{\partial x} (x,t) \left(x \Phi'\left(d_1(x,t)\right)- e^{-r(T-t)} K \Phi'\left(d_2(x,t)\right)\right)% \\ % =&\Phi(d_1(x,t))+\frac{x \Phi'\left(d_1(x,t)\right)- e^{-r(T-t)} K \Phi'\left(d_2(x,t)\right)}{x\sigma \sqrt{T-t}} .\end{align}\] Since \(\Phi'(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\) and \(d_1(x,t)=\frac{\log\left(\frac{x}{K}\right) + \left(r +\frac12\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\), we obtain \[\begin{align} \Phi'(d_2) & = \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{\left(d_1(x,t)-\sigma\sqrt{T-t}\right)^2}{2}}=\Phi'(d_1(x,t))\mathrm{e}^{\sigma\sqrt{T-t}d_1(x,t)-\frac{\sigma^2(T-t)}{2}}\\ & = \Phi'(d_1(x,t))\frac{x}{K\mathrm{e}^{-r(T-t)}}\end{align}\] Substituting into the expression above, we get \(\frac{\partial f}{\partial x}(x,t)=\Phi(d_1(x,t))\), i.e., \(\Delta=\Phi(d_1(S_t,t))\). \(\square\)

Beyond Black-Scholes:

The Black-Scholes theory relies on a number of assumptions about the way the market behaves:

To get more accurate prices, we need to relax assumptions, and construct more complicated models — for example: stochastic volatility models. \(\sigma\) is not a constant, but (e.g.) \(\sigma_t^2\) is the solution to: \[\mathrm{d}(\sigma_t^2) = \theta (\overline{\sigma}^2 - \sigma_t^2) \, \mathrm{d}t + \xi \sigma_t \, \mathrm{d}B_t.\]
In the Black-Scholes model, all risk was hedgeable. In practice, can’t hedge all risk, some risk will remain. Regulators and managers will care about the associated risk. How to model/adjust for this? What about counterparty risk?
Other important issues: stochastic interest rates, American-style options, market incompleteness, market imperfections (transaction costs, illiquid markets, taxes), energy & commodity markets, …

THE END

There is an extra concern here in continuous time, which we will not look at too closely, but which would be an issue if we were to try and make this rigorous: if we can choose our trading strategy without constraint, we might not choose a strategy for which the integrand is in \(\mathcal{V}\), and so our wealth process may not be a martingale. However, it is not economically meaningful to force us to choose trading strategies in \(\mathcal{V}\). Included in the class of possible trading strategies are ‘doubling’ strategies (recall Exercise sheet question Q2.4) which guarantee that we make money at a fixed time, provided we can sustain arbitrarily large losses: the way to ensure these are not possible is to restrict our trading strategies to those which keep the portfolio value above some arbitrary fixed loss, and this rules out doubling strategies, and essentially preserves the martingale property for our wealth.↩︎
Note that we know the value will be a function of the current asset value, and the time, since Brownian motion is a Markov process, and hence so too is \(S_t\). But the price of the option is \(\mathrm{e}^{-r(T-t)}\widetilde{\mathbb{E}}_{t} \left[ g(S_T) \right]\), which by the Markov property of \(S_t\), is a function of \(t\) and \(S_t\) only.↩︎