Self-Financing Portfolios and the Black–Scholes Equation

# Black Scholes differential


# Black Scholes differential












I'm studying a BS derivation and I don't understand one part .We have a portfolio consisting of $\Delta(t)S(t)+B(t)$ where the first term is risky and the second is a riskless bond. The part i don't understand is why when we take the differential of this portfolio we obtain: $\Delta dS +dB$. I understand the reason is related to the fact that this is the limit of a discrete model, so we can imagine $\Delta$ as a function of $t$ with more steps of lenght $\Delta t$ so in the limit $\Delta t \to0$. But this would mean that $\frac{\partial \Delta}{\partial S}=0$ that is $\frac{\partial^2 O}{\partial S^2}=0$ where $O$ is the derivative. Could anyone explain me this passage?

## Answer by Gordon (score 20)

https://quant.stackexchange.com/a/32187

In general, you won't be able to replicate the option by a portfolio of the form $\Delta_t S_t + B_t$, though it is possible to do so with a portfolio of the form $\Delta_t^1 S_t + \Delta_t^2B_t$; see Chapter 3 of this book. Here, $B_t=e^{rt}$ is the value of the money-market account, and $r$ is the risk-free interest rate.

On the other hand, you can create a locally risk-free self-financing portfolio of the form \begin{align*} X_t =\Delta_t^1 S_t + \Delta_t^2C_t, \tag{1} \end{align*} where $C_t$ is the option price (see also this question). Specifically, we assume that the stock price process $\{S_t, \, t > 0\}$ satisfies, under the probability measure $P$, an SDE of the form \begin{align*} dS_t = S_t(\mu dt + \sigma dW_t). \end{align*} As $X_t$ is self-financing, then \begin{align*} dX_t &= \Delta_t^1 dS_t + \Delta_t^2 dC_t\\ &= \Delta_t^1 S_t(\mu dt + \sigma dW_t) + \Delta_t^2\left(\frac{\partial C}{\partial t}dt + \frac{\partial C}{\partial S}dS_t + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2 dt\right)\\ &=\left[\mu \Delta_t^1 S_t + \Delta_t^2\left(\frac{\partial C}{\partial t} + \mu S_t \frac{\partial C}{\partial S} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2 \right)\right]dt \\ &\qquad\qquad\qquad\qquad\qquad + \left(\sigma\Delta_t^1 S_t + \sigma \Delta_t^2 S_t \frac{\partial C}{\partial S}\right)dW_t. \end{align*} Since $X_t$ is locally risk-free, we assume that $X_t$ earns the risk-free interest rate $r$, that is, \begin{align*} dX_t = r X_t dt, \tag{2} \end{align*} Then, \begin{align*} &\left[\mu \Delta_t^1 S_t + \Delta_t^2\left(\frac{\partial C}{\partial t} + \mu S_t \frac{\partial C}{\partial S} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2 \right)\right]dt \\ &\qquad\qquad\qquad\qquad\qquad + \left(\sigma\Delta_t^1 S_t + \sigma \Delta_t^2 S_t \frac{\partial C}{\partial S}\right)dW_t= r X_t dt. \end{align*} Consequently, \begin{align*} \Delta_t^1 + \Delta_t^2\frac{\partial C}{\partial S}=0, \tag{3} \end{align*} and \begin{align*} \mu \Delta_t^1 S_t + \Delta_t^2\left(\frac{\partial C}{\partial t} + \mu S_t \frac{\partial C}{\partial S} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2 \right) = r(\Delta_t^1 S_t + \Delta_t^2C_t), \end{align*} or, \begin{align*} \mu S_t\left(\Delta_t^1 +\Delta_t^2 \frac{\partial C}{\partial S}\right) + \Delta_t^2\left(\frac{\partial C}{\partial t} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2\right) &= r(\Delta_t^1 S_t + \Delta_t^2C_t)\\ &=r\Delta_t^2(-\frac{\partial C}{\partial S} S_t + C_t). \end{align*} That is, \begin{align*} \Delta_t^2\left(\frac{\partial C}{\partial t} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2\right) &= r\Delta_t^2(-\frac{\partial C}{\partial S} S_t + C_t). \tag{4} \end{align*} Canceling the term $\Delta_t^2$ from both sides of $(4)$, we obtain the Black–Scholes equation of the form \begin{align*} \frac{\partial C}{\partial t} + r S_t \frac{\partial C}{\partial S} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2 -rC = 0. \tag{5} \end{align*}

> Comments

Note that the quantity $\Delta_t^2$ is canceled from both sides of $(4)$, it is then tempting to set $\Delta_t^2=1$ and, consequently, $\Delta_t^1 = - \frac{\partial C}{\partial S}$. This is indeed adopted by some books, see, for example, Introduction to the Mathematics of Financial Derivatives, by Hirsa and Neftci, and Options, Futures, and Other Derivatives, by John Hull. However, it is easy to see that the strategy $\left(- \frac{\partial C}{\partial S}, 1\right)$ is not self-financing, and the replicating portfolio value, for a European exercise style vanilla call option, \begin{align*} X_t = - \frac{\partial C}{\partial S} S_t+C_t = -Ke^{-r(T-t)}N(d_2) \end{align*} does not satisfy $(2)$.

> Existence of Self-financing Strategy

From $(1)$ and $(3)$ above, we have that \begin{align*} \Delta_t^1 = -\frac{\frac{\partial C}{\partial S} X_t}{C_t - \frac{\partial C} {\partial S}S}, \quad \Delta_t^2 =\frac{X_t}{C_t - \frac{\partial C}{\partial S}S}. \end{align*} In order for $(2)$ to be satisfied, we set \begin{align*} \Delta_t^1 = -\frac{\frac{\partial C}{\partial S} B_t}{C_t - \frac{\partial C} {\partial S}S},\quad \Delta_t^2 =\frac{B_t}{C_t - \frac{\partial C}{\partial S}S}. \end{align*} Then, using the Black–Scholes equation $(5)$, it is easy to see that \begin{align*} dB_t &= \Delta_t^1 dS_t + \Delta_t^2 dC_t. \end{align*} That is, $(\Delta_t^1, \Delta_t^2)$ is a self-financing strategy.

In addition, we can derive the existence using the martingale representation theorem; see Shreve. Specifically, let $\lambda = (\mu-r)/\sigma$, and define the risk-neutral probability measure $Q$ such that \begin{align*} \frac{dQ}{dP}\big|_t = e^{-\frac{1}{2}\lambda^2 t - \lambda W_t}. \end{align*} Then $\tilde{W}= \{\tilde{W}_t, t \ge 0\}$, where $\tilde{W}_t = W_t + \lambda t$ is a standard Brownian motion under $Q$. Moreover, \begin{align*} dS_t = S_t(r dt + \sigma d\tilde{W}_t), \end{align*} or \begin{align*} d\left(\frac{S_t}{B_t}\right) = \frac{S_t}{B_t}\sigma d\tilde{W}_t, \end{align*}

Let $C_T$ be the option payoff at maturity $T$. Then \begin{align*} C_t = B_tE_Q\left(\frac{C_T}{B_T}\mid \mathcal{F}_t \right), \end{align*} where $E_Q$ is the expectation operator under the probability measure $Q$, and $\mathcal{F}_t$ is the information set at time $t$. Note that, $\{C_t/B_t, t \ge 0\}$ is a martingale. Therefore, by the martingale representation theorem, there exists an adapted process $\{\kappa_t, 0\le t \le T\}$ such that \begin{align*} \frac{C_t}{B_t} &= C_0 + \int_0^t \kappa_u d\tilde{W}_u\\ &=C_0 + \int_0^t \frac{B_u\kappa_u}{\sigma S_u} d\left(\frac{S_u}{B_u}\right) :\equiv C_0 + \int_0^t \gamma_u d\left(\frac{S_u}{B_u}\right), \end{align*} where $\gamma_t = \frac{B_t\kappa_t}{\sigma S_t}$, for $0\le t \le T$. Then \begin{align*} dC_t &= d\left(B_t \frac{C_t}{B_t} \right)\\ &=rC_t dt + B_t\gamma_t d\left(\frac{S_t}{B_t}\right)\\ &=r(C_t -\gamma_tS_t) dt + \gamma_t dS_t.\tag{6} \end{align*} We seek adapted processes $\{\Delta_t^1, 0\le t \le T\}$ and $\{\Delta_t^2, 0\le t \le T\}$ such that \begin{align*} B_t &=\Delta_t^1 S_t + \Delta_t^2C_t,\tag{7} \end{align*} and \begin{align*} dB_t &= \Delta_t^1 dS_t + \Delta_t^2dC_t\\ &=\Delta_t^1 dS_t + r \Delta_t^2 (C_t -\gamma_tS_t) dt + \Delta_t^2\gamma_t dS_t. \end{align*} Then \begin{align*} \Delta_t^1 + \Delta_t^2\gamma_t=0. \tag{8} \end{align*} From $(7)$ and $(8)$, \begin{align*} \Delta_t^1 = -\frac{\gamma_t B_t}{C_t - \gamma_t S_t}, \quad \Delta_t^2 =\frac{B_t}{C_t - \gamma_t S_t}. \end{align*} It is then clear that \begin{align*} \Delta_t^1 S_t + \Delta_t^2 C_t = B_t. \end{align*} Moreover, from $(6)$, \begin{align*} \Delta_t^1 dS_t + \Delta_t^2dC_t &=rB_t dt = dB_t. \end{align*} That is, $(\Delta_t^1, \Delta_t^2)$ is a self-financing trading strategy.