# 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?

• The reason is that we assume the quantity of the risk asset does not change during the infinitesimal interval $[t, t+dt]$, that is, it is self-financing. – Gordon Jan 30 '17 at 18:06
• Thank you, but the part i don't understand is this: in order to get BS equation we make the differential of the portfolio $d\Pi(t)$ and we set it equal to the differential of the derivative $dO(t)$ calculated by Ito's lemma. Now what i'm saying is that we are calculating a differential taking something constant, and this condition is not included in the differential $dO(t)$. So we can't say this object are equal. Am i wrong? – giuliok94 Jan 31 '17 at 12:05
• i discuss this extensively in my book "concepts and practice " – Mark Joshi Feb 1 '17 at 3:14

## 1 Answer

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.

• Thank you again. So i have no problem with the rest of the derivation after the self-financing condition. Is it right if i say this:"$\Delta(t)$ is a step function with step lenght $dt$ and it becomes continuous in the limit $dt\rightarrow0$"? According to mathematical formalism i mean. – giuliok94 Feb 3 '17 at 14:14
• You are basically right. The quantity $\Delta(t)$ is assumed to be a constant over the infinitesimal interval $[t, t+dt]$. – Gordon Feb 3 '17 at 14:19
• Ok but is there a more formal way to state this thing? Because everything i read about self-financiability doesn't give credit to the continuous limit we are doing. I mean if i show this to my teacher of math he wouldn't agree with me – giuliok94 Feb 3 '17 at 14:52
• @Gordon: \begin{align*} X_t = - \frac{\partial C}{\partial S} S_t+C_t = -Ke^{-r(T-t)}N(d_2) \end{align*} is self financing. The confusion arises due to simplification. Because by defintion self fiancing means no exogeneous captial infusion. And if you treat, $\Delta$ as a constant within the small time interval, $dX_t = \Delta dS - \frac{\partial C}{\partial S}$ and this satisfies the property of self-financing – kasa Aug 22 '17 at 14:23
• @kasa: For self-financing, you need to check that $dX_t = -\frac{\partial C}{\partial S} dS + dC$. Here, $X_t = -Ke^{-r(T-t)}N(d_2)$. You can compute all derivative and then see whether the self-financing condition is satisfied. See also the discussion in this question. – Gordon Aug 22 '17 at 14:42