Self-Financing Delta Hedging and the Black–Scholes PDE

# Derivation of BS PDE problem using Delta hedging


# Derivation of BS PDE problem using Delta hedging












I've always been confused with Delta hedging. It is well-known that for a (smooth enough) function of $(S,t)$ we have, due to Ito's lemma, that: \begin{eqnarray*} dC = \left(\frac{\partial C}{\partial t} (S,t) + \mu S \frac{\partial C}{\partial S} (S,t) + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{\partial S^2}(S,t)\right)dt + \sigma S \frac{\partial C}{\partial S} (S,t) dX \end{eqnarray*} (source here, where $X$ is the standard Brownian motion). The author then proceeds to show that if we choose $\Delta = -\frac{\partial C}{\partial S} (S,t)$, then we will have \begin{align*} d(C+\Delta S) &= \left(\frac{\partial C}{\partial t} (S,t) + \mu S \frac{\partial C}{\partial S} (S,t) + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{\partial S^2}(S,t) + \Delta \mu S\right) dt\\ &+ \sigma S \left(\frac{\partial C}{\partial S}+\Delta\right) dX\\ &=\left(\frac{\partial C}{\partial t}(S,t) + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{\partial S^2}(S,t)\right)dt \end{align*} (I've corrected the author's typo and replaced the term $ \Delta S (\frac{\partial C}{\partial S}+\Delta) dX$ in the formula above with the correct one $\sigma S (\frac{\partial C}{\partial S}+\Delta) dX$ )

I understand the usual procedure of $\Delta$, but I never understand how $\Delta$ can be defined as $-\frac{\partial C}{\partial S} (S,t)$. By definition a hedging strategy must be a predictable process, is there any justification that $\Delta$ given as above is predictable, not merely adapted, to the filtration generated by the stock process?

Even if we accept the definition of $\Delta$, the author also doesn't explain how to compute $d(C+\Delta S)$. But, as far as I know, if $\Delta$ is really the symbol for $-\frac{\partial C}{\partial S} (S,t)$, then we will have $$d(\Delta S)=Sd(\Delta)+\Delta dS+(dS)\cdot(d\Delta).$$ And as for $d\Delta$, I think it's just replacing $C$ by $\Delta$ in the expression of $dC$, and the resulting $d(\Delta S)$ will be super-complicated with $\partial^3 C/\partial S^3$ present. What I obtained in the end was a horribe mess, the risky term $dX$ wasn't eliminated and the non-random term $dt$ doesn't match that in the BS PDE.

Wherein lies the problem?

## Answer by Gordon (score 9, accepted)

https://quant.stackexchange.com/a/34059

This question has been asked many times and some clarifications appear needed.

- As pointed out in an answer to this question, the portfolio \begin{align*} \Delta_t^1 S_t + \Delta^2_t C, \end{align*} where $\Delta_t^1 = -\frac{\partial C}{\partial S}$ and $\Delta_t^2 =1$, is, generally, neither self-financing nor locally risk-free.

- To derive the Black-Scholes' PDE, we seek a portfolio $\Delta_t^1 S_t + \Delta^2_t C$ such that it is self-financing and locally risk-free. As is shown in answer to the above question, we derived the PDE \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{1} \end{align*} and the portfolio weights \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}, \tag{2} \end{align*} where $B_t=e^{rt}$ is the money-market account value. Note that \begin{align*} \Delta_t^1 S_t + \Delta^2_t C = B_t. \end{align*}

- Based on PDE $(1)$ and the portfolio weights give by $(2)$, it is easy to verify that \begin{align*} \Delta_t^1 dS_t + \Delta^2_t dC = dB_t. \end{align*} That is, the portfolio is indeed self-financing and locally risk-free.

- Though messy, based on PDE $(1)$ and the portfolio weights give by $(2)$, it is possible to show directly that \begin{align*} d\left(\Delta_t^1 S_t + \Delta^2_t C \right) = \Delta_t^1 dS_t + \Delta^2_t dC \end{align*} using Ito's lemma. Here, we assume that the respective partial derivatives, such as $\frac{\partial^2 C}{\partial t\partial S}$ and $\frac{\partial^3 C}{\partial S^3}$, exist.

$$ $$

> Below, we show the self-financing property using Ito's lemma, that is, \begin{align*} Sd\Delta_t^1 + C d\Delta_t^2 + d\langle \Delta_t^1,\, S \rangle_t +d\langle \Delta_t^2,\, C \rangle_t =0. \end{align*}

For weight $\Delta_t^1$, \begin{align*} \frac{\partial \Delta_t^1}{\partial t} &= -\frac{\left(\frac{\partial^2 C}{\partial t \partial S} B_t + r\frac{\partial C}{\partial S}B_t\right) \left(C_t - \frac{\partial C}{\partial S}S \right) - \left(\frac{\partial C}{\partial t} - \frac{\partial^2 C}{\partial t \partial S}S \right)\frac{\partial C}{\partial S} B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2}\\ &=-\frac{\frac{\partial^2 C}{\partial t \partial S} B_t C_t + r\frac{\partial C}{\partial S}B_t C_t - r \left(\frac{\partial C}{\partial S} \right)^2B_t S_t - \frac{\partial C}{\partial t}\frac{\partial C}{\partial S} B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2}\\ &=-\frac{\frac{\partial^2 C}{\partial t \partial S} B_t C_t + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \frac{\partial C}{\partial S}\sigma^2S_t^2 B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2} \qquad\qquad\qquad\qquad\qquad\qquad \mbox{(From Eqn $(1)$)}\\ &=-\frac{\frac{\partial^2 C}{\partial t \partial S} B_t C_t^2 + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \frac{\partial C}{\partial S}\sigma^2S_t^2 B_tC_t - \frac{\partial^2 C}{\partial t \partial S}\frac{\partial C}{\partial S} B_t C_t S - \frac{1}{2}\frac{\partial^2 C}{\partial S^2} \left(\frac{\partial C}{\partial S}\right)^2\sigma^2S_t^3 B_t}{\left(C_t - \frac{\partial C}{\partial S}S\right)^3},\\ \frac{\partial \Delta_t^1}{\partial S} &=-\frac{\frac{\partial^2 C}{\partial S^2} B_t \left(C_t - \frac{\partial C}{\partial S}S \right) - \left(\frac{\partial C}{\partial S} - \frac{\partial^2 C}{\partial S^2}S - \frac{\partial C}{\partial S}\right)\frac{\partial C}{\partial S} B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2}\\ &=-\frac{\frac{\partial^2 C}{\partial S^2} B_t C_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2},\\ \frac{\partial^2 \Delta_t^1}{\partial S^2} &=-\frac{\left(\frac{\partial^3 C}{\partial S^3} B_tC_t + \frac{\partial^2 C}{\partial S^2}\frac{\partial C}{\partial S} B_t \right) \left(C_t - \frac{\partial C}{\partial S}S \right)^2 +2\left(C_t - \frac{\partial C}{\partial S}S\right) \left(\frac{\partial^2 C}{\partial S^2}\right)^2S B_t C_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^4}\\ &=-\frac{\frac{\partial^3 C}{\partial S^3} B_tC_t^2 - \frac{\partial^3 C}{\partial S^3}\frac{\partial C}{\partial S} B_tC_t S + \frac{\partial^2 C}{\partial S^2}\frac{\partial C}{\partial S} B_t C_t - \frac{\partial^2 C}{\partial S^2}\left(\frac{\partial C}{\partial S}\right)^2 B_t S + 2\left(\frac{\partial^2 C}{\partial S^2}\right)^2S B_t C_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^3} \end{align*}

For weight $\Delta_t^2$, \begin{align*} \frac{\partial \Delta_t^2}{\partial t} &= \frac{rB_t \left(C_t - \frac{\partial C}{\partial S}S \right) - \left(\frac{\partial C}{\partial t} - \frac{\partial^2 C}{\partial t \partial S}S \right) B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2}\\ &=\frac{ rB_t C_t - r\frac{\partial C}{\partial S} B_t S - B_t \frac{\partial C}{\partial t} +\frac{\partial^2 C}{\partial t \partial S}S B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2}\\ &=\frac{\frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2 B_t +\frac{\partial^2 C}{\partial t \partial S}S B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2} \qquad\qquad\qquad\qquad\qquad\qquad \mbox{(From Eqn $(1)$)}\\ &=\frac{\frac{1}{2}\frac{\partial^2 C}{\partial S^2} \sigma^2S_t^2 B_tC_t +\frac{\partial^2 C}{\partial t \partial S}S B_tC_t - \frac{1}{2}\frac{\partial^2 C}{\partial S^2}\frac{\partial C}{\partial S} \sigma^2S_t^3 B_t -\frac{\partial^2 C}{\partial t \partial S}\frac{\partial C}{\partial S}S^2 B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2},\\ \frac{\partial \Delta_t^2}{\partial S} &=\frac{ - \left(\frac{\partial C}{\partial S} - \frac{\partial^2 C}{\partial S^2}S - \frac{\partial C}{\partial S}\right) B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2}\\ &=\frac{\frac{\partial^2 C}{\partial S^2} B_t S }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2},\\ \frac{\partial^2 \Delta_t^2}{\partial S^2} &=\frac{\left(\frac{\partial^3 C}{\partial S^3} B_t S + \frac{\partial^2 C}{\partial S^2} B_t \right) \left(C_t - \frac{\partial C}{\partial S}S \right)^2 +2\left(C_t - \frac{\partial C}{\partial S}S\right) \left(\frac{\partial^2 C}{\partial S^2}\right)^2S^2 B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^4}\\ &=\frac{\frac{\partial^3 C}{\partial S^3} B_tC_t S - \frac{\partial^3 C}{\partial S^3}\frac{\partial C}{\partial S} B_t S^2 + \frac{\partial^2 C}{\partial S^2} B_t C_t - \frac{\partial^2 C}{\partial S^2}\frac{\partial C}{\partial S} B_t S + 2\left(\frac{\partial^2 C}{\partial S^2}\right)^2S^2 B_t }{\left(C_t - \frac{\partial C}{\partial S}S\right)^3}. \end{align*}

Then \begin{align*} &\ d\langle \Delta_t^1,\, S \rangle_t +d\langle \Delta_t^2,\, C \rangle_t \\ =& \left(-\frac{\frac{\partial^2 C}{\partial S^2} B_t C_t \sigma^2 S^2}{\left(C_t - \frac{\partial C}{\partial S}S\right)^2} +\frac{\frac{\partial^2 C}{\partial S^2}\frac{\partial C}{\partial S}\sigma^2 B_t S^3 }{\left(C_t - \frac{\partial C}{\partial S}S\right)^2}\right)dt\\ =& \left(\frac{-\frac{\partial^2 C}{\partial S^2} B_t C_t^2 \sigma^2 S^2 + \frac{\partial^2 C}{\partial S^2}\frac{\partial C}{\partial S} B_t C_t \sigma^2 S^3 + \frac{\partial^2 C}{\partial S^2}\frac{\partial C}{\partial S}\sigma^2 B_t S^3 C_t - \frac{\partial^2 C}{\partial S^2}\left(\frac{\partial C}{\partial S}\right)^2\sigma^2 B_t S^4}{\left(C_t - \frac{\partial C}{\partial S}S\right)^3} \right)dt. \end{align*} Moreover, \begin{align*} Sd\Delta_t^1 + C d\Delta_t^2 &= S\left(\frac{\partial \Delta_t^1}{\partial t}dt + \frac{1}{2}\frac{\partial^2 \Delta_t^1}{\partial S^2}\sigma^2 S^2 dt + \frac{\partial \Delta_t^1}{\partial S}dS \right) \\ &\qquad + C\left(\frac{\partial \Delta_t^2}{\partial t}dt + \frac{1}{2}\frac{\partial^2 \Delta_t^2}{\partial S^2}\sigma^2 S^2 dt + \frac{\partial \Delta_t^2}{\partial S}dS \right)\\ &=S\left(\frac{\partial \Delta_t^1}{\partial t} + \frac{1}{2}\frac{\partial^2 \Delta_t^1}{\partial S^2}\sigma^2 S^2\right)dt + C\left(\frac{\partial \Delta_t^2}{\partial t} + \frac{1}{2}\frac{\partial^2 \Delta_t^2}{\partial S^2}\sigma^2 S^2 \right)dt. \end{align*} By combining all terms together, we can show that \begin{align*} Sd\Delta_t^1 + C d\Delta_t^2 + d\langle \Delta_t^1,\, S \rangle_t +d\langle \Delta_t^2,\, C \rangle_t =0. \end{align*} Then \begin{align*} d\left(\Delta_t^1 S_t + \Delta_t^2 C_t \right) = \Delta_t^1 dS_t + \Delta_t^2 dC_t, \end{align*} that is, $\left(\Delta_t^1, \, \Delta_t^2\right)$ constitutes a self-financing strategy.

## Answer by Mark Joshi (score 4)

https://quant.stackexchange.com/a/34028

this is the d on the delta problem. (In my book concepts I discuss this. Sections 5.6 and 5.7 )

Essentially we hold delta constant across time steps and so ignore the d on the delta because there is an implicit bond holding into which we are putting any residual cash so the value of our hedging portfolio only changes due to changes in the financial instruments rather that from changes in how we hold the money.

Changing Stocks into Bonds or Bonds into stocks will not make or lose money in a no-transaction cost world.

Regarding predictability, if everything is continuous which they are here, adapted and predictable are the same.

## Answer by zer0hedge (score 2)

https://quant.stackexchange.com/a/34053

Let $S(t)$ be a Geometrical Brownian Motion as there and let $M(t) = e^{rt}$ be the price of a share of the money market account with the continuously compounding risk-free interest rate $r$ at time $t$.

A trading strategy is specified by an initial capital $X_0$ and a pair of adapted processes:

- $\Delta(S(t),t)$, which denotes the number of shares of stock held at time $t$

- $\Gamma(S(t),t)$, which denotes the number of shares of the money market account held at time $t$

The portfolio value at time $t$ is then: $$X(S(t), t) = \Delta(S(t), t) S(t) + \Gamma(S(t),t)M(t)\tag{0} \label{portfolio_value}$$

As defined here, the trading strategy is self-financing if: $$ X(S(T), T) - X_0 = \int_0^T \Delta(S(u),u) dS(u) + \int_0^T\Gamma(S(u),u) dM(u) \tag{1} \label{self_financing}$$

> implying that the change in value over any interval in portfolio value is entirely due to gains from trade

in differential notation $\eqref{self_financing}$ can be re-written as: $$ dX(S(t),t) = \Delta(S(t),t) dS(t) + \Gamma(S(t),t)dM(t) \tag{2}\label{self_fin_dif}$$

if we apply Ito product rule to the right hand side of $\eqref{portfolio_value}$ and compare the result with $\eqref{self_fin_dif}$, we will immediately get what is called continuous-time self-financing condition here: $$ S d\Delta + dS d\Delta +Md\Gamma + dM d\Gamma = 0 \tag{3} \label{ctsfc}$$

Corollary If $X$ is a given adapted process, then $X_0 = X(0)$, $\Delta$, $\Gamma = \frac{X - \Delta S}{M}$ where $\Delta$ is the solution of $\eqref{self_fin_dif}$ is a self-financing trading strategy.

Now let $c(x, t)$ be the price of a call at some time $t$ if the stock price at that time is $S(t) = x$.

Form a trading strategy $X_0^* = c(S(0),0)$, $\Delta^*$, $\Gamma^* = \frac{c - \Delta^* S}{M}$ whose portfolio value satisfies $$X^*(S(t),t) = c(S(t),t) \tag{4}\label{port_opt}$$ By corollary it will be a self-financing trading strategy.

Substitution of $\eqref{port_opt}$ into $\eqref{self_fin_dif}$ gives us: $$ dc = \Delta^* dS + \Gamma^*dM = \Delta^* dS + \frac{(c-\Delta^* S)}{M}dM \tag{5} \label{delta}$$

Applying Ito-Doeblin formula on the left hand side of $\eqref{delta}$ and using the fact that $dM = rMdt$, we will get: $$ c_t dt +c_x dS + \frac{1}{2}c_{xx} \sigma^2 S^2 dt = \Delta^* dS + r(c-\Delta^* S)dt \label{bs} \tag{6} $$

From $\eqref{bs}$ it follows that $\Delta^*(S(t),t) = c_x \equiv \frac{\partial{c}}{\partial{S}}(S(t),t)$

And the last point. In my opinion, the calculation of $d(C+\Delta S)$ by author here is wrong and you did it correctly in your question.