option pricing - 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

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

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  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.

  
  


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  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 $$\begin{align*} \Delta_t^1 S_t + \Delta^2_t C = B_t. \end{align*}$$

  
  


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  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.

  
  
  


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  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.