Find BlackScholes PDE $\frac{\partial C}{\partial t} + rS\frac{\partial C}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 C}{\partial S^2}=rC$

Question

G'day, I have some basic knowledge about Ito Calculus and Brownian Motions. We assume that stock price S follows the GBM:

• 1) $()=_t_t+_tS_t_t$

The value of the change in portfolio is given by:

• 2) $dX_t = rX_tdt + \Delta_t(u - r)S_td_t + \Delta_t \sigma S_tdW_t$

Let C be a call option written on underlying value S. The differential of $C(t,S_t)$ is given by:

• 3) $dC(s,S_t) = \frac{\partial C}{\partial t}dt + \frac{\partial C}{\partial S}dS_t + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(dS_t)^2$

• 4) $= [\frac{\partial C}{\partial t} + uS_t\frac{\partial C}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 C}{\partial S^2}]dt + \sigma S_t\frac{\partial C}{\partial S}dW_t $

to get eventually the result:

• 5) $\frac{\partial C}{\partial t} + rS\frac{\partial C}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 C}{\partial S^2}=rC$

I am very much wondering how we get from step 3) to 4) doing the derivation. When I was calculating stochastic equations, my method was to define f(x) = $X^3$, f'(x)= $3x^2$, f''(x) = 6x and then substitute them in Ito's lemma. Can someone use a similar approach?

And with the icing on the cake: how do we eventually get to step 5).

Any help or tips is very much appreciated!

Linked: How to obtain $\frac{\partial C}{\partial \sigma}$ from Black-Scholes PDE?

Answer 1

You have a self-financing portfolio $X_t=B_t+\Delta_tS_t$ where $B_t$ is your risk-free rate bond.

You want $X_t+C(t,S_t)=0$, in other words, a perfect hedge.

You also have $$dX_t+dC(t,S_t)=0$$

By using the self-financing property of $X_t$ $$dX_t=dB_t+\Delta_tdS_t$$

Using Ito's lemma on $C$, we have $$dC(t,S_t) = \frac{\partial C}{\partial t}dt + \frac{\partial C}{\partial S}dS_t + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}d<S_t,S_t>$$

with $d<S_t,S_t>=(\sigma_tS_t)^2dt$

Using the third equation, we have $$dB_t+\Delta_tdS_t+\frac{\partial C}{\partial t}dt + \frac{\partial C}{\partial S}dS_t + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma_tS_t)^2dt=0$$

To get rid of the source of randomness $dS_t$, a natural choice is $$\Delta_t=-\frac{\partial C}{\partial S}$$

Therefore, $$dB_t+\frac{\partial C}{\partial t}dt + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma_tS_t)^2dt=0$$

The bond grows are the risk-free rate $r_t$, hence $$dB_t=r_tB_tdt$$ Thus, $$r_tB_tdt+\frac{\partial C}{\partial t}dt + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma_tS_t)^2dt=0$$

Using the first two equations and the definition of $\Delta$, we have $$B_t=-C(t,S_t)-\Delta_tS_t=-(C(t,S_t)-\frac{\partial C}{\partial S}S_t)$$

Finally you have (removing the dt)

$$-r_tC(t,S_t)+\frac{\partial C}{\partial S}r_tS_t+\frac{\partial C}{\partial t} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma_tS_t)^2=0$$

Answer 2

In order to obtain 4), you plug $_t=_t_t+_tS_t_t$ into

$$dC(t,S_t) = \frac{\partial C}{\partial t}dt + \frac{\partial C}{\partial S}dS_t + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(dS_t)^2$$

and use $(_t)^2 = (_tS_t_t)^2=\sigma_t^2S_t^2dt$ to get

$$dC(t,S_t) = \left[\frac{\partial C}{\partial t} + uS_t\frac{\partial C}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 C}{\partial S^2}\right]dt + \sigma S_t\frac{\partial C}{\partial S}dW_t $$

To arrive at 5). You construct the hedged portfolio with $-\frac{\partial C}{\partial t}$ unit of stocks and apply the non-arbitrage argument that any hedged portfolio is equivalent to the risk-free portfolio that grows at risk-free rate $r$, i.e.

$$dC(t,S_t) - \frac{\partial C}{\partial t}_t = rC(t,S_t) dt$$

which leads to the Black-Scholes PDE,

$$\frac{\partial C}{\partial t} + rS\frac{\partial C}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 C}{\partial S^2}=rC$$