I have to use the risk neutral version of the First Fundamental Theorem of Asset Pricing to derive a partial differential equation (PDE) that the price/value process, $V_t = F(t,S_t)$, of a self-financing Markovian portfolio has to satisfy.

Some context:

Let $W$ be a standard Browian motion. We are in a financial market consisting of a risky asset $S$ and a money-market account $B$ with:

$$dS_t = a(b - S_t)dt + \sigma S_tdW_t$$ $$dB_t = rB_tdt$$

where, $$B_0 = 1,\; S_0 = s_0, \;\sigma > 0 \; \text{and}\; a,b \; \text{are constants unequal to zero.}$$

Normally, we don't have to use the FFT and we use these two equations: $$V_t = \phi_tS_t + \psi_tB_t$$ $$dV_t = \phi_tdS_t + \psi_tdB_t $$

I know that the FFTAP tells us that under regularity conditions absence of arbitrage holds if and only if, for some numeraire $N$, there exists a probability measure $\mathbb{Q} = \mathbb{Q}_N$ such that:

  1. $\mathbb{Q} \sim \mathbb{P}$
  2. For any asset $A$ in the market, the discounted price process $A/N$ is a $\mathbb{Q}$-martingale, i.e. $$\frac{A_t}{N_t} = \mathbb{E_Q}\left[ \frac{A_T}{N_T} | \mathcal{F}_t \right]$$

Could someone help me get started, since I have no idea how to start. If I need to provide extra information let me know and I will try to do so.


1 Answer 1


In the answer to a related question of yours it was shown that under the risk-neutral measure $\mathbb Q$ the process $$ S_te^{-rt}=S_0e^{-\frac{\sigma^2t}{2}+\sigma W^{\mathbb Q}_t} $$ is a martingale. In other words, under the risk-neutral $\mathbb Q\,,$ the numeraire $N_t$ is the money market account $e^{rt}\,.$ From \begin{align} V_t=F(t,S_t)=e^{-(T-t)r}\mathbb E\big[F(T,S_T)\big|S_t\big]\, \end{align} it follows directly that $F(t,S_t)e^{-rt}$ is a martingale as well. Applying Ito's formula yields \begin{align} &e^{-rT}F(T,S_T)\\& \quad=F(0,S_0)+\int_0^Te^{-rt}\partial_TF(t,S_t)\,dt+\int_0^Te^{-rt}\partial_xF(t,S_t)\,dS_t\\&\quad\quad+\frac{1}{2}\int_0^Te^{-rt}\partial_x^2F(t,S_t)\,d\langle S\rangle_t\\ &\quad\quad-r\int_0^Te^{-rt}F(t,S_t)\,dt \\ &\quad=F(0,S_0)+\int_0^Te^{-rt}\partial_TF(t,S_t)\,dt+\int_0^Te^{-rt}\partial_xF(t,S_t)\,r\,S_t\,dt\\&\quad\quad+\int_0^Te^{-rt}\partial_xF(t,S_t)\,\sigma\,S_t\,dW^{\mathbb Q}_t\\&\quad\quad+\frac{1}{2}\int_0^Te^{-rt}\partial_x^2F(t,S_t)\,\sigma^2 S_t^2\,dt-r\int_0^Te^{-rt}F(t,S_t)\,dt\,. \end{align} From the martingale property we know that $F(0,S_0)=\mathbb E[e^{-rT}F(T,S_T)]$ holds. It follows that \begin{align} 0&=\mathbb E\Bigg[\int_0^Te^{-rt}\partial_TF(t,S_t)\,dt+\int_0^Te^{-rt}\partial_xF(t,S_t)\,r\,S_t\,dt\\&\quad+\frac{1}{2}\int_0^Te^{-rt}\partial_x^2F(t,S_t)\,\sigma^2 S_t^2\,dt-r\int_0^Te^{-rt}F(t,S_t)\,dt \Bigg]\,. \end{align} Consequently, the Black-Scholes PDE \begin{align} 0=\partial_TF(t,S_t)+\partial_xF(t,S_t)\,r\,S_t+\frac{1}{2}\partial_x^2F(t,S_t)\,\sigma^2 S_t^2-rF(t,S_t)\, \end{align} holds.

  • $\begingroup$ Thanks @Kurt G. $\endgroup$
    – George
    Commented Dec 14, 2021 at 9:35

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