8.7 Risk-neutral valuation and the Black–Scholes formula

Feynman–Kac and Girsanov#

Why

Let $M_u=e^{-r(u-t)}V(u,S_u)$ with $S$ following (8.20). Itô's lemma gives $dM_u=e^{-r(u-t)}\big[V_u+(r-q)SV_S+\frac12\sigma^2S^2V_{SS}-rV\big]du+e^{-r(u-t)}\sigma SV_S\,d\widetilde W_u$. The bracket is zero by (8.18), so $M$ is a local martingale; with growth conditions it is a true martingale, and $V(t,S)=M_t=\mathbb E[M_T]=e^{-r(T-t)}\mathbb E[h(S_T)]$. See Shreve II, Thm 6.4.1.∎

Girsanov's theorem (stated, not proved) The tilde process lives under a different probability measure. Let $W$ be a $\mathbb P$-Brownian motion on $[0,T]$ and $\theta$ a constant. Define $\mathbb Q$ by the density

Then $\widetilde W_t=W_t+\theta t$ is a $\mathbb Q$-Brownian motion, and $\mathbb Q$ and $\mathbb P$ agree on which events have probability zero. Choose $\theta=(\mu-(r-q))/\sigma$, the market price of risk: excess return per unit of volatility. Then

which is (8.20). Under $\mathbb Q$ the discounted gains process $e^{-(r-q)t}S_t$ is a martingale; this is the continuous-time version of the risk-neutral $p$ of 8.3.

Intuition. For one variable, the density $e^{-\theta x-\theta^2T/2}$ is exactly the likelihood ratio $\varphi_{-\theta T,T}(x)/\varphi_{0,T}(x)$ of $N(-\theta T,T)$ to $N(0,T)$. Reweighting the paths of $W$ by it shifts the mean of $W_T$ by $-\theta T$ and leaves the variance alone. Girsanov says this works path by path for the whole process: a change of measure can change the drift of a diffusion but never its volatility. That is why option prices depend on $\sigma$ (which both measures share) and not on $\mu$ (which the measure change removes). The proof is in Shreve II, Thm 5.2.3, or Karatzas and Shreve (1991), 3.5.

The Black–Scholes–Merton formula#

Proof

By (8.19)–(8.20), with $Z=(\widetilde W_T-\widetilde W_t)/\sqrt\tau\sim N(0,1)$ under $\mathbb Q$,

$$S_T=S\exp\!\big((r-q-\tfrac12\sigma^2)\tau+\sigma\sqrt\tau Z\big).$$

Step 1: the exercise region $S_T>K$ iff $\sigma\sqrt\tau Z>\ln(K/S)-(r-q-\frac12\sigma^2)\tau$ iff $Z>-d_2$. So $\mathbb Q(S_T>K)=\mathbb Q(Z>-d_2)=N(d_2)$.

Step 2: split the integral

$$\begin{aligned} c&=e^{-r\tau}\int_{-d_2}^{\infty}\Big(Se^{(r-q-\frac12\sigma^2)\tau+\sigma\sqrt\tau z}-K\Big)\varphi(z)\,dz\\ &= I_1 - Ke^{-r\tau}\int_{-d_2}^\infty\varphi(z)\,dz = I_1-Ke^{-r\tau}N(d_2). \end{aligned}$$

Step 3: complete the square in $I_1$ Since $\varphi(z)=e^{-z^2/2}/\sqrt{2\pi}$,

$$e^{-\frac12\sigma^2\tau+\sigma\sqrt\tau z}\varphi(z)=\frac{1}{\sqrt{2\pi}}\exp\!\Big(-\tfrac12\big(z^2-2\sigma\sqrt\tau z+\sigma^2\tau\big)\Big)=\varphi(z-\sigma\sqrt\tau).$$

Therefore

$$\begin{aligned} I_1&=Se^{-r\tau}e^{(r-q)\tau}\int_{-d_2}^{\infty}\varphi(z-\sigma\sqrt\tau)\,dz=Se^{-q\tau}\int_{-d_2-\sigma\sqrt\tau}^{\infty}\varphi(y)\,dy\\ &=Se^{-q\tau}N(d_2+\sigma\sqrt\tau)=Se^{-q\tau}N(d_1), \end{aligned}$$

using the substitution $y=z-\sigma\sqrt\tau$ and $1-N(-x)=N(x)$. This proves (8.23). For (8.24), apply parity (8.4): $p=c-Se^{-q\tau}+Ke^{-r\tau}=Ke^{-r\tau}(1-N(d_2))-Se^{-q\tau}(1-N(d_1))$.∎

Interpretation of $N(d_1)$ and $N(d_2)$ $N(d_2)$ is the risk-neutral probability that the call finishes in the money. $N(d_1)$ is the probability of the same event under the measure that uses the stock (with dividends reinvested) as numéraire; Step 3 is exactly that change of measure, with the density $S_T/\mathbb E^{\mathbb Q}[S_T]$ tilting $Z$ by $\sigma\sqrt\tau$. In the CRR closed form (8.12) these are the limits of the $p$-tail and the $p'$-tail. Neither is the real-world probability of exercise; that would use $\mu$ in place of $r-q$.

Limits and reference values#

Remark 8.7.3 (limits) As $\tau\downarrow0$, $d_{1,2}\to\pm\infty$ according to the sign of $\ln(S/K)$, and $c\to(S-K)^+$. As $\sigma\downarrow0$ with $\tau>0$, $d_{1,2}\to\pm\infty$ according to the sign of $\ln(F/K)$, and $c\to e^{-r\tau}(F-K)^+$: with no uncertainty the stock surely ends at the forward. As $\sigma\to\infty$, $d_1\to\infty$ and $d_2\to-\infty$, so $c\to Se^{-q\tau}$, the upper bound (8.6). The formula spans exactly the no-arbitrage band; this is the basis for 8.10.