Risk-Neutral Pricing

One-Period Binomial Model

Stock price $S_0$ at time 0 and $S_1(H) = uS_0$, $S_1(T) = dS_0$ at time 1, with $P(H) = p$, $P(T) = q = 1 - p$, money market rate $r$, and $0 < d < 1 + r < u$ (no arbitrage). To replicate a derivative paying $V_1$ at time 1 (e.g. a European call, $V_1 = (S_1 - K)^+$), start with wealth $X_0$ and hold $\Delta_0$ shares. The cash position $X_0 - \Delta_0 S_0$ grows by $1 + r$,

$$X_1 = (1 + r)(X_0 - \Delta_0 S_0) + \Delta_0 S_1$$

Matching $X_1(H) = V_1(H)$ and $X_1(T) = V_1(T)$,

$$\Delta_0 = \frac{V_1(H) - V_1(T)}{S_1(H) - S_1(T)}, \qquad X_0 = \frac{1}{1 + r}\big[\tilde{p}\, V_1(H) + \tilde{q}\, V_1(T)\big], \qquad \tilde{p} = \frac{1 + r - d}{u - d}, \quad \tilde{q} = \frac{u - 1 - r}{u - d}$$

So the price $V_0 = X_0$ is the discounted expectation under the risk-neutral probabilities $\tilde{p}, \tilde{q}$, which do not depend on $p$.

Girsanov's Theorem

For a single random variable see change of measure, $\frac{d\tilde{\mathbb{P}}}{d\mathbb{P}} = Z$.

For a process Let $(\Omega, \mathcal{F}, \mathbb{P})$ with filtration $\mathcal{F}_t$, $0 \le t \le T$, $Z > 0$ a.s., $E[Z] = 1$, $\tilde{\mathbb{P}}(A) = \int_A Z\, d\mathbb{P}$, and

$$Z(t) = E[Z | \mathcal{F}_t]$$

which is a martingale. For $0 \le s \le t \le T$ and $Y$ $\mathcal{F}_t$-measurable,

$$\tilde{E}[Y] = E[Y Z(t)], \qquad \tilde{E}[Y | \mathcal{F}_s] = \frac{1}{Z(s)} E[Y Z(t) | \mathcal{F}_s]$$

Girsanov Let $W(t)$ be a Brownian motion on $(\Omega, \mathcal{F}, \mathbb{P})$ with filtration $\mathcal{F}_t$, $\Theta(t)$ an adapted process, and

$$Z(t) = \exp\Big( -\int_0^t \Theta\, dW - \frac{1}{2}\int_0^t \Theta^2\, du \Big), \qquad \tilde{W}(t) = W(t) + \int_0^t \Theta\, du$$

Assume $E\int_0^T \Theta^2 Z^2\, du < \infty$. Then with $Z = Z(T)$, $E[Z] = 1$, and under $\tilde{\mathbb{P}}$ the process $\tilde{W}$ is a Brownian motion.

Risk-Neutral Measure

Let $dS = \alpha S\, dt + \sigma S\, dW$, with $\alpha$ the mean rate of return and $\sigma$ the volatility, equivalently

$$S(t) = S(0) \exp\Big( \int_0^t \sigma\, dW + \int_0^t \big(\alpha - \tfrac{1}{2}\sigma^2\big)\, ds \Big)$$

With interest rate $R(t)$, the money market $M(t) = e^{\int_0^t R\, ds}$ ($dM = RM\, dt$) and the discount process $D = 1/M$, $dD = -RD\, dt$. $D$ has zero quadratic variation, so by the product rule

$$d(DS) = (\alpha - R) DS\, dt + \sigma DS\, dW = \sigma DS\, [\Theta\, dt + dW], \qquad \Theta = \frac{\alpha - R}{\sigma}$$

$\Theta$ is the market price of risk, and $\alpha - R$ is the mean rate of return of the discounted stock price. Applying Girsanov with this $\Theta$, $d\tilde{W} = \Theta\, dt + dW$ and

$$d(DS) = \sigma DS\, d\tilde{W}$$

so the discounted stock price is a $\tilde{\mathbb{P}}$-martingale, and under $\tilde{\mathbb{P}}$

$$dS = RS\, dt + \sigma S\, d\tilde{W}, \qquad S(t) = S(0) \exp\Big( \int_0^t \sigma\, d\tilde{W} + \int_0^t \big(R - \tfrac{1}{2}\sigma^2\big)\, ds \Big)$$

Portfolios and Replication

Hold $\Delta(t)$ shares in a portfolio of total value $X(t)$, so $\Delta S$ is invested in the stock and $X - \Delta S$ in the money market. The capital gain is $\Delta(t)[S(t + \delta t) - S(t)]$ and the interest gain $(X - \Delta S)R\, \delta t$, so

$$dX = \underbrace{RX\, dt}_{\text{risk-free return}} + \underbrace{\Delta(\alpha - R)S\, dt}_{\text{risk premium}} + \underbrace{\Delta \sigma S\, dW}_{\text{risk}}, \qquad d(DX) = \Delta \sigma DS\, d\tilde{W}$$

so the discounted portfolio value is also a $\tilde{\mathbb{P}}$-martingale. If the portfolio replicates a derivative, $X(T) = V(T)$, its price is $V(t) := X(t)$, and

$$D(t)V(t) = \tilde{E}\big[ D(T)V(T) \,\big|\, \mathcal{F}_t \big]$$

(risk-neutral pricing formula). In particular $\tilde{E}[D(T)S(T)] = D(0)S(0)$.

Black-Scholes-Merton

For constant $r$ and $\sigma$, $\tilde{E}[S(T)] = S(0)e^{rT}$, and the call price is

$$c(0, S(0)) = e^{-rT}\tilde{E}\big[(S(T) - K)^+\big] = S(0)N(d_+) - Ke^{-rT}N(d_-), \qquad d_\pm = \frac{\log(S(0)/K) + (r \pm \frac{\sigma^2}{2})T}{\sigma\sqrt{T}}$$

Put-call parity: $C(0) - P(0) = S(0) - e^{-rT}K$.

The price $v(t, s)$ of a derivative whose payoff depends on $S(T)$ solves the BSM equation

$$\frac{\partial v}{\partial t} = rv - rs\frac{\partial v}{\partial s} - \frac{1}{2}\sigma^2 s^2 \frac{\partial^2 v}{\partial s^2}$$

Barrier Options

A down-and-out call with barrier $B \le K$ pays $(S(T) - K)^+$ unless $S(t) \le B$ for some $t \le T$, in which case it pays 0. The payoff depends on the path, but augmenting $S$ with

$$Y(t) = \begin{cases} 1 & S(u) > B \;\; \forall u \in [0, t] \\ 0 & S(u) \le B \;\; \exists u \in [0, t] \end{cases}$$

makes $(S, Y)$ Markov, with $V(T) = (S(T) - K)^+ \, Y(T)$, so

$$v(t, S(t), Y(t)) = \tilde{E}\big[ e^{-r(T - t)} V(T) \,\big|\, \mathcal{F}_t \big]$$

Once knocked out, $v(t, s, 0) = 0$. Let $v(t, s) := v(t, s, 1)$, which satisfies BSM on $[0, T] \times [B, \infty)$ with

$$v(t, B) = 0, \; 0 \le t \le T, \qquad v(T, s) = 0, \; B \le s \le K, \qquad v(T, s) = s - K, \; s \ge K$$

If $v(t, s)$ solves BSM, so does $s^{2\alpha} v(t, c/s)$ for any constant $c$, with $2\alpha = 1 - \frac{2r}{\sigma^2}$. With $c(t, s)$ the vanilla call,

$$v(t, s) = c(t, s) - \Big(\frac{s}{B}\Big)^{2\alpha} c\Big(t, \frac{B^2}{s}\Big)$$

At $s = B$ the two terms cancel, and at $T$ for $s \ge B$, $B^2/s \le B \le K$ so the second term is 0.