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Let $W_t$ be a Brownian Motion and let

$S_t= S_0e^{(rt- \frac{\sigma^2}{3!}t^3 +\int_{0}^{t}\sigma W_s ds )}$

Price and Hedge at time $t=0$ European call with maturity $T$ and strike price $K$, written on an underlying with price $S$.

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Step 1: Know your distribution

Since $\int_0^t W_s\mathrm{d}s\sim N\left(0,\frac{1}{3}t^3\right)$, we have \begin{align*} S_t &= S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \int_0^t W_s\mathrm{d}s \right) \\ &\overset{d}{=} S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \sqrt{\frac{1}{3}t^3} Z \right) \\ &\overset{d}{=} S_0 \exp\left( \left(r-\frac{1}{2}\left(\frac{1}{3}\sigma^2 t^2\right)\right)t + \sqrt{\frac{1}{3}\sigma^2t^2} W_t \right), \end{align*} where $Z\sim N(0,1)$, as shown here. In particular, the stock price is log-normally distributed for every time point $t$.

Step 2: Remember your toolkit

We'll use the following result: if $\ln(X)\sim N(m,s^2)$, then \begin{align*} \mathbb{E}[\max\{X-K,0\}] &= e^{m+\frac{1}{2}s^2}\Phi\left(\frac{m-\ln(K)+s^2}{s}\right)-K\Phi\left(\frac{m-\ln(K)}{s}\right). \end{align*} In your example, \begin{align*} \ln(S_T) &= \ln(S_0) + rT-\frac{1}{6}\sigma^2 T^3 + \sqrt{\frac{1}{3}\sigma^2 T^3} Z, \\ \implies \mathbb{E}[\ln(S_T)] &= \ln(S_0) + rT-\frac{1}{6}\sigma^2 T^3, \\ \implies \mathbb{V}\mathrm{ar}[\ln(S_T)] &= \frac{1}{3}\sigma^2 T^3. \\ \end{align*}

Step 3: Put everything together

Assuming the absence of arbitrage, the option price is then the discounted expected payoff. I'll assume that the above stock price dynamics are with respect to the risk-neutral measure. Then,

\begin{align*} V_0 &= e^{-rT} \mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}] \\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT+\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}\sigma^2T^3}}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ rT-\frac{1}{6}\sigma^2 T^3}{\sqrt{\frac{1}{3}\sigma^2T^3}}\right)\\ &= S_0\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r+\frac{1}{6}\sigma^2 T\right)T}{\sqrt{\frac{1}{3}\sigma^2T}\; T}\right)-Ke^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(r-\frac{1}{6}\sigma^2 T^2\right)T}{\sqrt{\frac{1}{3}\sigma^2T}\; T}\right). \end{align*}

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