Pricing a Call When Log Price Depends on Integrated Brownian Motion
Summary
The document derives a European call price for a price process whose log value includes the time integral of Brownian motion. It uses the fact that the integral is normally distributed with variance proportional to the cube of time, showing that the terminal price is lognormally distributed. From this, it identifies the mean and variance of the terminal log price and applies the standard expected-payoff formula for a lognormal variable.
The resulting valuation discounts the expected call payoff at the risk-free rate and assumes the given dynamics are specified under the risk-neutral measure. That assumption is essential to the pricing step. Although the question mentions hedging, the answer provides a price derivation and does not develop a hedge or discuss whether the model permits dynamic replication. The treatment is limited to this particular process and a European call at a single maturity; it offers no empirical evidence or calibration guidance.
Key ideas
- The integral of Brownian motion is normally distributed, with variance proportional to the cube of time.
- The specified price process therefore has a lognormal terminal distribution.
- A call value follows by applying the lognormal expected-payoff formula and discounting at the risk-free rate.
- The calculation assumes the dynamics are under the risk-neutral measure and does not derive a hedge.
Tags
Full text
# Brownian motion Price and Hedge problem
# Brownian motion Price and Hedge problem
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 $S$.
## Answer by Kevin (score 4, accepted)
https://quant.stackexchange.com/a/50556
### 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*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.