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Forward-Start Option Pricing Under Geometric Brownian Motion

Article Quant Q&A · Author: glork
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Summary

The answer explains why a forward-start option on a price ratio depends on the interval between its observation dates in the Black–Scholes setting. Assuming constant interest and dividend rates and volatility, the log price follows a drifted Brownian motion. Brownian increments over intervals of equal length have the same distribution, so the log return from one date to another is distributed according to the interval length, not its calendar location.

The ratio of future to current prices can therefore be expressed using the elapsed time and a standard normal random variable. The answer then gives a discounted call-price expression with normal cumulative distribution terms. Its conclusion relies on the stated geometric Brownian motion assumptions, including constant parameters and the model's increment properties; it does not establish the same time invariance under stochastic volatility, changing rates, or anticipated events.

Key ideas

  • Under geometric Brownian motion, log-price increments over equal-length intervals share a distribution.
  • The forward price ratio depends on the time elapsed between its dates under constant model parameters.
  • The option value is expressed using a discounted normal-distribution formula.
  • Time invariance may fail when rates, volatility, or other model assumptions vary.

Tags

OptionsDerivatives pricingVolatility
Full text
# forward option, stochastic calculus


# forward option, stochastic calculus












I encounter a problem to understand this:

The price of a forward option is : $C(K,t,T)=\mathbb{E}[((S_{T}/S_{t})-K)+]$ OK

The option should only depend on $T-t$ because the yield randomness (for a week) in 2 years should be the same as the yield randomness (for a week) in 3 years and that will be the same as the yield randomness (for a week) today if no known events are expected WHY ?

Then taking $X_{t}=ln(S_{t}/S_{0})$ we should have: for all $u$ and for all $a$, $X_{u+a}-X_{u}=X_{a}$ (equality in distribution) WHY ?

Thank you

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/17684

For the last question. We assume that \begin{align*} S_t = S_0 e^{(r-q-\frac{1}{2}\sigma^2)t + \sigma W_t}, \end{align*} where $W$ is a standard Brownian motion, $r$ is the interest rate, $q$ is the dividend yield, and $\sigma$ is the volatility. Then, \begin{align*} X_{u+a}-X_a &= (r-q-\frac{1}{2}\sigma^2)a + \sigma(W_{u+a}-W_u)\\ &\sim (r-q-\frac{1}{2}\sigma^2)a + \sigma W_a\\ &= X_a. \end{align*}

For the forward start option, note that \begin{align*} S_T/S_t &= e^{(r-q-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T- W_t)}\\ &= e^{(r-q-\frac{1}{2}\sigma^2)(T-t) + \sigma \sqrt{T-t}\xi}, \end{align*} where $\xi$ is a standard normal random variable. Then \begin{align*} C(K, t, T) &= e^{-rT} \mathbb{E}\big(S_T/S_t -K)^+ \big)\\ &= e^{-rT}\big[N(d_1) - KN(d_2) \big], \end{align*} where $N$ is the cumulative distribution function of a standard normal random variable, \begin{align*} d_{1} = \frac{\ln (1/K) + (r-q+ \frac{1}{2}\sigma^2 )(T-t)}{\sigma \sqrt{T-t}}, \end{align*} and \begin{align*} d_{2} = \frac{\ln (1/K) + (r-q- \frac{1}{2}\sigma^2 )(T-t)}{\sigma \sqrt{T-t}}. \end{align*}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.