Pricing a Floating Strike Lookback Call from the Brownian Minimum

Summary

The document considers a floating strike lookback call under the Black–Scholes model without dividends. Its payoff at maturity is the terminal stock price minus the lowest stock price observed over the option’s life. The special condition sets the risk-free rate equal to half the variance, simplifying the stock process to an exponential of Brownian motion.

The answer reframes the valuation as the expected terminal stock price minus the expected running minimum, with the latter expressed through the minimum of Brownian motion. It recommends calculating that expectation from the density of the Brownian minimum. The excerpt does not carry out the integration, give a closed-form option price, or explain discounting in detail; it presents the setup for the calculation rather than a complete derivation. The question’s proposed expectation formula is not verified in the answer.

Key ideas

• Under the stated rate condition, the stock can be represented as its initial value times an exponential Brownian term.
• The floating strike payoff is the terminal stock value less the path minimum.
• Linearity of expectation separates the payoff valuation into a terminal-value term and a minimum-value term.
• The Brownian minimum density provides a route to calculating the expected running minimum.
• The excerpt gives a valuation method outline but not the final integration or price.

Tags

# Floating Strike Lookback Call Option


# Floating Strike Lookback Call Option












Assume the risk-free bond $B_t$ and the stock $S_t$ follow the dynamics of the Black & Scholes model without dividends (with interest rate $r$, stock drift $\mu$ and volatility $\sigma$).

If $r=\frac{\sigma^2}{2}$. Compute the price at time $t = 0$ of the lookback call option with maturity $T$, that is the option with payoff $S_T-min_{t\in[0,T]}S_t$ at time $T$.

If I follow the PDE approach to computing the price of the lookback call with floating strike, how do I make use of the condition $r=\frac{\sigma^2}{2}$? Do they mean that then $S_t=S_0e^{\sigma W_t}$? This is what I got when I calculated $\mathbb{E}[min_{[0,T]}S_t]=2\mathbb{E}[S_T]\Phi(-\sigma\sqrt{t})$. How do I further use this result in computing the price of the floating lookback?

Would really appreciate all the help I can get!

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/49559

Under the condition $r=\frac{\sigma^2}{2}$, it is true that $S_t = S_0e^{\sigma W_t}$. Since \begin{align*} E\Big( S_T - \min_{0 \le t \le T} S_t\Big) = E\big( S_T\big) - E\Big(\min_{0 \le t \le T} S_t\Big), \end{align*} what you need is the expectation $E\big(\min_{0 \le t \le T} S_t\big)$. Note that \begin{align*} \min_{0 \le t \le T} S_t = S_0e^{\sigma \min_{0 \le t \le T} W_t}. \end{align*} Using the density function of $\min_{0 \le t \le T} W_t$, the expectation $E\big(\min_{0 \le t \le T} S_t\big)$ can then be computed directly.

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