Risk-Neutral Pricing of a Log-Price Payoff

Summary

The document shows how to value a European payoff equal to the logarithm of the stock price at maturity in the Black–Scholes model. Under the risk-neutral measure, the stock follows geometric Brownian motion with constant interest rate and volatility. Taking the logarithm of its terminal value yields a deterministic component plus a scaled Brownian term.

The Brownian term has zero expectation, so the expected payoff is the initial log price plus the risk-neutral expected log return over the period. Discounting that expectation at the constant risk-free rate gives the stated option value. This illustrates risk-neutral valuation and the use of the martingale framework for a nonstandard payoff. The result relies on the document’s assumptions of constant parameters, no jumps, and a lognormal stock process. It also takes the logarithm of the stock price directly; in applications, the payoff’s units or a normalization such as a reference price may need to be specified.

Key ideas

• Under geometric Brownian motion, the terminal log stock price is normally distributed.
• The expected Brownian increment is zero under the risk-neutral measure.
• The payoff value is the discounted risk-neutral expectation of the terminal log price.
• The derivation assumes constant volatility and interest rates and a diffusion-only model.

Tags

# How would you price an option with payout ln(St) where St is the stock price at time t


# How would you price an option with payout ln(St) where St is the stock price at time t












I know it has to be done through martingales, but I am not fully sure how to do this BSM pricing.

## Answer by Gordon (score 6)

https://quant.stackexchange.com/a/31026

We assume that, under the risk-neutral measure, the stock price process $\{S_t, \, t\ge 0\}$ satisfies an SDE of the form \begin{align*} dS_t = S_t(rdt + \sigma dW_t), \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t, \, t \ge 0\}$ is a standard Brownian motion. Then \begin{align*} S_T = S_0 e^{(r-\frac{1}{2}\sigma^2) T + \sigma W_T}. \end{align*} Moreover, the option payoff $\ln S_T$ has a value given by \begin{align*} e^{-rT} E\big(\ln S_T\big) &= e^{-rT} E\Big(\ln S_0+\Big(r-\frac{1}{2}\sigma^2\Big) T + \sigma W_T \Big)\\ &=e^{-rT} \Big [\ln S_0+\Big(r-\frac{1}{2}\sigma^2\Big) T\Big]. \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.