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Pricing a Call on the Stock’s Gain from Midpoint to Maturity

Article Quant Q&A · Author: Pandaaaaaaa
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Summary

The payoff is the positive part of the stock-price change between the midpoint of the contract period and maturity. The question proposes factoring out the midpoint stock price and treating it as a numeraire, then applying a Black–Scholes call formula to the subsequent stock-price ratio. The reply instead develops the pricing argument in a standard Black–Scholes market with a stock and money-market account, and changes from the bank-account numeraire to the stock numeraire.

Under the stock measure, the stock has a shifted drift, and the payoff can be rewritten in terms of the ratio of the earlier stock price to its maturity value. That ratio is lognormal, allowing the payoff expectation to be evaluated using a Black–Scholes-style formula. The response gives the measure-change setup and distributional expression rather than a final closed-form price. Its derivation assumes the stated standard model with constant volatility and a traded stock numeraire; the original proposal to use the stock at the midpoint is not directly justified, since a numeraire must be a tradable asset process over the relevant pricing horizon. The valuation also presumes the specified payoff and model assumptions.

Key ideas

  • The payoff depends on the stock’s increase from the midpoint date to maturity.
  • Factoring out a stock price motivates a change of numeraire, but the chosen numeraire must be tradable over the pricing horizon.
  • The response changes measure from the bank account to the stock numeraire in a Black–Scholes setting.
  • Under the stock measure, the stock process has a shifted drift while retaining its volatility.
  • A lognormal stock-price ratio permits a Black–Scholes-style evaluation, though no final price is reported.

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# Pricing a call option with pay-off function max{$S_T - S_{T/2}, 0$}


# Pricing a call option with pay-off function max{$S_T - S_{T/2}, 0$}












Pricing a call option with payoff function $C=\max\{S_T - S_{T/2}, 0\}$, where $S_T$ is geometric brownian motion. I appreciate any help! Please close this question if this is a duplicated question. Thanks all!

My approach is to take out $S_{T/2}$:

$$C = S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}$$

Then we can define a risk neutral measure:

$$\begin{align} E[C] & = E\left[S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \\[3pt] & = \tilde{E}\left[\max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \end{align}$$

where we use $S_{T/2}$ as the numeraire. Then plug into the call option Black-Scholes formula with $S=1$, $K=1$ and $r=0$.

Is this approach correct? My main concern is, can we use $S_{T/2}$ as the numeraire?

## Answer by Daneel Olivaw (score 2)

https://quant.stackexchange.com/a/42047

Let us assume we are in a standard Black-Scholes setting with 2 traded assets, a money market account $B_t$ and a stock $S_t$, such that: $$\begin{align} \text{d}B_t & = rB_t\text{d}t \\ \text{d}S_t &= rS_t\text{d}t+\sigma S_t\text{d}W^Q_t \end{align}$$ where $W^Q_t$ is a Brownian motion under the risk-neutral measure associated to the bank account $B_t$. Defining $\tilde{B}_t = B_t/S_t$, by Itô's Lemma and Girsanov theorem: $$\begin{align} \text{d}\tilde{B}_t&=\sigma^2\tilde{B}_t\text{d}t-\sigma\tilde{B}_t\text{d}W^Q_t \\ &=\sigma\tilde{B}_t\left(\sigma\text{d}t-\text{d}W^Q_t\right) \\ &=\sigma\tilde{B}_t\text{d}W^S_t \end{align}$$ where $W^S_t=\sigma t-W^Q_t$ is a Brownian Motion under the measure associated to the stock numéraire, thus: $$\begin{align} \text{d}S_t & = rS_t\text{d}t+\sigma S_t\text{d}W^Q_t \\ &=(r+\sigma^2)S_t\text{d}t-\sigma S_t\text{d}W^S_t \end{align}$$ Under measure $Q^S$ the stock price is lognormal with drift $r+\sigma^2$ and volatility $\sigma$. It comes by a change of measure: $$\begin{align} E^Q_t\left[\frac{B_t}{B_T}\left(S_T-S_{T/2}\right)^+\right]&=E^Q_t\left[\frac{B_tS_T}{B_T}\left(1-\frac{S_{T/2}}{S_T}\right)^+\right] \\ &=S_tE^S_t\left[\left(1-\frac{S_{T/2}}{S_T}\right)^+\right] \end{align}$$ Yet: $$ \frac{S_{T/2}}{S_T}=\exp\left\{\sigma\sqrt{\frac{T}{2}}Z-\left(r+\frac{\sigma^2}{2}\right)\frac{T}{2}\right\}$$ where $Z \sim (-Z) \sim \mathcal{N}(0,1)$ is a standard Normal random variable. You can now recycle the Black-Scholes formula to find the price of the option.

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