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Black–Scholes Pricing for a Call on the Stock’s Increase

Article Quant Q&A · Author: user18866
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Summary

The document derives Black–Scholes prices for two contingent claims under geometric Brownian motion. The first pays the positive part of the difference between the stock price at maturity and its price at an earlier observation date. Conditioning on the stock price at that date reduces the payoff to a call-style expression, which is then averaged to obtain the time-zero value using the normal cumulative distribution function.

Key ideas

  • The first payoff is equivalent to a call struck at the stock price observed at the intermediate date.
  • Conditioning on the intermediate information gives a Black–Scholes form for the conditional value.
  • The derivation discounts the conditional value back to time zero and averages over the earlier stock price.
  • A claim paying the greater of the two observed stock prices is decomposed into the earlier price plus the call-style payoff.

Tags

OptionsDerivatives pricingEquities
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# Black Scholes Geometric Brownian Motion Option Pricing


# Black Scholes Geometric Brownian Motion Option Pricing












I'm doing a past paper for one of my masters modules and I'm stuck on this

and I have no idea how to tackle such a thing. It's worth 30% of the exam so would be great if anyone here has any suggestions.

Thanks

## Answer by Gordon (score 2)

https://quant.stackexchange.com/a/22605

For Question (1), the payoff is given by \begin{align*} (S_T-S_{T_0})^+. \end{align*} Note that \begin{align*} E\left(e^{-r(T-T_0)} (S_T-S_{T_0})^+ \mid \mathscr{F}_{T_0}\right) &=S_{T_0}N(d_1)-e^{-r(T-T_0)}S_{T_0}N(d_2), \end{align*} where \begin{align*} d_1 &= \frac{(r+\frac{1}{2}\sigma^2)(T-T_0)}{\sigma \sqrt{T-T_0}}\\ &=(r+\frac{1}{2}\sigma^2)\sqrt{T-T_0}/\sigma, \end{align*} and \begin{align*} d_2 = d_1 - \sigma \sqrt{T-T_0}. \end{align*} The option price is then given by \begin{align*} C_0 &\equiv e^{-rT}E\left( (S_T-S_{T_0})^+ \right)\\ &=e^{-rT_0}E\left(E\left(e^{-r(T-T_0)} (S_T-S_{T_0})^+ \mid \mathscr{F}_{T_0}\right)\right)\\ &=e^{-rT_0}E\left(S_{T_0}N(d_1)-e^{-r(T-T_0)}S_{T_0}N(d_2) \right)\\ &=S_0 \big[N(d_1) - e^{-r(T-T_0)} N(d_2)\big].\tag{*} \end{align*}

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For Question (2), note that \begin{align*} \max(S_T,\, S_{T_0}) = S_{T_0} + (S_T-S_{T_0})^+. \end{align*} Then, from (*), the contingent price is given by \begin{align*} P_0 &\equiv e^{-rT}E\left( S_{T_0} + (S_T-S_{T_0})^+ \right)\\ &=e^{-r(T-T_0)}S_0 + S_0 \big[N(d_1) - e^{-r(T-T_0)} N(d_2)\big]\\ &=S_0 \big[N(d_1) + e^{-r(T-T_0)} (1-N(d_2))\big]\\ &=S_0 \big[N(d_1) + e^{-r(T-T_0)} N(-d_2)\big]. \end{align*}

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