in their paper "European Real Options: An intuitive algorithm for the Black and Scholes Formula" Datar and Mathews provide a proof in the appendix on page 50, which is not really clear to me. It's meant to show the equivalence of their formula $E_{o}(max(s_{T}e^{-\mu T}-xe^{-rT},0))$ and Black and Scholes.

They refer to Hull(2000), define $y=s_{T}e^{-\mu T}$, and then do the following transformation:

$E_{o}(max(s_{T}e^{-\mu T}-xe^{-rT},0))$ $=\intop_{-xe^{-rT}}^{\infty}(s_{T}*e^{-\mu T})g(y)dy$ $=E(s_{T}e^{-\mu T})N_{d_{1}}-xe^{-rT}N_{d_{2}}$

An addition: Actually, in the paper it says $E_{o}(max(s_{T}e^{-\mu T}-xe^{-rT}),0)$, so the 0 is outside the brackets. However, I am not sure, if that is a typo and should rather be $E_{o}(max(s_{T}e^{-\mu T}-xe^{-rT},0))$. I am not familiar with a function E(max(x),0)

$\mu$ and $r$ are two different discount rates, one being the WACC and the other one the riskless rate.

Could I substitute $V=s_{0}e^{-\mu T},K=xe^{-rT}$, go through the BS steps and re-substitute? In other words, under what constrains is $E\left[max(V-K,0)\right]=E(V)N(d_{1})-KN(d_{2})$ valid?

The research related to it is a comparison of different real option pricing method.

Could anybody help me out?

Thanks in advance.



1 Answer 1


I don't know what $\mu$ stands for in the model so let me just recall the standard Black-Scholes formalism. It's likely that everything can be extended with minor modifications to the model you're interested in.

The price of the vanilla call option with a strike $K$ is equal to the expectation of the discounted pay-off $$C_K=\mathbb E(e^{-rT}(S_T-K)_+),$$ where $(S_T-K)_+:=\max(S_T-K,0)$ and $\mathbb E$ is taken with respect to the risk-neutral measure $\mathbb P$. Assuming that $\mathbb P$ admits a continuous density $p(y)$, we have that $$\mathbb E(e^{-rT}(S_T-K)_+)=\int_{-\infty}^{\infty}e^{-rT}(S_T-K)_+p(S_T)dS_T.$$ Now, in the risk-neutral Black-Scholes world $$S_T=S_0\exp\left(rT-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}N(0,1)\right).$$ Recalling that the density of $N(0,1)$ is $\frac{1}{\sqrt{2\pi}}e^{-s^2/2}$, we get that $$C_K=\frac{e^{-rT}}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-s^2/2}\left(S_0\exp\left(rT-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}s\right)-K\right)_+ds.$$ The integrand is non-zero if and only if $$S_0\exp\left(rT-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}s\right)>K,$$ i.e. when $$s> a=\frac{\ln(K/S_0)+\sigma^2T/2-rT}{\sigma\sqrt{T}}.$$ Therefore $$C_K=\frac{e^{-rT}}{\sqrt{2\pi}}\int_{a}^{\infty}e^{-s^2/2}S_0\exp\left(rT-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}s\right)ds-\frac{e^{-rT}}{\sqrt{2\pi}}\int_{a}^{\infty}e^{-s^2/2}Kds,$$ which implies after some straightforward manipulations the standard Black-Scholes formula $$C_K=S_0N(d_1)-Ke^{-rT}N(d_2).$$ Note that $S_0=\mathbb E(e^{-rT}S_T)$ since the discounted stock price is a martingale under the risk-neutral measure.

  • $\begingroup$ The $\mu$ is a different discount factor, set to the WACC. I still have some problems with the integral transformation from dSt to ds with the use of N(). Could you elaborate a bit, or provide a reference for reading? Cheers $\endgroup$
    – Corn
    Commented May 13, 2011 at 12:44

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