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Is there a formula for Black and Scholes when we have expected payoff $\mathbb{E}[\max(se^{X}-K,0)]$ for $X$ having any normal distribution?

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Let $X\sim N(m,v^2)$ be normally distributed. Then, for all strikes $K>0$ and $\omega\in\{-1,1\}$, \begin{align*} \mathbb{E}[\max\{\omega(e^X-K),0\}]=\omega e^{m+\frac{1}{2}v^2}\Phi\left(\omega\frac{m-\ln(K)+v^2}{v}\right)-\omega K\Phi\left(\omega\frac{m-\ln(K)}{v}\right), \end{align*} where $\Phi$ is the standard normal cdf. The indicator $\omega$ is used to differentiate whether you have a call option ($\omega=1$) or a put option ($\omega=-1$).

It follows directly from integrating the log-normal density. Brigo and Mercurio call this ``a useful calculation''.

The Black-Scholes formula is a special case of this equation where $m=\ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)T$ and $v^2=\sigma^2T$.

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  • $\begingroup$ what about the $s$? $\endgroup$ – Nocturnal Aug 27 '20 at 14:58
  • $\begingroup$ @Nocturnal Note that $se^X=e^{\ln(s)+X}$. Thus, you simply add $\ln(s)$ to the mean $m$ but don't change the variance $v^2$. This is the reason why $m$ starts with $\ln(S_0)$ in the Black-Scholes case. $\endgroup$ – Kevin Aug 27 '20 at 15:39

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