# Black-Scholes formula given arbitrary value of $S_{T}$

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?

## 1 Answer

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 $$v^2=\sigma^2T$$.

• what about the $s$? – Nocturnal Aug 27 '20 at 14:58
• @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. – Kevin Aug 27 '20 at 15:39