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I need to find an algebraic expression for E(max{X-a,0}), where X has a lognormal distribution with mean mu and standard deviation sigma. So far, I have derived the following expression, but I could use some help to solve the first part of the last equation, I do think it should have somewhat of a form of a normal distribution pdf:

\begin{array}{l} E[X-a]^{+}=\int_{a}^{\infty}(x-a) h(x) d x=\int_{a}^{\infty}(x-a) \frac{1}{x \sigma \sqrt{2 \pi}} \exp \left(-\frac{1}{2}\left(\frac{\log (x)-\mu}{\sigma}\right)^{2}\right) d x\\ \text { substitute } y=\frac{\log (x)-\mu}{\sigma} \quad x=\exp (\sigma y+\mu)\\ d x=\sigma \exp (\sigma y+\mu) d y\\ E[x-a]^{+}=\frac{1}{\sqrt{2 \pi}} \int_{\frac{\log (a)-\mu}{\sigma}}^{\infty}(\exp (\sigma y+\mu)-a) \cdot \exp \left(-\frac{1}{2} y^{2}\right) d y\\ =\frac{1}{\sqrt{2 \pi}} \int_{\frac{\log (a)-\mu}{\sigma}}^{\infty} \exp \left(\sigma y+\mu-\frac{1}{2} y^{2}\right) d y-a \int_{\frac{\log (a)-\mu}{\sigma}}^{\infty} \frac{1}{\sqrt{2 \pi}} \exp \left(-\frac{1}{2} y^{2}\right) d y\\ =\frac{1}{\sqrt{2 \pi}} \int_{\frac{\log (a)-\mu}{\sigma}}^{\infty} \exp \left(\sigma y+\mu-\frac{1}{2} y^{2}\right) d y-a[\Phi(y)]_{\frac{\log (a)-\mu}{\sigma}}^{\infty}\\ =\frac{1}{\sqrt{2 \pi}} \int_{\frac{\log (a)-\mu}{\sigma}}^{\infty} \exp \left(\sigma y+\mu-\frac{1}{2} y^{2}\right) d y-a\left(1-\Phi\left(\frac{\log (a)-\mu}{\sigma}\right)\right) \end{array}

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Complete the square in the exponent in the remaining integral and rewrite terms such as $\exp(\sigma y + \mu - \frac{1}{2}y^2)$ as $e^{\mu}\exp(\sigma y - \frac{1}{2}y^2)$ before and after completing the square. Then you can write that integral as a $\Phi$ of some argument, too.

Also, I would suggest you look for questions/answers related to this in Math Stack Exchange

https://math.stackexchange.com/search?q=lognormal+black+scholes

The question you are asking is a standard element of the Black-Scholes theory for derivative pricing.

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    $\begingroup$ Thanks! If I follow your steps correctly, I find: \begin{aligned} & \int_{\frac{\log (a)-\mu}{\sigma}}^{\infty} \frac{1}{\sqrt{2 \pi}} \exp \left(\sigma y+\mu-\frac{1}{2} y^{2}\right) d y \\ =& \exp (\mu) \int_{\frac{\log (a)-\mu}{\sigma}}^{\infty} \frac{1}{\sqrt{2 \pi}} \exp \left(-\frac{1}{2}(y-\sigma)^{2}+\frac{1}{2} \sigma^{2}\right) d y \\ =& \exp \left(\mu+\frac{1}{2} \sigma^{2}\right)[\Phi(y-\sigma)]_{\frac{\log (a)-\mu}{\sigma}}^{\infty} \\ =& \exp \left(\mu+\frac{1}{2} \sigma^{2}\right)\left(1-\Phi\left(\frac{\log (a)-\mu-\sigma^{2}}{\sigma}\right)\right) \end{aligned} Correct? $\endgroup$
    – Jelle van Lin
    Commented Dec 30, 2021 at 19:42
  • $\begingroup$ That looks perfectly correct to me. $\endgroup$
    – Shannon Starr
    Commented Jan 1, 2022 at 14:14

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