If $Z$ is a standard normal distribution $Z\sim N(0,1)$ and $C> 0$ a constant, $$E[(e^{Z}-C)^+] \text{ where} ^+ \text{entails only where the variable is positive }$$ I need to prove the mean is equal to: $$e^{1/2}\phi(1-\log(C))-C\phi(-\log(C))$$ where $\phi$ is just the standard normal cumulative distribution function.
I am really struggling to find it out but I am really stuck in the variable changes. It is part of the Black-Scholes equation coefficients.
Thanks and regards for your help.
EDIT: Thanks to epartow for your answer.
I guess that my approach was totally wrong. Just to summarize:
- I try to get the probability density function:
\begin{align} f_x&=\frac{d}{dx}P_x[X<x]=P_x[e^z<x-C]=P_x[\ z < \log (x-C)] \\ &=\frac{d}{dx}\phi(\log (x-C))=\varphi(\log (x-C))\frac{d}{dx}(\log (x-C)) \\ &=\varphi(\log (x-C))\frac{1}{(x-C)}=\frac{1}{(x-C)}\frac{1}{\sqrt{2\pi}}e^{\frac{-\log^2(x-C)}{2}}. \end{align}
Since it should be integrated only when $e^z-C>0$, the integretation for the mean shold be done between $\log C$ and $\infty$.
So for the mean we have to add $x$ to the density function so the integration should be: $$\int_{\log C}^\infty\frac{x}{(x-C)}\frac{1}{\sqrt{2\pi}}e^{\frac{-\log^2(x-C)}{2}}dx.$$ Changing variable $\log (x-C) = y$ the integral should be: $$\int(e^y+C)\frac{1}{\sqrt{2\pi}}e^{\frac{-y^2}{2}}dy,$$ but the limits are: $$\int_{\log [\log(C) - C] }^\infty$$
Could you explain to me where is my failure and why epartow is correct? I can understand the normal standard property: $$1−ϕ(x)=ϕ(−x),$$ as well as all the maths after setting the integral, but I don't understand how you get the mean with this integral instead of the probability density function multiplied by $x$.
Thanks a lot for your support.
As already mentioned by Sassatelli Giulio, we will use the identity \begin{align} 1 - \phi(x) = \phi(-x), \end{align} i.e. symmetry of the density of the standard Gaussian. We calculate \begin{align} \mathbb{E}[(e^z - C)_+] = \int_\mathbb{R} (e^z - C)_+ \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \, dz = \int_{z > \log(C)} (e^z - C) \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \, dz. \end{align} We continue by splitting the latter integral into two summands. It holds \begin{align} \int_{z > \log(C)} (e^z - C) \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \, dz &= \int_{z > \log(C)} \frac{1}{\sqrt{2\pi}} e^{\frac{1}{2} - \frac{1}{2}(z-1)^2} \, dz - C \int_{z > \log(C)} \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \, dz \\ &= e^{\frac{1}{2}}\int_{z > \log(C)} \frac{1}{\sqrt{2\pi}} e^{- \frac{(z-1)^2}{2}} \, dz - C (1 - \phi(\log(C))) \\ &= e^{\frac{1}{2}} \int_{z > \log(C) - 1} \frac{1}{\sqrt{2\pi}} e^{- \frac{z^2}{2}} \, dz - C \phi(-\log(C)) \\ &= e^{\frac{1}{2}} (1 - \phi(\log(C) - 1)) - C \phi(-\log(C)) \\ &= e^{\frac{1}{2}} \phi(1 - \log(C)) - C \phi(-\log(C)). \end{align}