Dear Wilmott community,I have been wondering this particular problem for several days and tried several approaches to get the solution. This should be quite trivial and elementary. However, I am unable to derive the solution myself.How do you implement integration by parts to get the solution:[$]E[(S-K)+]=\int_{\infty }^{K}\left ( \frac{1}{\sigma \sqrt{2\pi }} \right )\int_{x}^{\infty }e^{{-(t-\mu)^2 }/2\sigma ^2} dtdx=\frac{\sigma }{\sqrt{2\pi }}e^{-(-\mu -K)^2/2\sigma ^2}+(\mu-K)\phi \left ( \frac{\mu-K}{\sigma } \right )[$]Proof given by texbook begins like this: For any [$]x\geq 0,(X-K)^+\geq x[$] if and only if [$] X-K\geq x [$] which is true if and only if [$]X\geq x+K[$]and[$]P((X-K)^+\geq x)=1-P((X-K)^+ < x )=1-P(X<K+x )=1-\int_{-\infty }^{K+x}f(t)dt)[$]Therefore we have:[$]E[(X-K)^+]=\int_{K }^{\infty}\left ( \int_{\infty }^{K+x} f(t)dt)\right )dx,u=t-K,\rightarrow \int_{K }^{\infty}\left (1- \int_{-\infty }^{x} f(t)dt)\right )dx=\int_{K }^{\infty}\left (\int_{x}^{\infty} f(t)dt)\right )dx[$]Now, if we apply this formula to X which is a normal random variable and integrate by parts using[$]E[(X-K)^+]=\int_{K }^{\infty}\left ( \int_{x }^{\infty} \frac{1}{\sigma \sqrt{2\pi }}e^{-(t-\mu)^2/2\sigma ^2 }dt\right )dx[$][$]u=\int_{x }^{\infty} \frac{1}{\sigma \sqrt{2\pi }}e^{-(t-\mu)^2/2\sigma ^2 }dt, v=x[$][$]du= {\frac{1}{\sigma \sqrt{2\pi }}e^{-(t-\mu)^2/2\sigma ^2}}, dv=dx[$]Next step is unclear to me:[$]E[(X-K)^+]=\frac{-K}{\sigma \sqrt{2\pi }}\int_{K}^{\infty}e^{-(t-\mu )^2 /2\sigma ^2}dt+\frac{1}{\sigma \sqrt{2\pi }}\int_{K}^{\infty}te^{-(t-\mu )^2 /2\sigma ^2}dt[$]How are you supposed to get this result with the substitutions?

Last edited by JSHellen on June 8th, 2014, 10:00 pm, edited 1 time in total.

Frankly, none of this makes any sense. If I understand correctly what you want, you need to compute this[$]\int_{\ln(K/S)}^{+\infty}\left(Se^t-K\right)\exp\left(-\frac{(t-\mu)^2}{2\sigma^2}\right)\frac{dt}{\sqrt{2\pi}\sigma}[$]There is no really need to integrate by parts, it is a standard integral.

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