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JSHellen
Topic Author
Posts: 13
Joined: July 10th, 2013, 6:42 am

Expectation of max(S-K;O

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.

EBal
Posts: 431
Joined: May 20th, 2005, 1:30 pm

Expectation of max(S-K;O

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.

JSHellen
Topic Author
Posts: 13
Joined: July 10th, 2013, 6:42 am

Thanks!