Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please I need help with the evaluation of this integral. I've tried with both mathematica and maple, but to no avail. Here is the integral:

$$ e^{-r(T-t)}\int_{-\infty}^{\infty}\frac{(Se^{x})^{n}}{\sqrt{2\pi\sigma^{2}(T-t)}}\exp\left[-\frac{\left\{ x-\left(r-\frac{1}{2}\sigma^{2}\right)\left(T-t\right)\right\} ^{2}}{2\sigma^{2}\left(T-t\right)}\right]\mbox{dx} $$ for $n\geq 2$.

Thanks.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Set $\mu = (r-\frac{1}{2}\sigma^2)(T-t)$, $\tau^2 = \sigma^2(T-t)$. Then we have $$e^{-r(T-t)} S^n \int_{-\infty}^\infty e^{nx} \frac{1}{\sqrt{2 \pi \tau^2} }e^{-(x-\mu)^2/2\tau^2}dx$$. The integral is thus $E[e^{nZ}]$ where $Z$ is a normal random variable with mean $\mu$ and variance $\tau^2$. Knowing the moment generating function for normal random variables says the integral equals $e^{\mu n + n^2\tau^2/2}$.

share|improve this answer
    
Thanks. That was very insightful. I really appreciate your help. I really like your approach, but I was wondering if there is another way. –  user8280 Apr 5 '11 at 3:32
1  
@Eric: you can complete the square in the final integral. –  Fabian Apr 5 '11 at 6:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.