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.

I would like to compute the expectation of the following expectation

$\mathbb{E}[\int_a^\infty e^{-rt}\min(x_t,c)\,dt]\,$

where a, r, c are constants, $dx_t = \mu x_t dt + \sigma x_t dW_t$ is a geometrical Brownian motion with $(\mu < r)$ and $\min(x_t,c)$ denotes the minimum of $x_t$ and c. Any help would be much appreciated!

share|improve this question
    
What do you know? What did you try? Where are you stuck? –  Did Jan 12 '12 at 6:15
    
I know that $\lim_{c \to \infty}$ the expectation of the integral is $\frac{x_a}{r-\mu}$ –  ant Jan 12 '12 at 14:26
add comment

1 Answer 1

Here are my two cents, you have :

1-$\mathbb{E}[\int_a^\infty e^{-rt}\min(x_t,c)dt]=\int_a^\infty e^{-rt}\mathbb{E}[\min(x_t,c)]dt$ (from Fubini but see why Fubini can be applied here ?)

2-$\mathbb{E}[\min(x_t,c)]=c+\mathbb{E}[\min(x_t-c,0)]=c+x_0.e^{-\mu.t}.(1-N[d])-c.(1-N[d'])$ which can simplified a little further. Where :
$d=\frac{Ln(c/x_0)-(1/2\sigma^2+\mu)t}{\sqrt{t}\sigma}$
$d'=\frac{Ln(c/x_0)+(1/2\sigma^2-\mu)t}{\sqrt{t}\sigma}$
and $N(x)=P(X<x)$ for standard normal random variable $X$.

Now integrating 1 from 2 with respect to time explicitly seems difficult, but this can be done numerically by a math-software.

Regards

share|improve this answer
    
Thank your very much for your help. Indeed, I was trying to get a closed-form solution since the value for the expectation when $c=\infty$ is so simple. But couldn't find any paper where this is discussed, although, I believe, this problem should be relevant in the field of mathematical finance. Possibly because as you say is difficult (if possible) to get a closed-form solution. –  ant Jan 12 '12 at 16:19
    
@ant : Well you know it depends on what you mean by closed form. For many people obtaining the integral of (2) of my post is considered to be a closed form. Regards. –  TheBridge Jan 13 '12 at 8:09
    
I think the answer in (2) of your post should be +μ.t, not -μ.t Do you think so? –  user28138 Apr 2 '12 at 12:37
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.