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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!

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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

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 :
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.


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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

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