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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!

  • 0
    I know that $\lim_{c \to \infty}$ the expectation of the integral is $\frac{x_a}{r-\mu}$2012-01-12

1 Answers 1

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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 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

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    I think the answer in (2) of your post should be +μ.t, not -μ.t Do you think so?2012-04-02