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- Thread starter Jason
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- Thread starter
- #1

- Feb 13, 2012

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In 'standard form' the SDE is written as...I've been at this for ages but I can't make sense of it. Can somebody help me out?

Use Ito's Lemma to solve the stochastic differential equation:

[tex]X_t=2+\int_{0}^{t}(15-9X_s)ds+7\int_{0}^{t}dB_s[/tex]

and find:

[tex]E(X_t)[/tex]

$\displaystyle d X_{t}= (15-9\ X_{t})\ dt + 7\ dW_{t}\ ,\ x_{0}=2$ (1)

The (1) is a

http://www.mathhelpboards.com/f23/unsolved-statistic-questions-other-sites-part-ii-1566/index2.html#post8411

... and its solution is...

$\displaystyle X_{t}= \varphi_{t}\ \{ x_{0} +\int_{0}^{t} \varphi_{s}^{-1}\ u_{s}\ ds + \int_{0}^{t} \varphi_{s}^{-1}\ v_{s}\ dW_{s} \}$ (2)

Here is $a_{t}=-9$ , so that is $\varphi_{t}=e^{-9 t}$,$u_{t}=15$, $v_{t}=7$ and $x_{0}=2$ so that (2) becomes...

$\displaystyle X_{t}= e^{- 9 t}\ \{ 2 + 15\ \int_{0}^{t} e^{9 s} ds + 7\ \int_{0}^{t} e^{9 s}\ dW_{s} \}= e^{-9 t}\ \{2 + \frac{5}{3}\ (e^{9 t}-1) + \frac{7}{9}\ (e^{9 W_{t}}-1) - \frac{7}{2}\ (e^{9t}-1) \} = $

$\displaystyle = - \frac{11}{6} + \frac {55}{18}\ e^{-9 t} + \frac{7}{9}\ e^{9\ (W_{t}-t)}$ (3)

Kind regards

$\chi$ $\sigma$