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I have to solve $dX_t=5\,dt+3X_t\,dW_t$


Let $$Y_t:=X_t\exp(-3W_t+\frac{9}{2}t)=X_t\cdot Z_t$$

Calculating differential of $Y_t$ we have

$$dY_t=X_t(9Z_t\,dt-3Z_t\,dW_t)+(5\,dt+3X_t\,dW_t)Z_t+(5\,dt+3X_t\,dW_t)(9Z_t\,dt-3Z_t\,dW_t)=X_t(9Z_t\,dt-3Z_t\,dW_t)+(5\,dt+3X_t\,dW_t)Z_t-9X_tZ_t\,dt=5Z_t\,dt=5\exp(-3W_t+\frac{9}{2}t)\,dt$$


$$X_t=Y_t\exp(3W_t-\frac{9}{2}t)=\exp(3W_t-\frac{9}{2}t)+5 \Big(\int_0^t \exp(-3W_s+\frac{9}{2}s)\,ds \Big)\exp(3W_t-\frac{9}{2}t)$$$$=\exp(3W_t-\frac{9}{2}t)+5\int_0^t \exp(-3(W_s-W_t)+\frac{9}{2}(s-t))\,ds$$

It seems ok to me, but I would like to assure myself before my exams.

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    $\begingroup$ To reassure your self..I would use Ito again to see if your solution returns the sde. $\endgroup$ – Chinny84 Feb 11 '15 at 20:34
  • $\begingroup$ hmm i think it returns only $5dt$ ;/ $\endgroup$ – luka5z Feb 11 '15 at 20:42
  • $\begingroup$ oh i forgot about $Y_0$ $\endgroup$ – luka5z Feb 11 '15 at 20:43
  • $\begingroup$ Now should be ok? $\endgroup$ – luka5z Feb 11 '15 at 20:45
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    $\begingroup$ Check below. You seem fine. $\endgroup$ – Chinny84 Feb 11 '15 at 20:58
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$$ \partial_tX_t = 5-\frac{9}{2}X_t\\ \partial_{W_t}X_t = 3X_t \\ \partial^2_{W_t}X_t = 9X_t $$ The first of the above comes from the fact $$ \frac{\partial}{\partial t}\int^t_0 g(t,W_t,W_s, s) ds = g(t,W_t,W_t, t) + \int^t_0 g' ds \tag{*} $$ Thus using Ito we have $$ dX_t = \left(5-\frac{9}{2}X_t +\frac{1}{2}9X_t\right)dt + 3X_t dW_t $$

So you were correct as far as I can see.

Ps forgive the lazy notation when computing the integral. In step (*)

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