I am trying to solve the following system of SDEs $$ \left[ \begin{array}{c}{d X_{1}} \\ {d X_{2}}\end{array}\right]=\left[ \begin{array}{l}{1} \\ {0}\end{array}\right] d t+\left[ \begin{array}{cc}{1} & {0} \\ {0} & {X_{1}}\end{array}\right] \left[ \begin{array}{c}{d W_{1}} \\ {d W_{2}}\end{array}\right] $$ I would appreciate your comment on my approach and if it is wrong or it can be further solved.

Here is my try: First, I rewrite it in the form $$ \left\{\begin{array}{l}{d X_{1}(t)=d t+d W_{1}(t)} \\ {d X_{2}(t)=X_{1} d W_{2}(t)}\end{array}\right. $$ For the first equation, I find $$ X_{1}(t)=W_{1}+t+X_{1}(0) $$ Then I rewrite the second equation as $$ d X_{2}(t)=(W_{1}+t+X_{1}(0)) d W_{2}(t) $$ I'm not exactly sure how to proceed from here. Should I just write $$ X_{2}(t)=X_{2}(0)+\int_{0}^{t} 0 d s+\int_{0}^{t} (W_{1}+t+X_{1}(0))d W_{2} $$ $$ X_{2}(t)=X_{2}(0)+(W_{1}+t+X_{1}(0)) W_{2} $$ ? I guess I'm more confused because of the different indexes and how to handle them. Thanks for your help!


You are almost there. Up to $$dX_2(t)=(W_1(t)+t+X_1(0))dW_2(t)$$ it looks good to me. All you have to do is to write this equation in its integral form (rather than getting rid of "d" from it which is not correct).

  • $\begingroup$ Oh, I missed $X_2(0)$. Does that integration look right? $\endgroup$ – Blade Apr 25 at 3:15
  • $\begingroup$ The integral form for $X_2$ looks fine to me, though not sure why you included there an integral of $0$. But the formula that follows is not correct. $\endgroup$ – Mdoc Apr 25 at 3:20

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