So, I need some help to move forward with this problem.

$$ \begin{cases} \frac{\partial F(t,x,y)}{\partial t}+\frac{1}{2}\frac{\partial^2 F(t,x,y)}{\partial x^2}+\frac{9}{2}\frac{\partial^2 F(t,x,y)}{\partial y^2}+\frac{\partial^2 F(t,x,y)}{\partial x \partial y}-F(t,x,y)=0 \ \ \ (t,x) \in[0,T) x R^2\\ F(T,x,y)=x^2y\\ \end{cases} $$

I have concluded that the C matrix defined as $\sigma*\sigma^T=\begin{matrix} 1 & 1\\ 1 & 3 \end{matrix}$

and that $\mu_{1}=\mu_{2}=0$

and that I should solve $$F(t,x,y)=E_{t,x,y}[e^{-(T-t)}x^2y]$$

but my problem is to figure out how $x$ and $y$ should look like. My attempt was $$dX(s)=dW_{1}(s)+dW_{2}(s)$$ $$dY(s)=3dW_{2}(s)+dW_{1}(s)$$ since they should be correlated given the problem, but I'm unsure if I have understood how to define $dX(s)$ and $dY(s)$ properly.

Well, if i preceed with $dX(s)$ and $dY(s)$ as defined above I get


which does not lead me to the correct answer. Anyone with some guiding for me?

  • $\begingroup$ Can you specify what the correct answer is supposed to be? $\endgroup$ – Daneel Olivaw Nov 24 '17 at 1:20
  • $\begingroup$ @DaneelOlivaw Well, i substituted the F(t,x,y) that I received into the PDE, which did not lead to 0.. My book does not provide solutions I'm afraid $\endgroup$ – Femman Nov 24 '17 at 11:09

Try with

\begin{align} dX(s)&=dW_{1}(s) \\ dY(s)&=dW_{1}(s)+\sqrt{2}dW_{2}(s) \end{align}

The covariance matrix of these differentials is $$\left(\begin{array}{cc} 1 & 1 \\ 1 & 3\end{array}\right) \; .$$

  • $\begingroup$ Thanks! I try to calculate $F(t,x,y)=E_{t,x,y}[e^{-(T-t)}x^2y]=e^{-(T-t)}E_{t,x,y}\bigg[\bigg(x+\big(W_{1}(T)-W_{1}(t)\big)\bigg)^2\bigg(y+\sqrt{2}\big(W_{2}(T)-W_{2}(t)\big)+\big(W_{1}(T)-W_{1}(t)\big)\bigg)\bigg]=e^{-(T-t)}[(x^2+x\big(W_{1}(T)-W_{1}(t)\big)+\big(W_{1}(T)-W_{1}(t)\big)^2)(y+\sqrt{2}\big(W_{2}(T)-W_{2}(t)\big)+\big(W_{1}(T)-W_{1}(t)\big)\bigg)]=e^{-(T-t)}[x^{2}y+x(T-t)+y(T-t)+\sqrt{2}x(T-t)]$ which is not correct. I use that \big(W_{.}(T)-W_{.}(t)\big)^2=(T-t) and that \big(W_{.}(T)-W_{.}(t)\big)\big(W_{-}(T)-W_{-}(t)\big)=(T-t) as well, but maybe the last one is wrong? $\endgroup$ – Femman Nov 27 '17 at 16:10
  • $\begingroup$ You forgot a factor 2 when working out the square. And the last term is not there because your two Brownian motions are independent. $\endgroup$ – Raskolnikov Nov 27 '17 at 16:51
  • 1
    $\begingroup$ Perfect! I misinterpreted that $W_{1}$ and $W_{2}$ were dependent, when it should be X and Y that are. Now I finally get it, thank you! (I also realized that C should be $$\begin{bmatrix} 1 & 1 \\ 1 & 9 \end{bmatrix}$$ giving me \begin{cases} dX&=dW_{1}(s)\\ dY&=dW_{1}(s)+2\sqrt{2}dW_{2}(s) \end{cases} $\endgroup$ – Femman Nov 27 '17 at 21:05

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