# Deriving SDE(s) and Expectation from Given PDE

We want to solve the PDE $u_t + \left( \frac{x^2 + y^2}{2}\right)u_{xx} + (x-y^2)u_y + ryu = 0$ where $r$ is some constant and $u(x,y,T) = V(x,y)$ is given. Write an SDE and express $u(x,y,0)$ as the expectation of some function of the path $X_t, Y_t$.

Attempt: I tried to use the multivariate backward equation (2 dimensional) to recover the original SDE's and ended up with $dX_t= \sqrt{x^2 + y^2} dW_t$ and $dY_t = (x-y^2)dt + \sqrt{x^2 + y^2} dW_t$.

The problem I have is recovering the expectation. I'm not too familiar with multidimensional Feynman-Kac, but judging by the $ryu$ term and extrapolating from the one-dimensional case, the desired expectation should have the form E[exp(riemann integral of Y_t)]. Can anyone shed some light on this? Thank you.

EDIT: Oops, wrote the forward equation incorrectly and made a typo, the SDE's have changed

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Is it really missing a $u_{yy}$ term? – Jon Dec 18 '11 at 10:18
@Bob : I think you should start from that fact that you would like to express $V(x,y)=E[F(X_T,Y_T)|X_t=x,Y_t=y]$ as a martingale for a regular function $F$ where $X_t$ and $Y_t$ are following unknwon SDE (so the coefficient are the unknowns) and use Itô on $F(X_T,Y_T)$ to retrieve your PDE by setting the drift on $dF$ equal to 0. I don't feel like doing the calculations but I think this should do the trick. Regards – TheBridge Dec 18 '11 at 12:22
@Bob : By the way the final condition giving $F$ is missing in your question. – TheBridge Dec 18 '11 at 13:14
– David Dec 19 '11 at 0:38
@Bob, thanks I'll think about it. Also u(x,y,T) = V(x,y) where V is some "payoff" function is the final condition I think. – David Dec 19 '11 at 0:41
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## 1 Answer

What do you think about the system of SDEs :
$$dX_t=\sqrt{X_t^2+Y_t^2}dW_t$$ $$dY_t=(X_t-Y_t^2)dt$$ And finally :

$$u(X_t,Y_t,t)=\mathbb{E}[V(X_T,Y_T).e^{-\int_t^TrY_s.ds}|X_t,Y_t]$$

You can check that $u$ is satisfying your PDE, but as always check my calculations as I am used to making errors.

The way I found this is the following : I set $r=0$, then looking for $u$ as an expectation of $V(X_T,Y_T)$ and deriving its SDE via Itô's lemma and looking for a null drift and then indentifying terms with the original PDE with those coming from the drift of $dV$ with $dX_t=a_1(X,Y,t)dt+b_1(X,Y,t)dW_t$ and $dX_t=a_2(X,Y,t)dt+b_2(X,Y,t)dB_t$ gives the solution for $a_1,a_2,b_1,b_2$ when $r=0$ ($B$ and $W$ are independent Brownian motions, which is coming from the intuitive fact that there is no $u_{xy}$ terms in the PDE).

Then two minutes of reflection gives that $F(X_T,Y_T,T)=V(X_T,Y_T).e^{-\int_T^Tr.Y_sds}$ respects the final condition and acts on the drift part of $dF$ by only multiplying the PDE's with a $e^{-\int_t^Tr.Y_s.ds}$ and adds the $rYV$ term which was missing in the solution with $r=0$.

Best regards

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Hmm, you're right, Y should not have a diffusion term. This was a mistake on my part in working from the backward equations. I now know the approach is to use Ito or equivalently, use the tower property and make a Taylor expansion on f. There are quite a few typos in my OP, but thanks for verifying my intuition. – David Dec 19 '11 at 21:17
– David Dec 19 '11 at 21:25
@Bob : Sasha gave you an excellent answer on your new posted question. By the way did I answered properly your question ? If so you might consider accepting it. Regards. – TheBridge Dec 20 '11 at 9:34
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