option pricing - Feynman Kac and choice of measure

I seem to be confused on this topic. So I write my SDE without a drift to make it simple: $$dX_t=dW_t$$ and before I get to any finance there is a relation that the solution to $$u_t+0.5u_{xx}-ru=0$$ can be written as an expectation $$\mathbb{E}[e^{-rT}f(X_T)]$$ at time 0. Expectation is written in a measure where $W_t$ is defined.

Now we look at finance and say that if we choose a BM under RN measure this expectation resembles RN formula! Did not change anything about the PDE, we just gave a name to a measure. But what if I started choosing $W_t$ in a different measure, say associated with a numerraire $N_t$ with $dN_t=adt+bdW_t$? Then I get from finance arguments the price of a derivative $$u(t,x)=\mathbb{E}^N[N(t)/N(T)f(X_T)]$$ There is no discounting anymore, so can I still apply FC and get a different PDE? So the pde derived using Feynman Kac formula looks different for different choice of measures?

Answer

1) Feynmann-Kac and Girsanov

First you should remember that the process $X$ is independent of the measure you are considering.

Now let's consider a change of measure from ${\mathbb{P}}$ to ${\mathbb{Q}}$. Let us assume $\mathbb{E}_t^{\mathbb{P}}[\tfrac{d{\mathbb{Q}}}{d{\mathbb{P}}}] = e^{\theta W_t^P - \frac{1}{2}\theta^2 t}$ for some constant $\theta$. The BM $W^{\mathbb{P}}$ under ${\mathbb{P}}$ is no longer a BM under ${\mathbb{Q}}$. But Girsanov tells us that $dW^{\mathbb{Q}} = dW^{\mathbb{P}} - d\langle W^{\mathbb{P}}_t,\log \mathbb{E}_t^P[\tfrac{d{\mathbb{Q}}}{d{\mathbb{P}}}]\rangle = dW^{\mathbb{P}} - \theta dt$ is a BM under ${\mathbb{Q}}$.

If you rewrite the SDE of $X$ in terms of this new BM, you see a drift term $d\langle X_t,\log \mathbb{E}_t^{\mathbb{P}}[\tfrac{d{\mathbb{Q}}}{d{\mathbb{P}}}]\rangle$ appear. In your case, this reads $$dX_t = \theta dt + dW^{\mathbb{Q}}_t$$ Now you can apply Feynman-Kac which tells you $$u^{\mathbb{Q}}(t,x) := \mathbb{E}^{\mathbb{Q}}[e^{-rT}f(X_T)|X_t = x]$$ is going to be solution of the PDE $$v_t + \theta v_x + \frac{1}{2}v_{xx} - rv = 0$$ This is a different function because expectation is taken under a different measure and it satisfies a different PDE than your original function $$u^{\mathbb{P}}(t,x) = \mathbb{E}^{\mathbb{P}}_t[e^{-rT}f(X_T)|X_t = x]$$

2) Derivative pricing and change of numeraire

Now if you are considering $$u(t,x)=\mathbb{E}^N_t[N(t)/N(T)f(X_T)]$$ This function does not depend on the numeraire $N$ you are using. In financial terms, the price does not depend on the currency or asset you are doing your accounting in.

In the case where $N_t = e^{\int_0^t \beta(X_u)\,du}$ for a deterministic function $\beta$, you end up with the usual function $$u(t,x)=\mathbb{E}^N[N(t)/N(T)f(X_T)|X_t = x]$$ being solution of $$u_t + \frac{1}{2}u_xx - \beta(x)u = 0$$ But in general, $N_t$ is not entirely determined by $X_t$ and you cannot apply FK directly. Remember that FK assumes you have a Markovian process driving everything. So you would still need some assumption like $(X,N)$ being Markovian for example and the conditional expectation should be taken with respect to the value of both $X$ and $N$ : $$u(t,x,n)=\mathbb{E}^N[N(t)/N(T)f(X_T)|N_t=n,X_t = x]$$ would then be solution of a PDE given by FK.

Hope that clarifies things a bit.