# 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?

• risk neutral measure is a trick, forget about it, you do not need risk neutral measure to write the pricing pde (where price means the initial cost of the replication strategy) Aug 25, 2016 at 7:59
• The Feynman-Kac formula (or actually the Kolmogorov-Backward equation on which it relies) should be thought of as a one to one link between PDEs and SDEs. When you change of measure, you change the SDE describing the dynamics of your underlying asset, Feynman-Kac then tells you that the PDE will change since the SDE has changed. So yes, different measures = different SDEs => different PDEs. Aug 25, 2016 at 9:07
• Also notice that the dynamics you provided for $N_t$ is not necessarily a valid one since a numéraire should be a tradable asset with positive price. Also remember that the option price does not depend on your choice of numéraire.. Aug 25, 2016 at 9:10
• ok, let me be specific, I started with RN measure and I want to change to forward measure with $N_t=B(t,T)$. Aug 25, 2016 at 14:18

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.

• Thanks. I can follow what you wrote. However, correct t me if I am wrong but FK is not about particular measure. So the very first BM I have written could be equal to ANY BM, under any of my favorite measures. As long as this is the dynamics of the X_t, the E is written under my chosen measure. Even though it looks like a typical RN E, it can be under any other measure. Aug 25, 2016 at 1:30
• So if I have two different pdes and get two different function values at time 0, which one is the price? Aug 25, 2016 at 2:10
• @AFK: if $\mathbb{P}$ is a RN measure, $u^P$ is the derivative price when I look at how the expectation is defined. But with a new measure $\mathbb{Q}$, which corresponds to another numerraire(because it is another measure), different from money market. I understand that if you define $v=u^Q$ as above it satisfies the pde, but what is that new expectation(nothing changed inside of expectation)? Is that a price? Then why is it? Aug 25, 2016 at 13:15
• Check again. This is a conditional expectation.
– AFK
Nov 28, 2016 at 7:41

You must separate the mathematical theory from the financial theory. The notion of numéraire specifically pertains to the latter.

[A] Mathematical perspective

You reach the following PDE (regardless of how you did it) $$u_t + \mu u_x + \frac{1}{2}\sigma^2 u_{xx} - ru = 0$$ Feynman-Kac then tells you that the unique solution can be written as $$u(t,x) = \mathbb{E}^\mathbb{X} \left[e^{-r(T-t)} f(X_T) \vert X_t = x\right]$$ where $(X_s)_{s \geq t}$ solves $\forall s \geq t$ $$dX_s = \mu dt + \sigma dW_s^\mathbb{X} ,\ \ \ X(t) = x$$ There is no question of numéraire here. $\mathbb{X}$ is a probability measure, it can be anything you like as long as $W_s^\mathbb{X}$ is a $\mathbb{X}$-Brownian motion it does not matter.

[B] Financial perspective

Consider a positive traded asset $N_t$. Arbitrage opportunities are precluded if any self-financing strategy expressed in terms of $N$ (numéraire) emerges as a $\mathbb{N}$-martingale i.e. $$\frac{V_t}{N_t} = \mathbb{E}^{\mathbb{N}}\left[ \frac{V_T}{N_T} \vert \mathcal{F}_t \right]$$ Should we introduce yet another numéraire $M$, the following would hold: $$V_t = N_t \mathbb{E}^{\mathbb{N}}\left[ V_T N_T^{-1} \vert \mathcal{F}_t \right] = M_t \mathbb{E}^{\mathbb{M}}\left[ V_T M_T^{-1} \vert \mathcal{F}_t \right]$$

It entails that if investing in the asset $X_t$ is a self-financing strategy, then we should have that, for any choice of numéraire $N$: $$\frac{X_t}{N_t} \text{ is a } \Bbb{N}\text{-martingale}$$

[Example] A $\to$ B

Consider a risk-free investment vehicle, with return $r$. Let $S_t$ denote a risky asset paying no dividends with dynamics $$dS_t = \delta dt + \sigma S_t dW_t^\mathbb{P} \tag{1}$$ This will be our working modelling assumption. Now let $V_t = V(t,S_t,...)$ denote a contingent claim written on $S_t$ which pays no coupons. Consider the self-financing strategy $\Pi_t$ which consists in holding both the contingent claim $V_t$ and a fraction $\alpha_t$ of the risky asset $S_t$: $$\Pi_t = V_t + \alpha_t S_t$$ Picking $\alpha_t = -\partial V/\partial S$ allows us to "delta hedge" the portfolio $\Pi_t$ so that it's infinitesimal P&L reads: \begin{align} d\Pi_t &= dV_t - \alpha_t dS_t \\ &= \frac{\partial V}{\partial t} dt + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} \sigma^2 S_t^2 dt \end{align} since the latter evolution does not depend the latent source of randomness $dW_t^\mathbb{P}$, the delta hedged portfolio should evolve at the risk-free rate by absence of arbitrage opportunity: \begin{align} d\Pi_t &= \Pi_t r dt \\ &= (V_t - \alpha_t S_t) r dt \end{align} hence the PDE: $$\frac{\partial V}{\partial t}+ r S \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} - rV = 0$$ with $V(t=T,S_T=S) = f(S)$ by absence of arbitrage, $f(S)$ denoting the contingent claim's payout. Feynman-Kac then tells us that the solution writes: $$V_t = \mathbb{E}^\mathbb{Q} \left[ e^{-r(T-t)} f(S_T) \vert \mathcal{F}_t \right] \tag{3}$$ where under $\mathbb{Q}$ $$dS_t = rS_t dt + \sigma S_t dW_t^\mathbb{Q} \tag{2}$$ You can now:

• Compare the dynamics of the risky asset $(1)$ and $(2)$. Notice how we've started from some modelling assumptions under $\mathbb{P}$ but saw that we could actually express the price using a mathematical trick under yet another measure (similar to what happens in the CRR binomial framework: the historical probability disappear from thee pricing equation)
• From $(2)$ we see that $S_t/B_t$ is a $\mathbb{Q}$-martingale, while from $(3)$ we see that $V_t/B_t$ is also a $\mathbb{Q}$-martingale. Thus $B_t$ is indeed the numéraire associated to the measure $\mathbb{Q}$.

[Example] B $\to$ A

See application of Itô's lemma discussed here and relevant references inside.

• I am comfortable with math part. However, assume I happen to have everything under the forward measure to start with, i.e. some state process $dX_t=adt+bdW_t^T$. Then, I start by writing the proper expectation which is the price of some derivative, i.e $V_t=\mathbb{E}_{t,x}^T[B(t,T)f(X(T)/B(T,T)]$, so all I have is an expectation using non arbitrage arguments. Now this E is the solution of some PDE. What is the PDE in this case? Aug 25, 2016 at 15:28
• So first, $a$ cannot be anything: $X_t$ being a traded asset, $X_t/B(t,T)$ should to emerge as a $T$-martingale. Then your question is what @AFK explains in the second part of his post: it depends on how your numéraire $B(t,T)$ is defined. As mentioned in the link given in last part of my own answer, if you have an explicit form for the dynamics of $B(t,T)$ (or the underlying short rate), you could try to apply Itô's lemma and see what terms you need to get rid of for $V_t/B(t,T)$ to emerge as a martingale. In other words compute $d(V_t/B(t,T)) = ... dt + ... dW_t^T$ and set drift = 0. Aug 25, 2016 at 15:39
• ok, but you mentioned above in @AFK answer $u$ and $v$ should give me the same value, you mean $u(t,x_0)=v(t,x_0)$? But how could this be the case if they satisfy two different PDEs? unless the terminal payoff function is different but the answer doesn't specify it. Aug 25, 2016 at 15:45
• I'm sorry for being unclear. The terminal condition is the same. Thus $u^P$ and $u^Q$ un AFK's answer are not the same since the dynamics unddr the 2 measurzs differ. I've deleted m'y comment which was confusing. Aug 25, 2016 at 15:57
• :ok, but which one is the price then? Aug 25, 2016 at 16:00