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?

  • $\begingroup$ 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) $\endgroup$ Aug 25, 2016 at 7:59
  • $\begingroup$ 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. $\endgroup$
    – Quantuple
    Aug 25, 2016 at 9:07
  • $\begingroup$ 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.. $\endgroup$
    – Quantuple
    Aug 25, 2016 at 9:10
  • $\begingroup$ ok, let me be specific, I started with RN measure and I want to change to forward measure with $N_t=B(t,T)$. $\endgroup$
    – Medan
    Aug 25, 2016 at 14:18

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Medan
    Aug 25, 2016 at 1:30
  • $\begingroup$ So if I have two different pdes and get two different function values at time 0, which one is the price? $\endgroup$
    – Medan
    Aug 25, 2016 at 2:10
  • $\begingroup$ @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? $\endgroup$
    – Medan
    Aug 25, 2016 at 13:15
  • $\begingroup$ Check again. This is a conditional expectation. $\endgroup$
    – 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.

  • $\begingroup$ 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? $\endgroup$
    – Medan
    Aug 25, 2016 at 15:28
  • $\begingroup$ 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. $\endgroup$
    – Quantuple
    Aug 25, 2016 at 15:39
  • $\begingroup$ 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. $\endgroup$
    – Medan
    Aug 25, 2016 at 15:45
  • $\begingroup$ 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. $\endgroup$
    – Quantuple
    Aug 25, 2016 at 15:57
  • $\begingroup$ :ok, but which one is the price then? $\endgroup$
    – Medan
    Aug 25, 2016 at 16:00

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