# Heath–Jarrow–Morton under real-world measure

In HJM model (framework), the drift of the forward is determined by its diffusion coefficient:

$$\mu(t,s) = \sigma(t,s)\int_t^s \sigma(t,v)^Tdv$$

My understanding, is that the change of measure under Grisanov theorem for continuous-time semi-martingales only affect the finite variation part (i.e. drift for HJM). Thus, if we start with an SDE under a risk-neutral-measure $$Q$$

$$df(t,s) = \mu^Q(t,s)dt + \sigma(t,s)dW_t^Q$$

and the change to the real-world measure $$P$$ changes this to

$$df(t,s) = \mu^P(t,s)dt + \sigma(t,s)dW_t^P$$

does this then mean that $$\mu^Q(t,s) = \mu^P(t,s)$$ since they are both functions of $$\sigma(t,s)$$?

Your statement at the beginning of the question is not correct. That's why you have the "contradiction" later. It should say: In HJM model (framework), the drift of the forward under the risk-neutral measure Q is determined by its diffusion coefficient: $$\mu^Q(t,s) = \sigma(t,s)\int_t^s \sigma(t,v)^Tdv.$$ That formula is not a general formula to obtain the drift under any probability measure, only applies to $$Q$$. Note that later under the forward measure $$Q^T$$ the drift is zero and the volatility is terms are not zero.
As you said, the change of measure only affects the finite variation part, which is the drift. It's not clear why this implies $$\mu^Q(t,s)=\mu^P(t,s)$$. These are the drifts under the two measures, so I don't think they have to be the same.
• It's not clear how we can start with an SDE under the risk neutral measure and apply Girsanov's theorem to change to the actual measure. How we get to the drift under the risk-neutral measure is that we assume the evolution of the forward curve has some drift $\alpha(t,T)$ and volatility $\sigma(t,T)$, then we derive that under the risk neutral measure, the drift is what you had above. Jun 5 '20 at 1:09
• In $P$ the expected value of the rate at $T$ is not necessarily equal to the forward rate, so I don’t think you can derive the same drift formula.