I have a question regarding the proof of the Musiela parametrization for the dynamics of the forward rate curve. If $T$ is the maturity, $\tau=T-t$ is the time to maturity, and $dF(t,T)$ defines the dynamics of the forward rate curve, then the Musiela parametrization defines the forward rate dynamics $$d\bar{F}(t,\tau)=dF(t,t+\tau)$$.

My question is regarding the next step in the working of the Musiela parametrization. All of the literature I've looked at explains the next line by simply stating that a "slight variation" of Ito is applied. The line reads:

$$d\bar{F}(t,\tau)=dF(t,T)+\frac{\partial F}{\partial T}dt$$

Can someone please clarify what variation of Ito is being used here? I'm not following. The parameters to $d\bar{F}$ do not include an Ito drift/diffusion process, so why is Ito being used?


1 Answer 1


$dF(t,T)$ describes the dynamics of the rate of a particular forward contract as time $t$ moves forward to a fixed expiration $T$.

$d\bar F(t,\tau)$ describes the dynamics of the rate at a particular point on the yield curve as time moves forward.

The differential $\frac{\partial F}{\partial T}dt$ is simply the difference between holding the expiration time $T$ constant in the case of $F$ and moving it ahead with time $t$ to stay at the same point $t+\tau$ on the yield curve in the case of $\bar F$.

Somewhere underlying all this is a drift-diffusion process, but it isn't stated explicitly in your equations.

$dF(t,t+\tau)$ is a "total" differential of $F$ with respect to a simultaneous change in both its arguments. This becomes the sum of a partial differential w.r.t. change in the first argument only, $dF(t,T)$, and a partial differential w.r.t. change in the second argument only, $\frac{\partial F}{\partial T}dt$, as time moves forward.

  • $\begingroup$ Thanks for the response. I think I follow your explanation, but I'm looking for more of a mathematical derivation of the first and second terms for $d\bar{F}(t;\tau)$. Mathematically, how would you explain going from the first equation I've listed in the original post to the second? Thanks for your time. $\endgroup$ Jul 1, 2012 at 16:24
  • $\begingroup$ I edited my answer to explain a little more. $\endgroup$
    – JL344
    Jul 1, 2012 at 17:16
  • $\begingroup$ Thanks for the revision. That's along the lines of what I was thinking, but I think I'm mis-interpreting the notation somehow. Here's how I understand breaking up the total differential into the sum of partials: $$dF(t,T)=\frac{\partial F}{\partial t} dt + \frac{\partial F}{\partial T} dT$$ $\endgroup$ Jul 1, 2012 at 18:38
  • $\begingroup$ T is a function of t, $T=t+\tau$, so applying the chain rule on the second term yields: $$dF(t,T)=\frac{\partial F}{\partial t} dt + \frac{\partial F}{\partial T} \frac{\partial T}{\partial t} dT$$ with $\frac{\partial T}{\partial t}=1$ $\endgroup$ Jul 1, 2012 at 18:49
  • $\begingroup$ As I wrote that last comment, I realized that I don't think I can write $dT$ as a differential since T is a function. So that explains the second term. For the first term though, your answer is implying that $\frac{\partial F}{\partial t}dt = dF(t,T)$. If that's correct, that's where I'm getting lost. Thanks again for your help with this. $\endgroup$ Jul 1, 2012 at 18:53

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