In Andersen and Piterbarg (2010), the authors study the short rate process under a HJM framework and derive the following expression (Section 4.4.3):

$$ r(t)=f(t,t)=f(0,t)+\int_0^t\sigma_f(u,t)^\intercal\int_u^t\sigma_f(u,s)\text{d}s\text{d}u+\int_0^t\sigma_f(u,t)^\intercal \text{d}W(u)$$


  • $r(t)$ is the short rate process;
  • $f(t,t')$, $t'\geq t$, is the instantaneous forward rate;
  • $\sigma_f(t,t')$ is the instantaneous volatility of $f(t,t')$;
  • $W(t)$ is a standard Brownian motion.

Note that $\sigma_f(t,t')$ and $W(t)$ are multi-dimensional. Still in Section 4.4.3, the authors then define:

$$ D(t) = \int_0^t\sigma_f(u,t)^\intercal \text{d}W(u)$$

They write $D(T)$, $T>t$, as follows:

$$ D(T) = D(t) + \underbrace{\int_t^T\sigma_f(u,T)^\intercal\text{d}W(u)}_{I_1}+\underbrace{\int_0^t\sigma_f(u,T)^\intercal\text{d}W(u)}_{I_2}-\underbrace{\int_0^t\sigma_f(u,t)^\intercal\text{d}W(u)}_{I_3}$$

which, according to them, proves that $D(t)$ is not Markovian (unless $I_2-I_3$ is either deterministic or a function of $D(t)$), namely:

$$ E^Q\left(D(T)|\sigma(D(t))\right) \not= E^Q\left(D(T)|\mathscr{F}_t\right)$$

where $\mathscr{F}_t$ is the model's filtration.

I am not sure I catch their reasoning. Assuming the volatility process is deterministic, in my view:

  • By independence of increments: $$ E^Q\left(I_1|\sigma(D(t))\right)=E^Q\left(I_1|\mathscr{F}_t\right)=0$$
  • Given $I_3=D(t)$: $$ E^Q\left(I_3|\sigma(D(t))\right)=E^Q\left(I_3|\mathscr{F}_t\right)=D(t)$$

Thus it is in $I_2$ where the difference must lie. However, I do not see how they conclude so quickly. I understand that by conditioning on the $\sigma$-algebra we know the value of this stochastic integral, but how do we know outright the 2 conditional expectations of $I_2$ are different?

  • $\begingroup$ Hi: This isn't my field at all but markovian just means that the process from $t $ to $T$, only depends on where one is at time $t$. Based on the 3 integral equation, the process from $t$ to $T$ depends on things that happen between $0$ and $t$ ( because the last two integrals are from $0$ to $t$ ) which means that the process is not markovian. I could be not understanding something crucial so hopefully someone else can confirm. $\endgroup$ – mark leeds Sep 17 '18 at 17:37
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    $\begingroup$ You may check Chapter 11 of the book Martingale Methods in Financial Modeling, where it is shown that the short rate is Markovian, then $\sigma_f(t, T) = h(t)g(T)$ for some functions $g$ and $h$. $\endgroup$ – Gordon Sep 17 '18 at 18:01

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