Why HJM Short Rates May Not Be Markovian

# Markovianity of the short rate process in the HJM framework


# Markovianity of the short rate process in the HJM framework












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):

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

where:

- $r(t)$ 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?