Why is the Stochastic Process in the HJM model non-Markovian?

Question

I want to understand exactly what my title asks "Why is the Stochastic Process for the short rate in the HJM model of interest rates non-Markovian?" That process is the following: $r(t)=F(0,t)+\int^{t}_0 \left[ \sigma(s,t,\Omega_s) \left(\int^{t}_s \sigma(s,\tau,\Omega_s)\ d\tau \right) \right] ds + \int^{t}_0 \sigma(s,t,\Omega_s)\ dB(s)$ where F(0,t) is deterministic and $\Omega_s$ is a vector of relevant information to determine the volatility $\sigma(s,t,\Omega_s)$. I believe the reason is because of this $\Omega_s$. Basically, it seems like since this can be say bond prices from the past, then the non-Markovian follows. However, I read that even if you essentially ignore this $\Omega_s$ the process will still be non-Markovian. So I need to know if my first assumption is correct and why then if we ignore that assumption, we still have non-Markov. Thanks for the help!

Answer 1

Here is a simple situation showing that the process $(\Omega_s)$ is irrelevant but that the process $(r(t))$ is usually not Markov.

Assume that $\sigma(s,t,\nu)=t-s$ for every $s\leqslant t$ and every $\nu$, then elementary computations yield $$r(t)=F(0,t)+\tfrac18t^4+X_t$$ with $$X_t=\int_0^tB_s\mathrm ds$$ hence the process $(r(t))$ is Markov if and only if the process $(X_t)$ is. But the process $(X_t)$ is not Markov (only the bivariate process $(B_t,X_t)$ is).