= One Brownian factor does not imply a Markov short rate
{c}
For the <Heath-Jarrow-Morton model> with $\sigma(t,T)=T-t$, the risk-neutral drift is $\alpha(t,T)=(T-t)^3/2$. Its <short rate> is $r_t=f(0,t)+t^4/8+\int_0^tW_sds$. The centered integrated Brownian component is not <Markov>: its past reveals its derivative $W_t$, which affects the conditional mean of future increments, whereas its current integral does not determine $W_t$. Indeed the conditional variance of $W_t$ given the current integral is $t/4$. A one-dimensional driving noise therefore does not force a one-dimensional Markov state.
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