 One Brownian factor does not imply a Markov short rate

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 $r_t=f(0,t)+t^4/8+{\int }_0^t{W}_{s}ds$. 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.