 Solution

An empty universe with $\Lambda =0$ has ${\Omega }_{r,0}={\Omega }_{m,0}={\Omega }_{\Lambda ,0}=0$. The present Friedmann equation then requires ${\Omega }_{k,0}=1$, soThe Friedmann acceleration equation gives $\ddot{a}=0$, hence $a\propto t$ after choosing its zero at $t=0$, andThis is the Milne universe. A larger past Hubble parameter does not require negative $\ddot{a}$: even constant $\dot{a}$ produces $H=\dot{a}/a=1/t$. Equivalently, deceleration parameter from the Hubble derivative gives $q=-1-\dot{H}/H^2=-1+1=0$. The age is $t_0=H_0^{-1}$.