I understand that Hubble's law states that:$\\$
$v=H_{0}D$
However, given that the Hubble time (an estimate of the age of the universe) is given by $\frac{1}{H_{0}}$, doesn't that mean...
$v=H_{0}D\quad \to \quad \frac{D}{v}=\frac{1}{H_{0}}$
Now, if we are saying that $t_{H}=\frac{1}{H_{0}}$, it's equivalent to saying that $t_{H}=\frac{D}{v}$; so wouldn't this mean that we are implying that the recessional velocity of all objects in the universe has been constant from the beginning of time?
Since D is increasing at a rate $v$ at any time, then $H_{0}D$ must also be increasing; then this would mean that $v$ isn't constant.
Based on the above argument, what is a better way to interpret the Hubble time?
