 Solution

For a homogeneous isotropic metric, radial light propagation covers comoving distance $\int cdt/a(t)$. The physical particle horizon is $d_p(t)=a(t){\int }_{t_i}^{t}cd{t}^{\prime }/a(t^{\prime })$, when this is finite. A future event horizon exists when $d_e(t)=a(t){\int }_t^{t_f}cd{t}^{\prime }/a(t^{\prime })$ is finite. The Hubble radius is the instantaneous scale $c/H$, with $H=\dot{a}/a$; it is not generally the same as either causal horizon.

For an Einstein-de Sitter universe, $H=2/(3t)$, so $c/H=3ct/2$, while $d_p=3ct$ from a beginning at zero. The future integral diverges, so there is no event horizon. For an eternal-past de Sitter spacetime, $H$ is constant and the particle-horizon integral from $-\infty$ diverges. Its event horizon and Hubble radius both equal $c/H$. Thus a Hubble radius is not automatically a particle horizon, even in these simple isotropic models.