Master equation for the Zipf’s law in the Urn Model with Triggering

Let |$n_k$| be the number of colours that appear at least once in the sequence |$\mathcal{S}$| and with |$k$| balls in the urn. It thus holds |$\sum_k n_k \,{=}\, D$|⁠, and we can write the following master equation (for |$k\,{\gt}\,1$|⁠, which implies |$k \,{\geq}\, \rho\,{+}\,1$|⁠): where |$N(t)$| is the total number of balls in the urn at time |$t$| and in the last equality we used the relations |$N(t) \,{=}\, N_0\,{+}\,\rho t \,{+}\, (\nu\,{+}\,1) D$| and |$ \frac{dD(t)}{dt} \,{=}\, \frac{N_0 \,{+}\, \nu D}{N_0\,{+}\,\rho t \,{+}\, (\nu\,{+}\,1) D}$|⁠. We now introduce the probability |$p_k$| that a given colour appears |$k$|-times in the urn, i.e., the corresponding normalised version of the number of occurrences |$n_k$|⁠. In order to have |$\sum p_k \,{=}\, 1$|⁠, we must choose |$n_k \,{=}\, Dp_k$|⁠. The idea is that, as the time runs, the probabilities |$p_k$| will tend to a stationary distribution, i.e., a distribution independent of |$t$|⁠. By substituting |$n_k \,{=}\, Dp_k$| in Eq. (H.1), and neglecting |$N_0$| in the large time limit, After simple algebra this can be written as: and solve by separation of variables. We finally obtain: which gives a frequency-rank distribution |$f(R)\,{\sim}\, R^{-\alpha}$|⁠, with |$\alpha\,{=}\,\rho/\nu$|⁠. Note that this result is valid in all the cases |$\rho \,{\gt}\, \nu $|⁠, |$\rho\lt \nu $|⁠, and |$\rho \,{=}\, \nu $|⁠. We finally note that the frequency rank for the balls in the urn that appear at least once in |$\mathcal{S}$| equals the frequency-rank of the balls in |$\mathcal{S}$|⁠.