I'm going through equations in this paper Structure of Growing Networks with Preferential Linking, I was not able to understand how they derived equation $[3]$ by summing up equation $[2]$.
eqn [2] $$p(q,s,t+1)=\sum_{l=0}^{m}\binom{m}{l}\left [\frac{ q-l+am}{(1+a)mt} \right ]^{l}\left [1-\frac{ q-l+am}{(1+a)mt} \right ]^{m-l}p(q-l,s,t)$$
substituting for sum over $p(q,u,t)$ using: $P(q,t)=\sum_{u=1}^{t}p(q,u,t)/t$
They obtain $eqn [3]$
$$(t+1)P(q,t+1)-p(q,t+1,t+1)=(1-\frac{q+am}{1+a})P(q,t)+(\frac{q-1+am}{1+a})P(q-1,t)+\mathcal{O}(\frac{P}{t})$$
My understanding is that $P(q,t+1)$ is the probability of having sites with $q$ in_links, this consists of two parts, one being the set of sites with in_degree $q-1$ at time $t$ that gain one additional link in time step $t+1$. Another set is due to loosing sites that already have $q$ in_links at time $t$ and gain a new links, thus no longer contribute to $P(q,t+1)$, therefore have a negative sign. These should correspond the second and first term on the right hand side of the equation $[3]$. What I don't understand is how to obtain the multiplying factor for each term.
I would be very grateful for any insight.
For the LHS of eqn. 2, observing that $$ \sum_{s=1}^{t}p(q,s,t+1)=\sum_{s=1}^{t+1}p(q,s,t+1)-p(q,t+1,t+1) $$ so that $$ \sum_{s=1}^{t}p(q,s,t+1)=(t+1)\sum_{s=1}^{t+1}\frac{p(q,s,t+1)}{t+1}-p(q,t+1,t+1) $$ and using $P(q,t)=\sum_{s=1}^t \frac{p(q,s,t)}{t}$, we have $$ LHS=(t+1)P(q,t+1)-p(q,t+1,t+1) $$
For the RHS, observe that we neglect the terms of order higher than $1/t$ so that you have to sum the first two terms for $l=0, 1$. So, calling $b(l)=\frac{ q-l+am}{(1+a)mt}$, we have $$\sum_{s=1}^t \sum_{l=0}^{m}\binom{m}{l}b^{l}(1-b)^{m-l}p(q-l,s,t)= \sum_{l=0}^{m}\binom{m}{l}b^{l}(1-b)^{m-l}\underbrace{\sum_{s=1}^t t\frac{p(q-l,s,t)}{t}}_{tP(q-l,t)}$$ and considering the first two term only, we obtain: $$ RHS=(1-b(0))^m tP(q,t)+m b(1)(1-b(1))^{m-1}tP(q-1,t)+\mathcal{O}(P/t) $$ Using the binomial approximation $(1+x)^n\approx 1+nx$ we obtain: $$ (1-b(0))^m \approx 1-m\frac{q-l+am}{(1+a)mt}=1-\frac{q-l+am}{(1+a)t} $$ and $$ m b(1)(1-b(0))^m \approx m b(1)=m\frac{q-l+am}{(1+a)mt}=\frac{q-l+am}{(1+a)t} $$
Putting all togheter we have $$ RHS=\left(1-\frac{q-l+am}{(1+a)t}\right) t P(q,t)+\frac{q-l+am}{(1+a)t}tP(q-1,t)+\mathcal{O}(P/t) $$ and finally $$ RHS=\left(t-\frac{q-l+am}{(1+a)}\right) P(q,t)+\frac{q-l+am}{(1+a)}P(q-1,t)+\mathcal{O}(P/t). $$ The LHS and RHS give us the eqn. 3 $$ (t+1)P(q,t+1)-p(q,t+1,t+1)=\left(t-\tfrac{q-l+am}{(1+a)}\right) P(q,t)+\tfrac{q-l+am}{(1+a)}P(q-1,t)+\mathcal{O}(P/t). $$