# Derive the Demographic Structure in the Steady State

I am reading a paper with following description on the demographics in their model: "... each (representative) agent lives for $$T$$ periods ... We assume that each individual has $$e^{f}$$ children at age $$B$$. Since we consider only steady states, we need to derive the stationary age distribution of this economy associated with this fertility rate. Our assumptions imply $$N(a, t)=e^{f} N(B, t-a)$$ and $$N\left(t^{\prime}, t\right)=0, t^{\prime}>T$$. It is easy to check that in the steady state $$N(a, t)=\phi(a) e^{\eta t}$$, where $$\phi(a)=\eta \frac{e^{-\eta a}}{1-e^{-\eta T}}$$ and $$\eta=f / B$$ is the growth rate of population."

I have no idea how is this steady state calculated?

• @Giskard Yes, I update the link. Please note that this demographic setting is rather a separate part and has nothing to do with the main part of the model. – Alalalalaki May 14 at 21:58

I don't know the paper nor the notation, so I am just guessing here. I gues $$N(a,t)$$ is the number of agents of age $$a$$ at time period $$t$$.
Let's follow the number of age $$B$$ accross generations: \begin{align*} N(B,t) &= e^f N(B, t- B),\\ &= e^{2f} N(B, t - 2 B),\\ &= \ldots,\\ &= e^{f t/B} N(B,0),\\ &= e^{\eta t} N(B, 0). \end{align*} Then using this in the definition of $$N(a,t)$$, we have: $$N(a,t) = e^f N(B, t- a) = e^f e^{\eta(t - a)}N(B,0).$$ Next, I assume that at period $$0$$ there is a mass of size 1 but nobody lives longer than $$T$$ periods, so integrating across all ages: \begin{align*} &\int_0^T N(a,0) da = 1,\\ \to &\int_0^T e^f e^{-\eta\, a}N(B,0) da = 1,\\ \to &e^f N(B,0) \left[-\frac{e^{-\eta a}}{\eta}\right]^T_0 = 1,\\ \to &e^f N(B, 0) \left[1 -e^{-\eta T}\right] = \eta,\\ \to &e^f N(B, 0) = \frac{\eta }{1 - e^{-\eta T}} \end{align*} So subsituting into the expression for $$N(a,t)$$ gives: $$N(a,t) = \eta\frac{e^{-\eta a}}{1 - e^{-\eta T}} e^{\eta t}$$