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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?

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  • $\begingroup$ @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. $\endgroup$ May 14 at 21:58
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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: $$ N(a,t) = \eta\frac{e^{-\eta a}}{1 - e^{-\eta T}} e^{\eta t} $$

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