[ \hatR = R_n-2 + \epsilon,\quad \epsilon \sim \mathcalN(0, \sigma^2),\ \sigma=0.03 ]
They compute expected marginal utility of an additional child: the hardest interview 2
If (\lambda = 0.1), threshold (p=0.2). If estimated (p < 0.2), they stop early. Families observe historical stops and national ratio changes. Using Bayesian learning, after several days they form a posterior on (\lambda). This influences future stopping. [ \hatR = R_n-2 + \epsilon,\quad \epsilon \sim
The fixed point (R^ ) satisfies (p(R^ ) = 0.5) → (R^* = 1). So long-term ratio tends to 1 even with feedback. Families compute (\Delta U) using their noisy (\hatR). For a family with ((b,g)): [ \hatR = R_n-2 + \epsilon
[ p_n = \frac11 + e^-k \cdot (R_n-1 - 1) ]
[ U = \frac\text# boys\text# girls - \lambda \cdot \text(total births) ]
[ \hatR = R_n-2 + \epsilon,\quad \epsilon \sim \mathcalN(0, \sigma^2),\ \sigma=0.03 ]
They compute expected marginal utility of an additional child:
If (\lambda = 0.1), threshold (p=0.2). If estimated (p < 0.2), they stop early. Families observe historical stops and national ratio changes. Using Bayesian learning, after several days they form a posterior on (\lambda). This influences future stopping.
The fixed point (R^ ) satisfies (p(R^ ) = 0.5) → (R^* = 1). So long-term ratio tends to 1 even with feedback. Families compute (\Delta U) using their noisy (\hatR). For a family with ((b,g)):
[ p_n = \frac11 + e^-k \cdot (R_n-1 - 1) ]
[ U = \frac\text# boys\text# girls - \lambda \cdot \text(total births) ]