何因扬清芬3 发表于 2016-8-20 16:40 
是呀 折腾了一天我已经在考虑直接生成平方项跑了回归再算边际效应了 不过你说的那样手动计算 具体应 ...
假设有一回归为:\[ n_{it}=\alpha_i+\beta_1 w_{it}+\beta_2 w_{it}^2+\beta_3 w_{it}\cdot k_{it}+\beta_4 ys_{it}+\epsilon_{it}\]根据 key paper,以 $w_{it}$ 为例,其对 $n_{it}$ 之边际效果 (marginal effects) 为(我省略下标):
\[ \frac{\partial n}{\partial w}=\underbrace{\beta_1}_{\mbox{main effect}}+\underbrace{2\beta_2 w}_{\mbox{squared term}}+\underbrace{\beta_3 k}_{\mbox{interaction}} \]
我必须重申,我并非完成赞同文之说/作法!而根据文献,我们以 $w,k$ 之平均值代入!对应之程序如下(根据你的 key paper,w 对 n 之不同效果):
- webuse abdata, clear
- gen w2 = w^2
- gen wk = w*k
- xtivreg n w w2 wk (ys = rec), fe
- // mean of w
- sum w if e(sample)
- scalar wm = r(mean)
- scalar list wm
- // mean of k
- sum k if e(sample)
- scalar km = r(mean)
- scalar list km
- // main effect
- scalar main = _b[w]
- scalar list main
- // squared term
- scalar square = 2*_b[w2]*wm
- scalar list square
- // interaction term
- scalar inter = _b[wk]*km
- scalar list inter
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