As well known to us, K.Y. Liang and S. Zeger proposed GEE for longitudinal data analysis in their famous paper[1]. At the appendix of the paper, authors show the proof of Theorem 2. I tried to reproduce that proof following their idea, but when I tried to derive $A\_{i}$, I failed. Here, I want to add two comments:


1. "... $\sum A\_{i}/K$ converges as K $\rightarrow \infty$ to $-\sum D\_{i}^{T}V\_{i}^{-1}D\_{i}/K$." 
I don't think it involves any asymptotics. That is, whatever $K$ is, it holds still. (Do you agree ?)
2. $A\_{i} = \partial U\_{i}\{\beta, \alpha^{\*}(\beta)\}/\partial \beta$, as stated above, $A\_{i}$ should be equal to $-D\_{i}^{T}V\_{i}^{-1}D\_{i}$, however, I failed to derive it. Note that, $U\_{i} = D\_{i}^{T}V\_{i}^{T}S\_{i}$, if we fix $D\_{i}$ and $V\_{i}$ as constant, that is, not related with $\beta$, then $\partial S\_{i}/\partial \beta = - A\_{i}\Delta\_{i}X\_{i} = - D\_{i}$. But, $D\_{i}$ and $V\_{i}$ are all related with $\beta$. So, I feel very confused about that!


GEE has been popularly used in longitudinal data analysis and has been further studied in more complicated circumstances by many statisticians. So, the results in Liang's paper should be correct, otherwise, there should be someone pointing it out. Do you have any idea about the proof ?


Thank you very much!


[1]: K.Y. Liang, S. Zeger, longitudinal data analysis using generalized linear models, <https://academic.oup.com/biomet/article/73/1/13/246001>.



Deriving asymptotic variance of generalized estimating equation estimator (GEE)
The key point is Law of Large Numbers.


Answer:


$\partial \frac{1}{K}\sum A\_{i}/\partial \beta = \frac{1}{K}\sum \partial A\_{i}/\partial \beta = \frac{1}{K}\sum \{[\partial(D\_{i}^{T}V\_{i}^{-1})/\partial \beta \times S\_{i}] + [D\_{i}^{T}V\_{i}^{-1} \times \partial S\_{i}/\partial \beta]\}$. 


Let $E\_{1} = \frac{1}{K}\sum [\partial(D\_{i}^{T}V\_{i}^{-1})/\partial \beta \times S\_{i}]$, apply LLN to $E\_{1}$, we have $E\_{1} \rightarrow \mathbb{E}[\partial(D\_{i}^{T}V\_{i}^{-1})/\partial \beta \times S\_{i}] = \partial(D\_{i}^{T}V\_{i}^{-1})/\partial \beta \times \mathbb{E}[S\_{i}] = 0$, as $K \rightarrow \infty$.


Let $E\_{2} = \frac{1}{K}\sum [D\_{i}^{T}V\_{i}^{-1} \times \partial S\_{i}/\partial \beta] = -\frac{1}{K}\sum D\_{i}^{T}V\_{i}^{-1}D\_{i}$.


Done.


