When reading a paper, i find it's difficult to derive the following equation.
In the paper, authors define:
$f_A=(1-\omega)+\omega[(k-1)q_{A|A}*a+[(k-1)q_{B|A}+1]*b]$
$f_B=(1-\omega)+\omega[(k-1)q_{A|B}*c+[(k-1)q_{B|B}+1]*d]$
$g_A=(1-\omega)+\omega[[(k-1)q_{A|A}+1]*a+(k-1)q_{B|A}*b]$
$g_B=(1-\omega)+\omega[[(k-1)q_{A|B}+1]*c+(k-1)q_{B|B}*d]$
where $\omega \rightarrow 0$
Then, they say:
$$ \frac{1} {N} p_B\sum_{k_A=0}^k\left(\begin{array}{l} k \\ k_A \end{array}\right) q_{A|B}^{k_A}q_{B|B}^{k_B} \frac{k_A f_A}{k_A f_A+k_Bf_B} -\frac{1} {N} p_A\sum_{k_A=0}^k\left(\begin{array}{l} k \\ k_A \end{array}\right) q_{A|A}^{k_A}q_{B|A}^{k_B} \frac{k_B g_B}{k_A g_A+k_B g_B} $$
where $k_A+k_B=k, p_A+p_B=1, q_{A|X}+q_{B|X}=1, p_{XY}=q_{X|Y}*p_Y, p_{AB}=p_{BA}$
equals to
$\omega \frac{k-1} {N} p_{AB}(I_aa+I_bb-I_cc-I_dd)+O(w^2)$
where
$I_a = \frac{k-1}{k}q_{A|A}(q_{A|A}+q_{B|B})+\frac{1}{k}q_{A|A}$
$I_b = \frac{k-1}{k}q_{B|A}(q_{A|A}+q_{B|B})+\frac{1}{k}q_{B|B}$
$I_c = \frac{k-1}{k}q_{A|B}(q_{A|A}+q_{B|B})+\frac{1}{k}q_{A|A}$
$I_d = \frac{k-1}{k}q_{B|B}(q_{A|A}+q_{B|B})+\frac{1}{k}q_{B|B}$
The detailed steps are omitted in the paper. So i wonder how to derive this equation. I have tried so many times, here is some of my thoughts: The equation includes a Binomial distribution of r.v. $k_A$, therefore, does the sum symbol in the equation can be seen as an expection? But the denominator item also contains $k_A$, how can i deal with it?
I donot major in math, so i wonder is there any tricks or theorems that can be used to derive or simplify this equation?
The detailed information can be found in the supplementary in the paper link: https://www.nature.com/articles/nature04605