everyone. I meet some problems on the simplification of the formular of Transfer Entropy and I'm not sure whether I am right.
Firstly we have 3 variables : $X=\{x_{i}\}_{i=1}^{n}, Y=\{y_{i}\}_{i=1}^{n}, X'=\{x'_{i}\}_{i=1}^{m}$. And $X$ and $Y$ are pair-wised, while $X'$ is related to $X$ (because $X$ is related to $Y$ so $X'$ is also related to $Y$).
Now assume that we have the original formulation of Transfer Entropy : $TE = H(Y|X') - H(Y|X',X)$
Then I tried to simplified them individually:
\begin{align} H(Y|X') &= -\sum_{x'\in X'} \sum_{y \in Y} p(x', y)\log p(y|x') \\ &=-\sum_{y \in Y} \sum_{x'\in X'} p(x'|y)p(y)\log p(y|x') \\ &=-\sum_{y \in Y} p(y) [\sum_{x'\in X'} p(x'|y)\log p(y|x')] \\ &=-\mathbb{E}_{y\in Y} [\sum_{x'\in X'} p(x'|y)\log p(y|x')] \\ &when ~y~is~sampled,~then ~x'\in~p(x'|y),~since ~x'~is~related~to~X~and~Y\\ &=-\mathbb{E}_{y\in Y}[\mathbb{E}_{x' \sim p(x'|y)} (\log p(y|x'))] \end{align}
\begin{align} H(Y|X',X) &= -\sum_{x\in X} \sum_{x'\in X'} \sum_{y \in Y} p(x, x', y)\log p(y|x,x') \\ &=-\mathbb{E}_{y\in Y} [\sum_{x\in X}\sum_{x'\in X'} p(x,x'|y)\log p(y|x,x')] \\ &=-\mathbb{E}_{y\in Y} [\sum_{x\in X} p(x|y) \sum_{x'\in X'} p(x'|x,y) \log p(y|x,x')] \\ &when ~y~is~sampled,~then ~x~is~ensured,~since ~x~is~pairwith~Y\\ &=-\mathbb{E}_{y\in Y} [\mathbb{E}_{x \sim p(x|y)}[\sum_{x'\in X'} p(x'|x,y) \log p(y|x,x')]]\\ &when ~y~and~x~are~sampled,~then ~x'\in~p(x'|x,y)\\ &=-\mathbb{E}_{y\in Y} [\mathbb{E}_{x \sim p(x|y)}[\mathbb{E}_{x' \sim p(x'|x,y)} \log p(y|x,x')]]\\ \end{align}
My question is : is my simplification correct? Thanks~