I'm a student studying bayesian inference.
When I read this paper(https://arxiv.org/pdf/1312.6114.pdf), I had one question in eq2.
$$log_{p_\theta}(x^{(i)}) = D_{KL}(q_\phi(z|x^{(i)})||p_{\theta}(z|x^{(i)}))\ + \mathcal{L}(\theta, \phi; x^{(i)})$$
$$log_{p_{\theta}}(x^{(i)}) \ge \mathcal{L}(\theta, \phi; x^{(i)}) = \mathop{\mathbb{E}}_{q_{\phi}(z|x)}[-log_{q_{\phi}}(z|x)\ + log_{p_{\theta}}(x, z)]$$
$$\mathcal{L}(\theta, \phi;x^{(i)}) = -D_{KL}(q_{\phi}(z|x^{(i)})||p_{\theta}(z)) + \mathop{\mathbb{E}}_{q_{\phi}(z|x^{(i)}\ )} [log_{p_{\theta}}(x^{(i)}|z)]$$
Above 3 equations are eq1, eq2, and eq3 each.
I could understand and accept eq1 and eq3, for each data point i.
But in eq2, why are superscripts i missed all in the RHS formula suddenly?
In the first time, I thought the RHS formula meant the expectation value of all data sets in ideal case.
If there is the ideal case, it means that $q_{\phi}(z|x)$ equals to $p_{\theta}(z|x)$.
So I thought it is possible to write the formula below.
$\sum^{N}_{i=1} log_{p_{\theta}}(x^{(i)}) = \sum^{N}_{i=1}\mathcal{L}(\theta, \phi; x^{(i)})$ (because $D_{KL}$ part vanishes.)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sum^{N}_{i=1}\mathop{\mathbb{E}}_{q_{\phi}(z|x)}[-log_{q_{\phi}}(z|x)\ +\ log_{p_{\theta}}(x, z)]$ (in eq2)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sum^{N}_{i=1}\mathop{\mathbb{E}}_{q_{\phi}(z|x)}[log_{p_{\theta}}(x)]$ (because $q_{\phi}$ and $p_{\theta}$ are equal.)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sum^{N}_{i=1}\mathop{\mathbb{E}}[log_{p_{\theta}}(x)]$ (because $z$ in $q_{\phi}(z|x)$ and $x$ in $log_{p_{\theta}}(x)$ are independent.)
The flow is reasonable for me, like below formula.
$$ n_1\ +\ n_2\ +\ n_3\ +\ ...\ + n_k = \sum^{k}_{i=1}\mathop{\mathbb{E}}[n] $$
But in this case, I could not accept eq3, because of the difference between $\mathop{\mathbb{E}}_{q_{\phi}(z|x)}$ in eq2 and $\mathop{\mathbb{E}}_{q_{\phi}(z|x^{(i)}\ )}$ in eq3.
So I'm very confuse. why can we write the RHS formula missing all superscript i in eq2?