Need intuition - how do they simplify the Q function for gaussian mixture EM? Background - I'm trying to follow section 7.2.4 in this EM tutorial. Basically the setup is I have a vector with 10 points $x$, and each of them can be assigned ($y$) to either Gaussian 1 or to Gaussian 2, and I want to estimate the parameters $\theta$ = (mean, std) of both Gaussians and assign which point belongs to which Gaussian.

I understand how do they arrive to 
$Q(\theta|\theta^t) = \sum_y{p(y|x,\theta^t)logP(x,y|\theta)}$
Should it be read as: an expectation over $y$ = all possible configurations of the entire sample of all 10 points ($2^{10}$ configurations), each with its own probability (under the old parameters $\theta^t$), of the complete log likelihood of the sample and the assignments, under the new parameter $\theta$.
If this is the correct way to read this, I fail to understand the trick that helps us simplify the Q function to regarding one point at a time, and one Gaussian assignment at a time! What's the core idea that leads to the simplification:
$Q(\theta|\theta^t) = \sum_i\sum_j{p(y_{ij} = 1|x_i,\theta^t)logP(x_i,y_i = j|\theta)}$
Sorry if this is trivial, but the derivation just doesn't sit right for me and if someone could give me an intuition for this it would be a great help.
 A: I'll use the example @Pegah quoted from nature. 
We are presented with 5 sets of 10 coin tosses. Each set was done with one of two imbalanced coins $A$ or $B$. We want to use EM in order to estimate the parameter of each coin $\theta_A$ and $\theta_B$ (the probability of heads in each coin), and to assign each set to the likely coin.
Now the general formulation of $Q$ is referring to vectors not specific instances, as it does not assume our instances are IID. So in our example $\vec{y}$ is the assignment of all 5 sets to either $A$ or $B$ and when we say:
$p(\vec{y}|\vec{x},\theta^t)$
we mean: the probability of a specific assignment $\vec{y}$ (e.g. $y_1 = A,y_2=A,y_3=B,y_4=A,y_5=B$) being the true assignment, given that we saw $\vec{x}$ which is the number of heads in each 10-tosses set (e.g. $x_1=4, x_2=3, x_3 = 7, x_4=2, x_5=1$) and under the current parameters $\theta^t_A$ and $\theta^t_B$.
So $Q$ is the expectation over all $2^5$ such $\vec{y}$ of the complete log-likelihood:
$Q(\theta|\theta^t) = \sum_{\vec{y}}{p(\vec{y}|\vec{x},\theta^t)logP(\vec{x},\vec{y}|\theta)}$
Not easy to compute with all those annoying vectors, huh?

Luckily in our case every instance (set of tosses) is IID from the other sets. So we simplify the complete log-likelihood:
$p(\vec{x}, \vec{y}|\theta_t) = \prod_i \prod_{j=\{A,B\}} p(x_i, y_i=j|\theta_t)^{y_{ij}}$
where ${y_{ij}}$ is an indicator variable which is 1 if $y_i = j$ and 0 otherwise.
so we can rewrite $Q$ as:
$Q(\theta|\theta^t) = \sum_{\vec{y}}{p(\vec{y}|\vec{x},\theta^t)} \sum_i\sum_j{y_{ij}logP(x_i,y_i = j|\theta)}$
we're still stuck with the vector, but now we can rearrange:
$Q(\theta|\theta^t) = \sum_i\sum_j logP(x_i,y_i = j|\theta) \sum_{\vec{y}} y_{ij} p(\vec{y}|\vec{x},\theta^t)$
and for a given $i, j$ the sum over all $\vec{y}$ is a sum over all $\vec{y}$ in which $y_i = j$ as in the others $y_{ij} = 0$, so according to the law of total probability it's just 
$\sum_{\vec{y}} y_{ij} p(\vec{y}|\vec{x},\theta^t) = p(y_{ij} = 1|\vec{x},\theta^t)$ 
and because they are IID, that simplifies again to 
$p(y_{ij} = 1|x_i,\theta^t)$
because the coins assigned to the other sets don't matter to $y_i$.
so we arrived to this nice form:
$Q(\theta|\theta^t) = \sum_i\sum_j logP(x_i,y_i = j|\theta) p(y_{ij} = 1|x_i,\theta^t)$
Thanks to EG for explaining this to me.
