# Is it acceptable in mathematical saying that the E-step is equal to posterior probability

I am studying EM-algorithm for mixture data. I read that some authors said that, the E-step is equivalent to the calculation of the posterior probability (I think this come from Bayesian rule). So, my question is, is it acceptable to says that the E-step is the same as or can be interpreted as a posterior probability? I meant will I be wrong if I said that the E-step can be viewed as a posterior probability that the point $$s$$ belongs to the $$k^{th}$$ component.

It is indeed incorrect in that the EM algorithm usually operates outside the Bayesian paradigm and hence posterior probabilities are not defined there. What is correct is that the E step for mixtures is equivalent to computing the conditional probability of an allocation of observation $$i$$ to component $$j$$: $$\mathbb{E}[\mathbb{I}_j(Z_i)|X_i=x_i,\theta]=\mathbb{P}(Z_i=j|X_i=x_i,\theta)=\dfrac{p_j f(x_i|\theta_j)}{\sum_\nu p_\nu f(x_i|\theta_\nu)}$$ This probability is both conditional on $$X_i=x_i$$ and on the value of the parameter $$\theta$$. The posterior probability that $$Z_i=j$$ would be conditional on the sample $$(X_1,\ldots,X_n)$$ only, integrating $$\theta$$ out.