Great question! Let's break this down clearly, tying back to the lower bounds you've outlined in your problem:

1. 理论收敛保证：不会被边缘化破坏 #

First, remember that EM's core convergence promise holds regardless of how many latent variables you use (as long as you implement the steps correctly): every E-M iteration will monotonically increase the log-likelihood $\log p(x|\theta)$, until the algorithm converges to a local (or sometimes global) maximum of the log-likelihood.

Marginalizing part of the latent variables (e.g., going from $z_1,z_2$ to just $z_1$ by integrating out $z_2$) doesn't break this guarantee. Both the single-latent-variable bound $\mathcal{L}_1$ and the two-latent-variable bound you started with will still act as valid lower bounds that drive the log-likelihood upward with each iteration.

2. 收敛速度：通常变慢，但有实用例外 #

The biggest impact is on convergence speed (how many iterations it takes to reach a target log-likelihood):

• Tighter bounds = faster iteration speed: As your equations show, the lower bound using both $z_1$ and $z_2$ is strictly tighter (or equal to) $\mathcal{L}_1$. A tighter bound means each M-step maximization brings you closer to the true log-likelihood, so you'll need fewer iterations to converge when keeping all relevant latent variables.
• Computation vs. iteration tradeoff: That said, marginalizing a latent variable can drastically simplify the M-step. If the computational cost of handling two latent variables per iteration is way higher than handling one, the total wall-clock time might be lower with the single latent variable setup—even if it takes more iterations. This isn't an improvement in algorithmic convergence, but a practical efficiency win.
• Redundant latent variables: If the marginalized variable adds almost no information about $x$ (e.g., it's nearly independent of $x$ and $z_1$), marginalizing it will barely loosen the bound but cut down on computation. In this case, you get similar iteration speed with lower overhead.

3. 收敛到的极值点质量 #

Marginalizing latent variables can change the local maximum the algorithm converges to. A model with more latent variables has a richer parameterization, so it might find a higher-log-likelihood local optimum than the marginalized version. Of course, if the marginalized variable is redundant, this difference will be negligible.

内容的提问来源于stack exchange，提问作者alberto