# What's going on in the equation of the variational lower bound? I don't really understand what this equation is saying or what the purpose of the ELBO is. How does it help us find the true posterior distribution?

• Can you specify the source of the image?
– nbro
Oct 21 '19 at 11:34

From this document, as you found here, $$X$$ is an observed variable and $$Z$$ is a hidden variable; $$p(X)$$ is the density function of $$X$$. The posterior distribution of the hidden variables can then be written as follows using the Bayes’ Theorem:

$$p(Z|X) = \frac{p(X|Z)p(Z)}{p(X)} = \frac{p(X|Z)p(Z)}{\int_Zp(X,Z)}$$

Now base on what you post, if we denote that $$L= \mathbb{E}_q [\log p(X, Z)] + H[Z]$$ ($$q(Z)$$ is a distribution we use to approximation the true posterior distribution $$p(Z|X)$$ in VB and $$H[Z] = -\mathbb{E}_q [\log q(Z)]$$), then it is obvious that $$L$$ is a lower bound of the log probability of the observations. As a result, if in some cases we want to maximize the marginal probability (the log probability of the observations), we can instead maximize its variational lower bound $$L$$. As a real example, you can follow the "Multiple Object Recognition with Visual Attention" example in the referenced document.

Moreover, the term $$L$$ will be presented in KL-divergence that will be used to measuring the similarity of two distributions. Be aware that there is progress on the bound in this paper (Fixing a Broken ELBO).

• You do not explain how $q(Z)$ is related to the last term. What is $H[Z]$? You also do not explain why we would want to maximize $p(X)$.
– nbro
Oct 21 '19 at 12:45
• @nbro well noted. It is updated.
– OmG
Oct 21 '19 at 13:00
• Your answer seems correct, but I think it will still be difficult for a newbie to understand the ELBO and its motivation. Maybe later I will add a more complete answer.
– nbro
Oct 21 '19 at 13:12
• what does log of the P(X) get us? Why do they log it? Oct 21 '19 at 19:30
• @DylanY It is a marginal probability. Indeed, it is used in the concept of KL-Divergence. There, it is related to the concept of entropy that there is a $\log$ in the definition of the entropy.
– OmG
Oct 21 '19 at 19:56

The use of KL provides a more intuitive way of what the ELBO is attempting to maximize.

Basically, we want to find a posterior approximation such that $$p(z\mid x) \approx q(z)\in\mathcal{Q}$$

$$KL(q(z)\parallel p(z\mid x)) \rightarrow \min_{q(z)\in\mathcal{Q}}$$

As a result of this, while finding this optimal posterior approximation, we maximize the probability of all the observed data $$x$$. Note that the evidence is usually intractable. Thus, can express the $$KL$$ as follows:

\begin{align*} \log p(x) &= \int q(z) \log p(x)dz \\ &= \int q(z) \log\frac{p(x,\theta)}{p(\theta\mid x)}dz \\ &= \int q(z) \log\frac{p(x,z)q(z)}{p(\theta\mid x)q(z)}dz\\ &= \int q(z) \log\frac{p(x,z)}{q(z)}dz + \int q(z) \log\frac{q(z)}{p(z\mid x)}dz \\ &= \mathcal{L}(q(z)) + KL(q(z)\parallel p(z\mid x)) \end{align*}

In this case, KL just gives us the difference between $$q$$ and $$p$$. We want to make this difference close to zero meaning that $$q=p$$. So, minimizing the KL is the same as maximizing the ELBO, and as a result, we obtain the lower bound in your expression. If you expand your bound, you can find a nice interpretion:

\begin{align*} \mathcal{L}(q(z)) &= \int q(z) \log\frac{p(x,z)}{q(z)}dz \\ &= \mathbb{E}_{q(z)} \log p(x\mid z) - KL(q(z)\parallel p(z)) \end{align*}

When we optimize this expression, we want to find a $$q$$ that fits our data properly and also is really close to true posterior. Thus, $$\mathbb{E}_{q(z)} \log p(x\mid z)$$ act as a data term and $$KL(q(z)\parallel p(z))$$ as a regularizer.

• I think that the best way to motivate the ELBO is to start from the minimization problem of the KL divergence (which is the actual problem you want to solve), that is, the problem of minimizing the difference between the variational distribution (a supposedly tractable distribution) and the posterior (that you want to approximate with the variational distribution), then derive the ELBO. The motivation should not be that you want to maximize $p(x)$.
– nbro
Oct 21 '19 at 15:13