This post is the first of a series on variational inference, a tool that has been widely used in machine learning. My goal here is to define the problem and then introduce the main characters at play: evidence, Kullback-Leibler (KL) divergence, and Evidence Lower BOund (ELBO). Those three quantities have a central role in variational inference and it is therefore necessary to have a clear understanding of how they are interconnected. I have followed a step-by-step approach in the disentanglement of the mathematical derivations, which should help you to keep the flow. Enjoy the reading!

## Definition of the problem

We are dealing with latent variable models. By definition, latent variables are hidden and they cannot be directly observed. However, we can infer latent variables by exploiting the observed variables. Examples of latent variables are intelligence and health. We cannot directly measure “intelligence” but we can infer its value by using appropriate tests (Raven’s Progressive Matrices, SAT scores, etc). Same story for “health”, we can measure various observable quantities (e.g. blood tests, pressure, etc) and from them we can infer an overall value. Here, I will refer to the latent variable as $$z$$ and to the observed variable as $$x$$. I will use the bold notation $$\mathbf{x}$$ and $$\mathbf{z}$$ to represent collection of variables

$\mathbf{x} = \{x_{1}, \dots, x_{N} \} \quad \text{and} \quad \mathbf{z} = \{z_{1}, \dots, z_{M} \}.$

Keep in mind that the observed variables $$\mathbf{x}$$ are often called just data. We suppose that the latent variable causes the observed variable, therefore the probabilistic graphical model would be $$z \rightarrow x$$. For instance, “health” causes specific values to be recorded in the blood tests. In a Bayesian treatment we imply that latent variables govern the distribution of the data. In particular, in a Bayesian model we draw the latent variables from a prior density $$p(\mathbf{z})$$ and then relates them to the observations through the likelihood $$p(\mathbf{x} \vert \mathbf{z})$$.

Inference. Our goal is to perform inference, that is estimate some hidden variables given some observations (data, observed variables). For instance, given various medical examinations we want to infer “health”, or given cognitive tests infer “intelligence”. Formally, we want to estimate the following posterior

$p(\mathbf{z} \vert \mathbf{x}) = \frac{p(\mathbf{z}, \mathbf{x})}{p(\mathbf{x})}.$

However, estimating the posterior in such a way is not possible.

Problem: intractable posterior. Let’s give a closer look at $$p(\mathbf{z} \vert \mathbf{x})$$. The numerator of the posterior is the joint distribution over the observed and latent variables $$p(\mathbf{z}, \mathbf{x})$$ which is generally efficient to compute. The denominator $$p(\mathbf{x})$$ of the posterior is called marginal likelihood or evidence and can be estimated as follows

$p(\mathbf{x}) = \int p(\mathbf{z}, \mathbf{x}) d \mathbf{z}.$

Here is the problem: in many cases this integral is intractable. Well, what does it mean intractable? Let’s try to understand it with a concrete example. In a previous post I have introduced Gaussian Mixture Models (GMMs). To estimating the evidence in GMMs we need to integrate (sum up) all configurations of the latent variables. This has an exponential cost in the number of latent variables, and it can rapidly become computationally intractable.

## Variational Bayesian methods

We can exploit variational Bayesian methods to solve our problem. In particular, since the exact estimation of the integral over $$p(\mathbf{x})$$ is intractable we can try approximating the true posterior $$p(\mathbf{z} \vert \mathbf{x})$$ using a variational distribution (sometimes called guide) over the latent variables $$q(\mathbf{z})$$. As a shorthand, I will often use just $$p$$ to refer to the true distribution and $$q$$ to refer to the approximated distribution.

The variational distribution. Let’s define a family of distributions $$\mathcal{Q}$$ from which we are going to pick the ideal candidate $$q_{\ast} \in \mathcal{Q}$$ that better fits the true distribution $$p$$. Generally, we are going to choose a friendly distribution family, such as Gaussians, to simplify our life. The parameters describing $$q$$ (e.g. mean and standard deviation of the Gaussians) are grouped in a vector $$\boldsymbol{\theta}$$, therefore in practice we are looking for $$\boldsymbol{\theta}_{\ast}$$ the optimal set of parameters of the ideal distribution $$q_{\ast}$$. The parameters of the variational distribution $$\boldsymbol{\theta}$$ and the latent variables $$\mathbf{z}$$ are often grouped together and called unobserved variables. The problem now is how to find the best candidate $$q_{\ast}$$. We need a measure of similarity between $$p$$ and $$q$$ that we can use as a metric during our search. The Kullback-Leibler (KL) divergence is what we are looking for.

The Kullback-Leibler (KL) divergence. The KL divergence can be used to measure the similarity between two distributions. For instance, given our distributions $$p$$ and $$q$$ we define

$\text{KL} \big( q(\mathbf{z}) || p(\mathbf{z} | \mathbf{x}) \big) = \int q(\mathbf{z}) \log \frac{q(\mathbf{z})}{p(\mathbf{z} | \mathbf{x})} d\mathbf{z} = - \int q(\mathbf{z}) \log \frac{p(\mathbf{z} | \mathbf{x})}{q(\mathbf{z})} d\mathbf{z}.$

In our particular case, we want to find the best candidate distribution $$q_{\ast}$$ that minimizes the KL divergence

$q_{\ast}(\mathbf{z}) = \text{argmin} \ \text{KL}\big(q(\mathbf{z}) \vert\vert p(\mathbf{z} \vert \mathbf{x})\big).$

Short note on KL divergence. The KL divergence is non-negative, that is $$\text{KL} \geq 0$$. Moreover, it is not symmetric and it does not satisfy the triangle inequality, meaning that it is not a proper distance metric. Asymmetry implies that

$\text{KL}(p \vert\vert q) \neq \text{KL}(q \vert\vert p),$

with $$\text{KL}(p \vert\vert q)$$ called the forward and $$\text{KL}(q \vert\vert p)$$ the backward (or reverse) KL divergence. The forward KL is mode averaging, whereas the backward is mode fitting. To remember this difference I came up with a mnemonic trick. In the forward KL we have $$\text{KL}(p \vert\vert q)$$ with the first term being $$p$$, the true distribution. The bump of the letter p points forward (left to right). In the backward KL instead the first distribution is $$q$$, the bump of the letter q points backward (right to left). To distinguish between mode averaging and mode fitting I use a similar trick. Imagine a bi-modal distribution with two symmetric modes, something like this _/\___/\_. In $$\text{KL}(p \vert\vert q)$$ the bumps of p and q point towards the center meaning that the resulting approximation will be mode averaging (positioned in between the two modes). In $$\text{KL}(q \vert\vert p)$$ the bumps q and p point outwards, meaning that the resulting approximation will be mode fitting (fitting one of the two modes).

Problem: intractable KL divergence. You may have noticed something shady going on here. Our goal was to find a way to avoid the estimation of $$p(\mathbf{z} \vert \mathbf{x})$$ because its evidence $$p(\mathbf{x})$$ was intractable. To do this we have introduced variational inference and the KL divergence $$\text{KL}\big(q(\mathbf{z}) \vert\vert p(\mathbf{z} \vert \mathbf{x})\big)$$. An obvious question is: how are we going to estimte this KL divergence if we cannot estimate $$p(\mathbf{z} \vert \mathbf{x})$$ in the first place? This question gets to the heart of the problem, our troubles are not ended. It turns out that using a variational approach by exploiting the KL divergence has the same intractability issues. Let’s try to decompose the KL divergence to see how far we can go before we find an obstacle

\begin{aligned} \text{KL} \big( q(\mathbf{z}) || p(\mathbf{z} | \mathbf{x}) \big) &= \int q(\mathbf{z}) \log \frac{q(\mathbf{z})}{p(\mathbf{z} | \mathbf{x})} d\mathbf{z} &\text{(1.1)}\\ &= \int q(\mathbf{z}) \big( \log q(\mathbf{z}) - \log p(\mathbf{z} | \mathbf{x}) \big) d\mathbf{z} &\text{(1.2)}\\ &= \int q(\mathbf{z}) \log q(\mathbf{z}) - q(\mathbf{z}) \log p(\mathbf{z} | \mathbf{x}) d\mathbf{z} &\text{(1.3)}\\ &= \int q(\mathbf{z}) \log q(\mathbf{z}) d\mathbf{z} - \int q(\mathbf{z}) \log p(\mathbf{z} | \mathbf{x}) d\mathbf{z} &\text{(1.4)}\\ &= \mathbb{E}_{q} \big[ \log q(\mathbf{z}) \big] - \mathbb{E}_{q} \big[ \log p(\mathbf{z} | \mathbf{x}) \big] & \text{(1.5)}\\ &= \mathbb{E}_{q} \big[ \log q(\mathbf{z}) \big] - \mathbb{E}_{q} \bigg[ \log \frac{p(\mathbf{x}, \mathbf{z}) }{p(\mathbf{x})} \bigg] &\text{(1.6)}\\ &= \mathbb{E}_{q} \big[ \log q(\mathbf{z}) \big] - \mathbb{E}_{q} \big[ \log p(\mathbf{x}, \mathbf{z}) - \log p(\mathbf{x}) \big] &\text{(1.7)}\\ &= \mathbb{E}_{q} \big[ \log q(\mathbf{z}) - \log p(\mathbf{x}, \mathbf{z}) \big] + \mathbb{E}_{q} \big[ \log p(\mathbf{x}) \big] &\text{(1.8)}\\ &= \mathbb{E}_{q} \big[ \log q(\mathbf{z}) - \log p(\mathbf{x}, \mathbf{z}) \big] + \underbrace{\log p(\mathbf{x})}_{\text{intractable}} &\text{(1.9)}\\ \end{aligned}

Give a closer look to the last form in $$\text{(1.9)}$$. The intractable term $$p(\mathbf{x})$$ is still there as we suspected. It is necessary to find another way to reach our goal, but first I would like to give a step-by-step explanation of what I have done above:

$$(1.1 \rightarrow 1.2)$$ Breaking the ratio by applying the property of logarithms

$\log \frac{f(x)}{g(x)} = \log f(x) - \log g(x).$

$$(1.2 \rightarrow 1.3)$$ Multiplying the two terms with $$q(\mathbf{z})$$ and expanding.

$$(1.3 \rightarrow 1.4)$$ Applying the property of integrals

$\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx.$

$$(1.4 \rightarrow 1.5)$$ Applying the definition of expectation over the integrals.

$$(1.5 \rightarrow 1.6)$$ Straightforward application of the rule of probabilities

$p(\mathbf{z} | \mathbf{x}) = \frac{p(\mathbf{x}, \mathbf{z}) }{p(\mathbf{x})}.$

$$(1.6 \rightarrow 1.7)$$ Applying again the rule of logarithms to split the ratio.

$$(1.7 \rightarrow 1.8)$$ Applying the rule of expectations (linearity)

$\mathbb{E}[A + B] = \mathbb{E}[A] + \mathbb{E}[B],$

to rearrange the terms, ending up with two different expectations.

$$(1.8 \rightarrow 1.9)$$ There is no expected value for $$\mathbb{E}_{q} \big[ \log p(\mathbf{x}) \big]$$ since we are considering an expectation $$\mathbb{E}_{q}$$ over $$q(\mathbf{z})$$ and not over $$p(\mathbf{x})$$.

## Evidence Lower BOund (ELBO)

Let’s go back to $$\text{(1.9)}$$ where we got

$\text{KL} \big( q(\mathbf{z}) \vert\vert p(\mathbf{z} \vert \mathbf{x}) \big) = \mathbb{E}_{q} \big[ \log q(\mathbf{z}) - \log p(\mathbf{x}, \mathbf{z}) \big] + \log p(\mathbf{x}).$

This decomposition is problematic because we still have two intractable terms $$\text{KL} ( q(\mathbf{z}) \vert\vert p(\mathbf{z} \vert \mathbf{x}) )$$ and $$\log p(\mathbf{x})$$. However, rearranging those quantities by moving the intractable terms on the same side and switching signs, we get

$\mathbb{E}_{q} \big[\log p(\mathbf{x}, \mathbf{z}) - \log q(\mathbf{z}) \big] = \log p(\mathbf{x}) - \text{KL} \big( q(\mathbf{z}) || p(\mathbf{z} | \mathbf{x}) \big),$

where the sign of the expectation has been changed by switching the sign of the terms inside the brackets. Here is the idea: what if we maximize the quantity on the left side? This would be great because the terms on the left are tractable. Is this legit? Here is the interesting part, by maximizing the quantity on the left we are simultaneously (i) maximizing the evidence $$p(\mathbf{x})$$, and (ii) minimizing the KL divergence between our variational distribution $$q(\mathbf{z})$$ and the true posterior $$p(\mathbf{z} \vert \mathbf{x})$$, that is what we wanted to achieve with this machinery. Crucially, since the KL divergence is non-negative $$\text{KL} \geq 0$$ the left term is a lower-bound over the log-evidence $$p(\mathbf{x})$$ called the Evidence Lower BOund (ELBO)

$\text{ELBO}(q) = \mathbb{E}_{q} \big[\log p(\mathbf{x}, \mathbf{z}) - \log q(\mathbf{z}) \big] = \mathbb{E}_{q} \bigg[ \log \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})} \bigg],$

where I have used the rule of logarithms to get the ratio in the last step.

Unpacking the ELBO. The ELBO is the last trick in our arsenal. The idea is to avoid minimizing the KL divergence by maximizing an alternative quantity which is equivalent up to an added constant. The ELBO is such a quantity. The ELBO is also known as the variational lower bound or negative variational free energy. Here, we are going to unpack the ELBO until we get a decomposition that is easy to manage:

\begin{aligned} \mathrm{ELBO}(q) &=\mathbb{E}_{q} \big[\log p(\mathbf{x}, \mathbf{z}) - \log q(\mathbf{z}) \big] &\text{(2.1)}\\ &=\mathbb{E}_{q}[\log p(\mathbf{z}, \mathbf{x})] -\mathbb{E}_{q}[\log q(\mathbf{z})] &\text{(2.2)}\\ &=\mathbb{E}_{q}[\log \big( p(\mathbf{x} \vert \mathbf{z}) p(\mathbf{z}) \big) ] -\mathbb{E}_{q}[\log q(\mathbf{z})] &\text{(2.3)}\\ &=\mathbb{E}_{q}[\log p(\mathbf{x} \vert \mathbf{z})] + \mathbb{E}_{q}[\log p(\mathbf{z})] - \mathbb{E}_{q}[\log q(\mathbf{z})] &\text{(2.4)}\\ &=\mathbb{E}_{q}[\log p(\mathbf{x} \vert\mathbf{z})] + \mathbb{E}_{q}[\log p(\mathbf{z}) - \log q(\mathbf{z})] &\text{(2.5)}\\ &=\mathbb{E}_{q}[\log p(\mathbf{x} \vert \mathbf{z})] + \int q(\mathbf{z}) \log \frac{p(\mathbf{z})}{q(\mathbf{z})} d\mathbf{z} &\text{(2.6)}\\ &=\mathbb{E}_{q}[\log p(\mathbf{x} \vert \mathbf{z})]- \text{KL}(q(\mathbf{z}) \| p(\mathbf{z})) &\text{(2.7)} \end{aligned}

Let’s take a closer look at $$\text{(2.7)}$$. The first term describes the probability of the data given the latent variable $$p(\mathbf{x} \mid \mathbf{z})$$. When we maximize the ELBO we also maximize this quantity which translate in picking those models $$q(\mathbf{z})$$ in the variational family $$\mathcal{Q}$$ that better predict the data $$\mathbf{x}$$. The second term, is the negative KL divergence between our variational model $$q(\mathbf{z})$$ and the prior over the latent variables $$p(\mathbf{z})$$. When we maximize the ELBO this term is pushed towards zero (because of the negative sign) meaning that the two distributions are forced to be close (identical if $$\text{KL}=0$$). In other words, the variational distribution is forced to be similar to the prior. For this reason, this form of the ELBO is sometimes called the prior-contrastive. Let’s break down what has been done above:

$$(2.1 \rightarrow 2.2)$$ Exploiting the linearity of expectation to separate the two terms. This form is often used in the literature to highlight $$\mathbb{E}_{q}[\log p(\mathbf{z})]$$, which is the entropy of the variational distribution. Note that, maximizing the ELBO implies the minimization of this entropy.

$$(2.2 \rightarrow 2.3)$$ Factorizing the joint probability: $$p(\mathbf{z}, \mathbf{x}) = p(\mathbf{x} \vert \mathbf{z}) p(\mathbf{z}).$$

$$(2.3 \rightarrow 2.4)$$ Breaking the product by applying the property of logarithms, then splitting the expectation

$\mathbb{E}_{q}[\log \big( p(\mathbf{x} \vert \mathbf{z}) p(\mathbf{z}) \big)] = \mathbb{E}_{q}[\log p(\mathbf{x} \vert \mathbf{z}) + \log p(\mathbf{z})] = \mathbb{E}_{q}[\log p(\mathbf{x} \vert \mathbf{z})] + \mathbb{E}_{q}[\log p(\mathbf{z})].$

$$(2.4 \rightarrow 2.5)$$ Applying the properties of expectation (linearity).

$$(2.5 \rightarrow 2.6)$$ Rewriting the expectation as an integral (just for clarity)

$\mathbb{E}_{q}[\log p(\mathbf{z}) - \log q(\mathbf{z})] = \int q(\mathbf{z}) \big( \log p(\mathbf{z}) - \log q(\mathbf{z}) \big) d\mathbf{z} = \int q(\mathbf{z}) \log \frac{p(\mathbf{z})}{q(\mathbf{z})} d\mathbf{z}.$

$$(2.6 \rightarrow 2.7)$$ The integral is equivalent to the negative KL divergence.

## Other routes to the ELBO

We are dealing with three quantities:

• the log-evidence $$\log p(\mathbf{x})$$
• the KL divergence $$\text{KL} ( q(\mathbf{z}) \vert\vert p(\mathbf{z} \vert \mathbf{x}) )$$
• the ELBO $$-\mathbb{E}_{q} \big[ \log q(\mathbf{z}) \big] + \mathbb{E}_{q} \big[ \log p(\mathbf{x}, \mathbf{z}) \big]$$

As we saw above, those quantities are strictly intertwined. In previous sections I have used the KL-divergence as starting point for the derivation of the ELBO and the evidence. However, it is also possible to use the evidence as starting point. In particular, there are two possible derivations, the first based on the Jensen’s inequality shows why the ELBO is a lower bound over the evidence, the second starts from the evidence to return the KL divergence and the ELBO. Let’s dive into those two derivations.

Derivation via Jensen’s inequality. This is the most popular derivation of the ELBO, which clearly shows why the ELBO is a lower bound over the evidence. The idea is to exploit the Jensen’s inequality to reorganize some of the terms. The Jensen’s inequality relates the value of a convex function of an integral (or expectation) to the integral of the convex function in this way

$g(\mathbb{E}[X]) \leq \mathbb{E}[g(X)],$

where $$g$$ is a convex function, and $$X$$ an integrable real-valued random variable. The inequality similarly applies to a concave function (with inequality reversed), which is what we care about here, since we are dealing with log-probabilities

$\log \mathbb{E}[X] \geq \mathbb{E}[\log X].$

That’s it. The log of the expectation of a random variable is greater than or equal to the expectation of the log of that random variable. Keep in mind this inequality, we are going to use it at some point in the decomposition below:

\begin{aligned} \log p(\mathbf{x}) &=\log \int p(\mathbf{x}, \mathbf{z}) d\mathbf{z} &\text{(3.1)}\\ &=\log \int p(\mathbf{x}, \mathbf{z}) \frac{q(\mathbf{z})}{q(\mathbf{z})} d\mathbf{z} &\text{(3.2)}\\ &=\log \mathbb{E}_{q} \bigg[ \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})} \bigg] &\text{(3.3)}\\ &\geq \underbrace{\mathbb{E}_{q} \bigg[ \log \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})} \bigg]}_{\text{ELBO}} &\text{(3.4)}\\ \end{aligned}

In the last step I have used the Jensen’s inequality such that

$\log p(\mathbf{x}) \geq \text{ELBO},$

and therefore the ELBO is a lower bound over the (log)evidence. Let’s breakdown the various steps:

$$(3.1)$$ Rewriting $$p(\mathbf{x})$$ as a marginalization of the joint distribution over $$\mathbf{z}$$.

$$(3.1 \rightarrow 3.2)$$ Including a new term for convenience. I will state the obvious: the new quantity is equal to 1 and has no effect on the product, therefore this is a legit operation.

$$(3.2 \rightarrow 3.3)$$ The new integral can be rewritten as an expectation.

$$(3.3 \rightarrow 3.4)$$ Finally, we apply the Jensen’s inequality to move the logarithm inside the brackets.

Alternative derivation. Starting from the evidence we should be able to retrieve both the ELBO and the KL-divergence. An alternative derivation can be used to achieve this goal. This derivation is not so popular as the previous one but it is helpful, since it gives a broader view over the connection between the three quantities of interest. For completeness, I will write it down here:

\begin{aligned} \log p(\mathbf{x}) &=\mathbb{E}_{q}[\log p(\mathbf{x})] &\text{(4.1)}\\ &=\mathbb{E}_{q} \bigg[ \log \frac{p(\mathbf{x}, \mathbf{z})}{p(\mathbf{z} \vert \mathbf{x})} \bigg] &\text{(4.2)}\\ &=\mathbb{E}_{q} \bigg[ \log \frac{p(\mathbf{x}, \mathbf{z})}{p(\mathbf{z} \vert \mathbf{x})} \frac{q(\mathbf{z})}{q(\mathbf{z})} \bigg] &\text{(4.3)}\\ &=\mathbb{E}_{q} \bigg[ \log \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})} \frac{q(\mathbf{z})}{p(\mathbf{z} \vert \mathbf{x})} \bigg] &\text{(4.4)}\\ &=\underbrace{\mathbb{E}_{q} \bigg[ \log \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})} \bigg]}_{\text{ELBO}} + \underbrace{\mathbb{E}_{q} \bigg[ \log \frac{q(\mathbf{z})}{p(\mathbf{z} \vert \mathbf{x})} \bigg]}_{\text{KL}} &\text{(4.5)}\\ \end{aligned}

As you can see this derivation is different from the one based on the Jensen’s inequality, since it returns two terms: the ELBO and the KL divergence. Here is a quick explanation of what has been done above:

$$(4.1)$$ The expectation over $$q(\mathbf{z})$$ has no effect on $$p(\mathbf{x})$$ because

$\mathbb{E}_{q}[\log p(\mathbf{x})] = \int q(\mathbf{z}) \log p(\mathbf{x}) d\mathbf{z} = \log p(\mathbf{x}) \int q(\mathbf{z}) d\mathbf{z} = \log p(\mathbf{x}).$

$$(4.1 \rightarrow 4.2)$$ Applying a refactoring based on

$p(\mathbf{x}) = \frac{p(\mathbf{x}, \mathbf{z})}{p(\mathbf{z} \vert \mathbf{x})}.$

$$(4.2 \rightarrow 4.3)$$ Including a new term for convenience, this has no effect on the product.

$$(4.3 \rightarrow 4.4)$$ Rearranging by switching the denominator of the two terms.

$$(4.4 \rightarrow 4.5)$$ Applying the rule of logarithms to turn a product into a sum, then splitting the expectation based on the linearity property.

## Conclusion

In this post I have provided an introduction to variational inference introducing the three main characters at play: the evidence, the KL-divergence, and the ELBO. I have tried to be as compact as possible, explaining every step along the way. Variational inference focuses on optimisation instead of integration, it can be applied to many probabilistic models (e.g. non-conjugate, high-dimensional, directed and undirected), it is numerically stable, fast to converge, and easy to train on GPUs. Below you can find some additional resources if you want to know more about variational inference.

## Resources

• “Variational Inference”, A.L. Popkes [PDF]
• “Variational Inference: A Review for Statisticians”, D. Blei, A. Kucukelbir, and J.D. McAuliffe [arXiv]
• Shakir Mohamed’s tutorials, e.g. [PDF-1] and [PDF-2]
• Yarin Gal’s thesis [PDF]
• “Pattern Recognition and Machine Learning”, Chapter 10, C. Bishop