# Laplacian Approximation

Posted by GwanSiu on May 15, 2021

## 1. Laplacian Approximation

We will talk about the Laplacian approximation in this post, which has been widely used in machine learning research. The core idea of Laplacian approximation is to use a well-defined unimodal function to approximate the integral value of a sophisticated function with a Gaussian density function.

Suppose we consider a function $g(x)\in \mathcal{L}^{2}$ and this function achieves its maximum value at the point $x_{0}$. The integral value of this function in $[a, b]$ is computed as follows:

$$$\displaystyle{\int_{a}^{b}}g(x)\mathrm{d}x.$$$

Let $h(x)=\log g(x)$. We can re-write the Eq.(1) as follows:

$$$\displaystyle{\int_{a}^{b}\exp(h(x))}\mathrm{d}x.$$$

We will take the Taylor series approximation of $h(x)$ at the point $x_{0}$, and have

$$$\displaystyle{\int_{a}^{b}}\exp(h(x))\mathrm{d}x \approx \displaystyle{\int_{a}^{b}\exp(h(x_{0}))+h^{\prime}(x_{0})(x-x_{0})+\frac{1}{2}h^{\prime\prime}(x_{0})(x-x_{0})}\mathrm{d}x.$$$

Because the function $g(x)$ achieves its maximum at the point $x_{0}$ and $\log(\cdot)$ is a monotonic function, we have $h^{\prime}(x_{0})=0$. Therefore, Eq.(3) can be simplified as follows:

$$$\displaystyle{\int_{a}^{b}}\exp(h(x))\mathrm{d}x \approx \displaystyle{\int_{a}^{b}}\exp(h(x_{0}) + \frac{1}{2}h^{\prime\prime}(x_{0})(x-x_{0})^{2})\mathrm{d}x$$$

It is found that the term $h(x_{0})$ does not depend on $x$, so that it can be pulled outside the integral. In addition, we can rearrange the term $\frac{1}{2}h^{\prime\prime}(x_{0})(x-x_{0})^{2}$, and have

$$$\displaystyle{\int_{a}^{b}}\exp(h(x))\mathrm{d}x \approx \exp(h(x_{0})) \displaystyle{\int_{a}^{b}}\exp\left(-\frac{1}{2}\frac{(x-x_{0})^{2}}{-h^{\prime\prime}(x_{0})^{-1}}\right)\mathrm{d}x.$$$

It is observed that the term $\displaystyle{\int_{a}^{b}}\exp\left(-\frac{1}{2}\frac{(x-x_{0})^{2}}{-h^{\prime\prime}(x_{0})^{-1}}\right)\mathrm{d}x$ is proportional to a Gaussian distribution with the mean $x_{0}$ and the variance $-h^{\prime\prime}(x_{0})^{-1}$.

Let $\Phi(x\vert \mu, \sigma^{2})$ be a cumulative function for a Gaussian distribution $\mathcal{N}(x\vert \mu, \sigma^{2})$ with the mean $\mu$ and the variance $\sigma^{2}$. We can re-write Eq.(5) as follows:

\begin{align} \displaystyle{\int_{a}^{b}\exp(h(x))}\mathrm{d}x &\approx \exp(h(x_{0})) \displaystyle{\int_{a}^{b}}\exp\left(-\frac{1}{2}\frac{(x-x_{0})^{2}}{-h^{\prime\prime}(x_{0})^{-1}}\right)\mathrm{d}x \\ &=\exp(h(x_{0}))\sqrt{\frac{2\pi}{-h^{\prime\prime}(x_{0})}}\left[\Phi(b\vert x_{0}, -h^{\prime\prime}(x_{0})^{-1}) - \Phi(a\vert x_{0}, -h^{\prime\prime}(x_{0})^{-1})\right] \end{align}

Given that $\exp(h(x_{0}))=g(x_{0})$. If $b=\infty, a=-\infty$, the term inside the bracket is equal to 1. By these condistions, the Laplacian approximation is equal to the value $g(x_{0})$ mutiplied by a constant term $\sqrt{\frac{2\pi}{-h^{\prime\prime}(x_{0})}}$, which depends on the curvature of $h(x)$. It means that the function $h(x)$ should be twice differentiable.

## 2. Application in Machine Learning.

In machine learning, we usually need to compute the mean of a posterior distribution from the observed data. Let $x$ denote the observed data. We use the notations $p(x\vert \theta)$ and $p(\theta)$ to represent the likelihood distribution and the prior distribution, respectively. The mean of the posterior distribution $p(\theta\vert x)$ is computed as follow:

\begin{align} \mathbb{E}[\theta] &= \displaystyle{\int}\theta p(\theta\vert x)\mathrm{d}\theta \\ &= \frac{\displaystyle{\int}\theta p(x\vert \theta)p(\theta)\mathrm{d}x}{\displaystyle{\int} p(x\vert \theta)p(\theta)\mathrm{d}\theta} \\ &= \frac{\displaystyle{\int}\theta \exp\left(\log(p(x\vert \theta)p(\theta))\right)\mathrm{d}\theta}{\displaystyle{\int} \exp\left(\log(p(x\vert \theta)p(\theta))\right)\mathrm{d}\theta} \end{align}

Let $h(\theta)=\log(p(x\vert \theta)p(\theta))$, and the function $\log(\cdot)$ is monotonic function. Therefore, the function $h(x)$ is proportional to the posterior density function. It means that the function $h(x)$ achieves its maximum value at the posterior mode. We assume $\hat{\theta}$ is the posterior mode. According to the description of Laplacian approximation, we have

\begin{align} \displaystyle{\int}\theta p(\theta\vert y)\mathrm{d}\theta &= \frac{\displaystyle{\int} \theta\exp(h(\hat{\theta})+\frac{1}{2}h^{\prime\prime}(\hat{\theta})(\theta-\hat{\theta})^{2})\mathrm{d}\theta}{\displaystyle{\int} \exp(h(\hat{\theta})+\frac{1}{2}h^{\prime\prime}(\hat{\theta})(\theta-\hat{\theta})^{2})\mathrm{d}\theta}, \\ &=\frac{\displaystyle{\int} \theta\exp(\frac{1}{2}h^{\prime\prime}(\hat{\theta})(\theta-\hat{\theta})^{2})\mathrm{d}\theta}{\displaystyle{\int} \exp(\frac{1}{2}h^{\prime\prime}(\hat{\theta})(\theta-\hat{\theta})^{2})\mathrm{d}\theta}, \\ &= \frac{\displaystyle{\int} \theta \sqrt{ \frac{2\pi}{-h^{\prime\prime}(\hat{\theta})}} \mathcal{N}(\theta)\vert \hat{\theta}, -h^{\prime\prime}(\hat{\theta})^{-1} \mathrm{d}\theta }{ \displaystyle{\int} \sqrt{ \frac{2\pi}{-h^{\prime\prime}(\hat{\theta})}} \mathcal{N}(\theta)\vert \hat{\theta}, -h^{\prime\prime}(\hat{\theta})^{-1} \mathrm{d}\theta }, \\ &= \hat{\theta}. \end{align}

It is found that the Laplacian approximation to the posterior mean is the posterior mode. Laplacian approximation works well when the posterior distribution is unimodal and relatively symmetric around the model. Intuitively, if the mass of the posterior distribution mainly concentrates around $\hat{\theta}$, the approximation will be better.

## Reference

All materials of this post come from the reference shown as follows: