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Bayesian Parameter Estimation with examples

A slecture by ECE student Yu Wang

Loosely based on the ECE662 Spring 2014 lecture material of Prof. Mireille Boutin.

Introduction: Bayesian Estimation

Suppose that we have an observable random variable$ X $ for an experiment, that takes values in a set S. Suppose that distribution of $ X $ depends on a parameter $\theta$ taking values in a parameter space $ \Theta $. We will denote the probability density function of $ X $ for a given value of $ \theta $ by $ f( '''x''' \mid \theta) $ for $ x \in S $ and $ \theta \in S $. Of course, our data variable X is almost always vector-valued. The parameter $ \theta $ may also be vector-valued.

In Bayesian analysis, named for the famous Thomas Bayes, we treat the parameter $ \theta $ as a random variable, with a given probability density function $ h(\theta) $ for $ \theta \in \Theta $. The corresponding distribution is called the prior distribution of $ \theta $ and is intended to reflect our knowledge (if any) of the parameter, before we gather data. After observing $ x \in S $, we then use Bayes' theorem, to compute the conditional probability density function of $ \theta $ given $ \bs X=\bs x $.

First recall that the joint probability density function of $ (\bs X,\theta) $ is the mapping on $ S \times \Theta $ given by

$ (x, \theta) \mapsto h(\theta) f(x \mid \theta) $

Next recall that the (marginal) probability density function f of $ X $ is given by \[ $ f(x) = \sum_{\theta \in \Theta} h(\theta) f(x | \theta), \quad x \in S $ \] if the parameter has a discrete distribution, or \[ $ f(x) = \int_\Theta h(\theta) f(x| \theta) \, d\theta, \quad \x\in S $ \] if the parameter has a continuous distribution. Finally, the conditional probability density function of $\theta$ given $ X= x $ is \[ $ h(\theta \mid x) = \frac{h(\theta) f(x \mid \theta)}{f(x)}; \quad \theta \in \Theta, \; x\in S $ \] The conditional distribution of $ \theta $ given $ X= x $ is called the \textit{posterior} distribution, and is an updated distribution, given the information in the data. Finally, if $ \theta $ is a real parameter, the conditional expected value $ \mathbb{E}(\theta \mid X) $ is the Bayes' estimator of $ \theta $. Recall that $ \mathbb{E}(\theta \mid X) $is a function of X and, among all functions of X, is closest to $ \theta $ in the mean square sense. Thus, once we collect the data and observe $ X= x $, the estimate of $ \theta $ is $ \mathbb{E}(\theta \mid X) $.

Bayesian Parameter Estimation: General Theory

We first start with a generalized approach which can be applied to any situation in which the unknown density can be parameterized. The basic assumptions are as follows:

1. The form of the density $ p(x|\theta) $ is assumed to be known, but the value of the parameter vector $ \theta $ is not known exactly.

2. The initial knowledge about $ \theta $ is assumed to be contained in a known a priori density $ p(\theta) $.

3. The rest of the knowledge about $ \theta $ is contained in a set $ \mathcal{D} $ of n samples $ x_1, x_2, ... , x_n $ drawn independently according to the unknown probability density $ p(x) $.

Accordingly, already know:

$ p(x|D) = \int p(x|\theta)p(\theta|D)d\theta $

and By Bayes Theorem,

$ p(\theta|D) = \frac{p(D|\theta)p(\theta)}{\int p(D|\theta)p(\theta|D)d\theta} $


Now, since we are attempting to transform the equation to be based on samples $ x_k $, by independent assumption,

$ p(D|\theta) = \prod_{k = 1}^n p(x_k|\theta) $

Hence, if a sample $ \mathcal{D} $ has n samples, we can denote the sample space as: $ \mathcal{D}^n = \{x_1, x_2, ... x_n\} $.

Combine the sample space definition with the equation above:


$ p(D^n|\theta) = p(D^{n-1}|\theta)p(x_n|\theta) $

Using this equation, we can transform the Bayesian Parameter Estimation to:

$ p(\theta|D^n) = \frac{p(x_n|\theta)p(\theta|D^{n-1})}{\int p(x_n|\theta)p(\theta|D^{n-1})d\theta} $



Bayesian Parameter Estimation: Gaussian Case

The Univariate Case: $ p(\mu|\mathcal{D}) $

Consider the case where $ \mu $ is the only unknown parameter. For simplicity we assume:

$ p(x|\mu) \sim N(\mu, \sigma^2) $
and
$ p(\mu) \sim N(\mu_0, \sigma_0^2) $

From the previous section, the following expression could be easily obtained using Bayes' formula:

$ p(\mu|D) = \alpha \prod_{k = 1}^n p(x_k|\mu)p(\mu) $

Where $ \alpha $ is a factorization factor independent of $ \mu $.

Now, substitute $ p(x_k|\mu) $ and $ p(u) $ with:

$ p(x_k|\mu) = \frac{1}{(2\pi\sigma^2)^{1/2}}exp[-\frac{1}{2}(\frac{x_k-\mu}{\sigma})^{2}] $
$ p(u) = \frac{1}{(2\pi\sigma_0^2)^{1/2}}exp[-\frac{1}{2}(\frac{\mu-\mu_0}{\sigma_0})^{2}] $

The equation has now become:

$ p(\mu|D) = \alpha \prod_{k = 1}^n \frac{1}{(2\pi\sigma^2)^{1/2}}exp[-\frac{1}{2}(\frac{x_k-\mu}{\sigma})^{2}] \frac{1}{(2\pi\sigma_0^2)^{1/2}}exp[-\frac{1}{2}(\frac{\mu-\mu_0}{\sigma_0})^{2}] $
$ p(\mu|D) = \alpha \prod_{k = 1}^n \frac{1}{(2\pi\sigma^2)^{1/2}} \frac{1}{(2\pi\sigma_0^2)^{1/2}}exp[-\frac{1}{2}(\frac{\mu-\mu_0}{\sigma_0})^{2} -\frac{1}{2}(\frac{x_k-\mu}{\sigma})^{2}] $

Update the scaling factor to $ \alpha' $ and $ \alpha'' $ correspondingly,

$ p(\mu|D) = \alpha' exp \sum_{k=1}^n(-\frac{1}{2}(\frac{\mu-\mu_0}{\sigma_0})^{2} -\frac{1}{2}(\frac{x_k-\mu}{\sigma})^{2}) $
$ p(\mu|D) = \alpha'' exp [-\frac{1}{2}(\frac{n}{\sigma^2} + \frac{1}{\sigma_0^2})\mu^2 -2(\frac{1}{\sigma^2}\sum_{k=1}^nx_k + \frac{\mu_0}{\sigma_0^2})\mu] $

With the knowledge of Gaussian distribution:

$ p(u|D) = \frac{1}{(2\pi\sigma_n^2)^{1/2}}exp[-\frac{1}{2}(\frac{\mu-\mu_n}{\sigma_n})^{2}] $

Finally, the estimate of $ u_n $ can be obtained:

$ \mu_n = (\frac{n\sigma_0^2}{n\sigma_0^2 + \sigma^2})\bar{x_n} + \frac{\sigma^2}{n\sigma_0^2 + \sigma^2}\mu_0 $

Where $ \bar{x_n} $ is defined as sample means and $ n $ is the sample size.

In order to form a Gaussian distribution, the variance $ \sigma_n^2 $ associated with $ \mu_n $ could also be obtained correspondingly as:

$ \sigma_n^2 = \frac{\sigma_0^2 \sigma^2}{n\sigma_0^2 + \sigma^2} $


Observation:

With $ N \to \infty $,
$ \sigma_D \to 0 $

$ p(\mu|D) $ becomes more sharply peaked around $ \mu_D $

The Univariate Case: $ p(x|\mathcal{D}) $

Having obtained the posteriori density for the mean $ u_n $ of set $ \mathcal{D} $, the remaining of the task is to estimate the "class-conditional" density for $ p(x|D) $.

Based on the text Duda's chatpter #3.4.2 and Prof. Mimi's notes:


$ p(x|\mathcal{D}) = \int p(x|\mu)p(\mu|\mathcal{D})d\mu $
$ p(x|\mathcal{D}) = \int \frac{1}{\sqrt{2 \pi } \sigma} \exp[{-\frac{1}{2} (\frac{x-\mu}{\sigma})^2}] \frac{1}{\sqrt{2 \pi } \sigma_n} \exp[{-\frac{1}{2} (\frac{\mu-\mu_n}{\sigma_n})^2}] d\mu $


$ p(x|\mathcal{D}) = \frac{1}{2\pi\sigma\sigma_n} exp [-\frac{1}{2} \frac{(x-\mu)}{\sigma^2 + \sigma_n^2}]f(\sigma,\sigma_n) $


Where $ f(\sigma, \sigma_n) $ is defined as:


$ f(\sigma,\sigma_n) = \int exp[-\frac{1}{2}\frac{\sigma^2 + \sigma_n^2}{\sigma^2 \sigma_n ^2}(\mu - \frac{\sigma_n^2 x+\sigma^2 \mu_n}{\sigma^2+\sigma_n^2})^2]d\mu $

Hence, $ p(x|D) $ is normally distributed as:

$ p(x|D) \sim N(\mu_n, \sigma^2 + \sigma_n^2) $

References

[1]. Mireille Boutin, "ECE662: Statistical Pattern Recognition and Decision Making Processes," Purdue University, Spring 2014.

[2]. R. Duda, P. Hart, Pattern Classification. Wiley-Interscience. Second Edition, 2000.

Questions and comments

If you have any questions, comments, etc. please post them on this page.

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