Bayersian Parameter Estimation - Rhea

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Bayesian Estimation

As stated in Duda's book, the conceptual difference between maximum likelihood estimation and Bayersian learning is that in MLE $ \theta $ is a fixed vector while in Bayersian estimation $ \theta $ is considered to be a random variable.

By definition, given samples $ \mathcal{D} $ Bayer's formula is defined as

$ P(w_i|x,D) = \frac{p(x|w_i,D)P(w_i|D)}{\sum_{j = 1}^c p(x|w_j,D)P(w_j|D)} $

Furthermore, $ p(x|D) $ can be computed as:

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

Bayesian Parameter Estimation: General Theory'

It is important to know that:

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, based on:

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

and 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_i $, by independent assumption,

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

$ 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} $

Investigation of Estimator's Accuracy of Bayesian Parameter Estimation: Gaussian Case

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

Assumptions:

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

$ p(\mu) \sim N(\mu_0, \sigma_0^2) $

From the previous section, the following expression could be easily obtained:

$ 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}] $

Hence, accordingly,

$ 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 $ \beta $,

$ p(\mu|D) = \beta 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) = \gamma 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] $

Furthermore, since

$ 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 $ is defined as:

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

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 by \textbf{Duda's},

$ p(x|\mathcal{D}) = \int p(x|\mu)p(\mu|\mathcal{D})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) $

\subsection{Experiment of Bayesian Parameter Estimation}

\paragraph{Design}

Assume n samples were obtained from the class $ \mathcal{D} $ of unknown mean $ \mu $ (known $ \sigma $). Assume,

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

$ p(\mu) \sim N(\mu_0, \sigma_0^2) $

While $ \sigma = \sigma_0 = constant $, and $ \mu_0 = 0 $ (It does not matter what $ \mu_0 $ it was assumed to be, this will be verified shortly after). Based on the sample data $ x_i \in \mathcal{D}, i = 1,2,3,...,n $, $ \mu $ is desired to be estimated.

The following results will be obtained: \begin{enumerate} \item The impact of $\mu_0$ on estimated $\hat{\mu}$ \item The impact of sample size $n$ have on estimation accuracy \end{enumerate}

\paragraph{Results} \begin{center} \includegraphics[scale=1]{BPE_1.png}

Figure 21. The impact of $\mu_0$ on estimated $\hat{\mu}$ averaged over 50 samples

\includegraphics[scale=1]{BPE_2.png}

Figure 22. The impact of $\mu_0$ on the variance of estimated $\hat{\mu}$ over 50 samples

\end{center}

The estimated mean is shifting up with $\mu_0$ increasing. \textbf{Based on the experiment it can be concluded that the most 'accurate' estimate could be obtained if $ \mu_0 = \mu $. But, according to the plot, even if the $\mu_0$ is different different from $\mu$, the error of estimation is still acceptable. (In our case, within [-0.1,+0.06] region)} However, the variance of estimated mean could be assumed to be identical as the \textbf{real empirical mean}.

\begin{center} \includegraphics[scale=0.7]{e23456.png}

Figure 23. The impact of sample size $n$ have on estimation shape accuracy (sample sizes = 2,3,4,5,6)

\includegraphics[scale=0.6]{ece662_14.png}

Figure 24. The impact of sample size $n$ have on estimation shape accuracy (sample sizes = 4,10,20,50,100)

\paragraph{Conclusion} Figure 23. and Figure 24. have demonstrated that with insufficient sample size the result would be really poor regarding prediction of points distribution.

Fig 3: Summary diagram of whitening and coloring process.