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		<title>Handbook of Markov Chain Monte Carlo</title>
		<link>http://thethong.wordpress.com/2012/01/11/893/</link>
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		<pubDate>Wed, 11 Jan 2012 09:45:07 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Bayesian]]></category>
		<category><![CDATA[Markov Chains]]></category>
		<category><![CDATA[Monte Carlo]]></category>
		<category><![CDATA[Probability]]></category>
		<category><![CDATA[RJMCMC]]></category>
		<category><![CDATA[Adaptive MCMC]]></category>
		<category><![CDATA[Books]]></category>
		<category><![CDATA[Handbook of Markov Chain Monte Carlo]]></category>
		<category><![CDATA[Likelihood-Free MCMC]]></category>
		<category><![CDATA[MCMC]]></category>
		<category><![CDATA[Particle MCMC]]></category>
		<category><![CDATA[pMCMC]]></category>
		<category><![CDATA[Reversible Jump MCMC]]></category>
		<category><![CDATA[Riemann Manifold MCMC]]></category>
		<category><![CDATA[Trans-dimensional MC]]></category>

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		<description><![CDATA[Today I received my copy of &#8220;Handbook of Markov Chain Monte Carlo&#8220;. Up to now I have consulted &#8220;Monte Carlo Strategies in Scientific Computing&#8220; of Jun S.Liu. In this post I will give my first thought on the new book. The book of Liu is indeed awesome. The author himself has made many original contributions to [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=893&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Today I received my copy of &#8220;<a href="http://www.crcpress.com/product/isbn/9781420079418">Handbook of Markov Chain Monte Carlo</a>&#8220;. Up to now I have consulted &#8220;<a href="http://www.amazon.com/Monte-Carlo-Strategies-Scientific-Computing/dp/0387952306">Monte Carlo Strategies in Scientific Computing</a>&#8220; of Jun S.Liu. In this post I will give my first thought on the new book.</p>
<p>The book of Liu is indeed awesome. The author himself has made many original contributions to sampling algorithms, and many of his ideas was explained in the book. The chapters on Gibbs Sampler (chapter 6),General Conditional Sampling (chapter 7) and multi-chain MCMC (chapter 10, 11) were excellent. But given that the book was written 10 years ago, many recent developments is missing. Some algorithms were not given enough spaces (in particular, Reversible Jump MCMC was given only  2 pages!).</p>
<p><span id="more-893"></span></p>
<p>The handbook covers many recent developments such as likelihood-free MCMC, Adaptive MCMC&#8230;, and treats Reversible Jump MCMC with the details it deserves (20 pages). Likelihood-free MCMC is one computational method for Approximate Bayesian Computation (ABC), which recently gains many attention. Instead of computing the likelihood ratio in the acceptance probability of the Metropolis-Hasting (MH) algorithm, one tries to simulate this ratio by generating simulated data.  Adaptive MCMC is a general name for a class of MCMC which parameters can be tuned automatically during the search. For example, consider a MCMC kernel that consists of a Gibbs sampler and a MH sampler. With probability <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> one will perform the Gibbs  move, and with probability <img src='http://s0.wp.com/latex.php?latex=1-+%5Calpha&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1- &#92;alpha' title='1- &#92;alpha' class='latex' /> one will perform the MH move. The problem is <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> had to be fixed, while we often want to change <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> along the course of the search, for example larger <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> at the beginning in order to make large transitions around the parameter space to quickly identify promising areas, and smaller <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> later to better explore the local landscape around good candidates. One can not change <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> carelessly, since there <em>was</em> no guarantee that the resulted MCMC kernel would converge to the desired distribution.</p>
<p>The book start with theoretical chapters discussed many aspects of MCMC. Each chapter is written by distinguished researchers in the corresponding field. One can expect to learn from experiences and perspectives of many experts in just one book. The chapters that pique my interest right now are Reversible Jump MCMC by Fan and Sisson, Adaptive MCMC by Jeffrey Rosenthal, Hamilton MCMC by Radford Neal, and Likelihood-free MCMC by Sisson and Fan.</p>
<p>The second part of the book consists of applications and case studies, with examples range from educational research to high-energy astrophysics. This is certainly richer than examples from Liu&#8217;s book, which were biased toward Liu&#8217;s research field. But to tell the truth I am not that interested in reading about MCMC in educational research or astrophysics.</p>
<p>On a final note, the book did <em>not</em> cover all the new developments in MCMC up to 2011. The biggest missing part is perhaps Particle MCMC, which is gaining more and more attentions. Some other notable algorithms that I  really wish they had been mentioned are the equi-energy sampler and the Riemann Manifold MCMC.  The equi-energy sampler was a simple-yet-powerful algorithm introduced in 2004. The sampler has a large memory size to remember where it has been, and use this information to speed up convergence. I think a detail discussion will help beginners like me understand better strong&amp;weak points of it, compared to Population MCMC. About the Particle MCMC and Riemann Manifold MCMC, I think intuitive introductory discussions may help the non-experts like me to gather enough courage to delve into those equation-packed papers.</p>
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		<title>Formulating the PCA</title>
		<link>http://thethong.wordpress.com/2012/01/11/formulating-the-pca/</link>
		<comments>http://thethong.wordpress.com/2012/01/11/formulating-the-pca/#comments</comments>
		<pubDate>Tue, 10 Jan 2012 17:06:28 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Bayesian]]></category>
		<category><![CDATA[Probability]]></category>
		<category><![CDATA[Supplementary Math]]></category>
		<category><![CDATA[Write up]]></category>
		<category><![CDATA[Bayesian Lasso]]></category>
		<category><![CDATA[Elastic Net]]></category>
		<category><![CDATA[Lasso]]></category>
		<category><![CDATA[linear regression problem]]></category>
		<category><![CDATA[PCA]]></category>
		<category><![CDATA[Principal Component Analysis]]></category>
		<category><![CDATA[Ridge Regression]]></category>
		<category><![CDATA[Sparse PCA]]></category>
		<category><![CDATA[SPCA]]></category>

		<guid isPermaLink="false">http://thethong.wordpress.com/?p=757</guid>
		<description><![CDATA[Today I have thought about how one can formulate the Principal Component Analysis (PCA) method. In particular I want to reformulate PCA as a solution for a regression problem. The idea of reformulation PCA as a solution for some regression problem is useful in Sparse PCA , in which a regularization term is inserted into [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=757&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Today I have thought about how one can formulate the Principal Component Analysis (PCA) method. In particular I want to reformulate PCA as a solution for a regression problem. The idea of reformulation PCA as a solution for some regression problem is useful in Sparse PCA , in which a <img src='http://s0.wp.com/latex.php?latex=L_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='L_1' title='L_1' class='latex' /> regularization term is inserted into a ridge regression formula to enforce spareness of the coefficients (i.e. elastic net). There are at least two equivalent ways to motivate PCA. In this post I will first give a formulation of PCA based on orthogonal projection, and then discuss a regression-type reformulation of PCA.</p>
<p><span id="more-757"></span></p>
<p>1. Orthogonal projection onto optimal hyperplane:</p>
<p>Given <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' />-dimensional column vectors <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7By%7D_1%2C%5Ccdots%2C+%5Cmathbf%7By%7D_n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{y}_1,&#92;cdots, &#92;mathbf{y}_n' title='&#92;mathbf{y}_1,&#92;cdots, &#92;mathbf{y}_n' class='latex' /> (assumed pre-processing is done <img src='http://s0.wp.com/latex.php?latex=%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%5Cmathbf%7By%7D_i+%3D+%5Cmathbf%7B0%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sum_{i=1}^{n}&#92;mathbf{y}_i = &#92;mathbf{0}' title='&#92;sum_{i=1}^{n}&#92;mathbf{y}_i = &#92;mathbf{0}' class='latex' />)</p>
<p>Suppose that we want to project these vectors orthogonally onto a unit vector <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1' title='&#92;mathbf{a}_1' class='latex' />, so that the projected vectors will be:</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bx%7D_i+%3D+%5Cmathbf%7Ba%7D_1%5Cmathbf%7Ba%7D_1%5ET%5Cmathbf%7By%7D_i+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{x}_i = &#92;mathbf{a}_1&#92;mathbf{a}_1^T&#92;mathbf{y}_i ' title='&#92;mathbf{x}_i = &#92;mathbf{a}_1&#92;mathbf{a}_1^T&#92;mathbf{y}_i ' class='latex' /></p>
<p>If we think of the projected vectors <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bx%7D+_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{x} _i' title='&#92;mathbf{x} _i' class='latex' /> as an approximation of the original <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7By%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{y}_i' title='&#92;mathbf{y}_i' class='latex' />, then a measure of how good the overall approximation is is the sum of lengths of the discrepancies <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7By%7D_i+-+%5Cmathbf%7Bx%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{y}_i - &#92;mathbf{x}_i' title='&#92;mathbf{y}_i - &#92;mathbf{x}_i' class='latex' />. So it is natural to adopt the following sum of square as the objective function, and judge <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1' title='&#92;mathbf{a}_1' class='latex' /> base on how small it can make the objective function be.</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=L_1%3D%5Cdfrac%7B1%7D%7B2%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%28%5Cmathbf%7By%7D_i+-+%5Cmathbf%7Bx%7D_i%29%5ET+%28%5Cmathbf%7By%7D_i+-+%5Cmathbf%7Bx%7D_i%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='L_1=&#92;dfrac{1}{2}&#92;sum_{i=1}^{n}(&#92;mathbf{y}_i - &#92;mathbf{x}_i)^T (&#92;mathbf{y}_i - &#92;mathbf{x}_i)' title='L_1=&#92;dfrac{1}{2}&#92;sum_{i=1}^{n}(&#92;mathbf{y}_i - &#92;mathbf{x}_i)^T (&#92;mathbf{y}_i - &#92;mathbf{x}_i)' class='latex' /></p>
<p>Expanding <img src='http://s0.wp.com/latex.php?latex=L_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='L_1' title='L_1' class='latex' /> yields</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=L_1+%3D+%5Csum_%7Bi%3D1%7D%5En%5Cmathbf%7By%7D_i%5ET%5Cmathbf%7By%7D+-+%5Cmathbf%7Ba%7D_1%5ET%5Csum_%7Bi%3D1%7D%5En%5Cmathbf%7By%7D_i%5Cmathbf%7By%7D_i%5ET%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='L_1 = &#92;sum_{i=1}^n&#92;mathbf{y}_i^T&#92;mathbf{y} - &#92;mathbf{a}_1^T&#92;sum_{i=1}^n&#92;mathbf{y}_i&#92;mathbf{y}_i^T&#92;mathbf{a}_1' title='L_1 = &#92;sum_{i=1}^n&#92;mathbf{y}_i^T&#92;mathbf{y} - &#92;mathbf{a}_1^T&#92;sum_{i=1}^n&#92;mathbf{y}_i&#92;mathbf{y}_i^T&#92;mathbf{a}_1' class='latex' /></p>
<p>Let <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7B%5COmega%7D+%3D+%5Cdfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5En%5Cmathbf%7By%7D_i%5Cmathbf%7By%7D_i%5ET&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{&#92;Omega} = &#92;dfrac{1}{n}&#92;sum_{i=1}^n&#92;mathbf{y}_i&#92;mathbf{y}_i^T' title='&#92;mathbf{&#92;Omega} = &#92;dfrac{1}{n}&#92;sum_{i=1}^n&#92;mathbf{y}_i&#92;mathbf{y}_i^T' class='latex' />. Choosing <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1' title='&#92;mathbf{a}_1' class='latex' /> to minimize <img src='http://s0.wp.com/latex.php?latex=L_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='L_1' title='L_1' class='latex' /> is equivalent to choosing <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1' title='&#92;mathbf{a}_1' class='latex' /> to maximize <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1%5ET%5Cmathbf%7B%5COmega%7D+%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1^T&#92;mathbf{&#92;Omega} &#92;mathbf{a}_1' title='&#92;mathbf{a}_1^T&#92;mathbf{&#92;Omega} &#92;mathbf{a}_1' class='latex' /> with the condition <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D%5ET%5Cmathbf%7Ba%7D+%3D+1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}^T&#92;mathbf{a} = 1' title='&#92;mathbf{a}^T&#92;mathbf{a} = 1' class='latex' />. With a change of variable, one can prove that the optimal <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1' title='&#92;mathbf{a}_1' class='latex' /> is the (normalized) eigenvector <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1' title='&#92;mathbf{v}_1' class='latex' /> of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7B%5COmega%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{&#92;Omega}' title='&#92;mathbf{&#92;Omega}' class='latex' />corresponding to the largest eigenvalue of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7B%5COmega%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{&#92;Omega}' title='&#92;mathbf{&#92;Omega}' class='latex' />.</p>
<p>Suppose after we have fixed <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1' title='&#92;mathbf{a}_1' class='latex' /> at <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1' title='&#92;mathbf{v}_1' class='latex' />, another unit vector <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2' title='&#92;mathbf{a}_2' class='latex' /> with the property <img src='http://s0.wp.com/latex.php?latex=a_2+%5Cperp+a_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a_2 &#92;perp a_1' title='a_2 &#92;perp a_1' class='latex' /> is given, and we are asked how to choose the best <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2' title='&#92;mathbf{a}_2' class='latex' /> if the same orthogonal projection is carried out onto <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2' title='&#92;mathbf{a}_2' class='latex' />. With the same reasoning, the best <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2' title='&#92;mathbf{a}_2' class='latex' /> is the one that maximizing <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2%5ET%5Cmathbf%7B%5COmega%7D%5Cmathbf%7Ba%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2^T&#92;mathbf{&#92;Omega}&#92;mathbf{a}_2' title='&#92;mathbf{a}_2^T&#92;mathbf{&#92;Omega}&#92;mathbf{a}_2' class='latex' /> with subject to <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2%5ET%5Cmathbf%7Ba%7D_2+%3D+1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2^T&#92;mathbf{a}_2 = 1' title='&#92;mathbf{a}_2^T&#92;mathbf{a}_2 = 1' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2%5ET%5Cmathbf%7Ba%7D_1+%3D+0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2^T&#92;mathbf{a}_1 = 0' title='&#92;mathbf{a}_2^T&#92;mathbf{a}_1 = 0' class='latex' />. Again with a change of variable, it can be proved that the best <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2' title='&#92;mathbf{a}_2' class='latex' /> is the (normalized) eigenvector <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_2' title='&#92;mathbf{v}_2' class='latex' /> of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7B%5COmega%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{&#92;Omega}' title='&#92;mathbf{&#92;Omega}' class='latex' />corresponding to the second largest eigenvalue of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7B%5COmega%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{&#92;Omega}' title='&#92;mathbf{&#92;Omega}' class='latex' />.</p>
<p>A natural question to ask is whether <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1' title='&#92;mathbf{v}_1' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_2+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_2 ' title='&#92;mathbf{v}_2 ' class='latex' /> are still the best if we consider orthogonal projection onto the plane spanned by two orthogonal unit vectors <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1' title='&#92;mathbf{a}_1' class='latex' />,<img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_2' title='&#92;mathbf{a}_2' class='latex' />.</p>
<p>In other words, we consider the following objective function</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=L_%7Bplane%7D+%3D+%5Cdfrac%7B1%7D%7B2%7D%5Csum_%7Bi%3D1%7D%5En%28%5Cmathbf%7By%7D_i+-+%5Cmathbf%7Ba%7D_1%5Cmathbf%7Ba%7D_1%5ET%5Cmathbf%7By%7D_i+-+%5Cmathbf%7Ba%7D_2%5Cmathbf%7Ba%7D_2%5ET%5Cmathbf%7By%7D_i%29%5ET%28%5Cmathbf%7By%7D_i+-+%5Cmathbf%7Ba%7D_1%5Cmathbf%7Ba%7D_1%5ET%5Cmathbf%7By%7D_i+-+%5Cmathbf%7Ba%7D_2%5Cmathbf%7Ba%7D_2%5ET%5Cmathbf%7By%7D_i%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='L_{plane} = &#92;dfrac{1}{2}&#92;sum_{i=1}^n(&#92;mathbf{y}_i - &#92;mathbf{a}_1&#92;mathbf{a}_1^T&#92;mathbf{y}_i - &#92;mathbf{a}_2&#92;mathbf{a}_2^T&#92;mathbf{y}_i)^T(&#92;mathbf{y}_i - &#92;mathbf{a}_1&#92;mathbf{a}_1^T&#92;mathbf{y}_i - &#92;mathbf{a}_2&#92;mathbf{a}_2^T&#92;mathbf{y}_i)' title='L_{plane} = &#92;dfrac{1}{2}&#92;sum_{i=1}^n(&#92;mathbf{y}_i - &#92;mathbf{a}_1&#92;mathbf{a}_1^T&#92;mathbf{y}_i - &#92;mathbf{a}_2&#92;mathbf{a}_2^T&#92;mathbf{y}_i)^T(&#92;mathbf{y}_i - &#92;mathbf{a}_1&#92;mathbf{a}_1^T&#92;mathbf{y}_i - &#92;mathbf{a}_2&#92;mathbf{a}_2^T&#92;mathbf{y}_i)' class='latex' /></p>
<p style="text-align:left;">and ask whether the following statement holds or not</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1%2C%5Cmathbf%7Bv%7D_2+%3D+%5Ctext%7Bargmax%7D_%7B%5Cmathbf%7Ba%7D_1%2C%5Cmathbf%7Ba%7D_2%7D+L_%7B%5Ctext%7Bplane%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1,&#92;mathbf{v}_2 = &#92;text{argmax}_{&#92;mathbf{a}_1,&#92;mathbf{a}_2} L_{&#92;text{plane}}' title='&#92;mathbf{v}_1,&#92;mathbf{v}_2 = &#92;text{argmax}_{&#92;mathbf{a}_1,&#92;mathbf{a}_2} L_{&#92;text{plane}}' class='latex' />with subject to <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1%5ET%5Cmathbf%7Ba%7D_1+%3D+%5Cmathbf%7Ba%7D_%7B2%7D%5ET%5Cmathbf%7Ba%7D_2+%3D+1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1^T&#92;mathbf{a}_1 = &#92;mathbf{a}_{2}^T&#92;mathbf{a}_2 = 1' title='&#92;mathbf{a}_1^T&#92;mathbf{a}_1 = &#92;mathbf{a}_{2}^T&#92;mathbf{a}_2 = 1' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1%5ET%5Cmathbf%7Ba%7D_2+%3D+0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1^T&#92;mathbf{a}_2 = 0' title='&#92;mathbf{a}_1^T&#92;mathbf{a}_2 = 0' class='latex' /></p>
<p>By Pythagoras theorem, this is indeed true. So in fact the plane spanned by <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1%2C%5Cmathbf%7Bv%7D_2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1,&#92;mathbf{v}_2' title='&#92;mathbf{v}_1,&#92;mathbf{v}_2' class='latex' /> is the optimal plane consider all <img src='http://s0.wp.com/latex.php?latex=2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='2' title='2' class='latex' />-dimensional plane.</p>
<p>This property can be generalized to the case of <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' />-dimensional hyperplane (<img src='http://s0.wp.com/latex.php?latex=q+%5Cle+p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q &#92;le p' title='q &#92;le p' class='latex' />). The optimal (in the orthogonal projection sense) hyperplane in this case is spanned by the first <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' /> eigenvectors <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1%2C%5Ccdots%2C%5Cmathbf%7Bv%7D_q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1,&#92;cdots,&#92;mathbf{v}_q' title='&#92;mathbf{v}_1,&#92;cdots,&#92;mathbf{v}_q' class='latex' /> of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7B%5COmega%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{&#92;Omega}' title='&#92;mathbf{&#92;Omega}' class='latex' />(in the order of descending eigenvalues). If we define <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7BV%7D_q+%3D+%5B%5Cmathbf%7Bv%7D_1+%5C+%5Cmathbf%7Bv%7D_2+%5C+%5Ccdots+%5C+%5Cmathbf%7Bv%7D_q%5D+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{V}_q = [&#92;mathbf{v}_1 &#92; &#92;mathbf{v}_2 &#92; &#92;cdots &#92; &#92;mathbf{v}_q] ' title='&#92;mathbf{V}_q = [&#92;mathbf{v}_1 &#92; &#92;mathbf{v}_2 &#92; &#92;cdots &#92; &#92;mathbf{v}_q] ' class='latex' /> then the projection matrix onto the optimal hyperplane is <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7BV%7D_%7Bq%7D+%5Cmathbf%7BV%7D_q%5ET&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{V}_{q} &#92;mathbf{V}_q^T' title='&#92;mathbf{V}_{q} &#92;mathbf{V}_q^T' class='latex' /> . (One can prove this generalized property by considering an arbitrary orthonormal basis <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Ba%7D_1%2C%5Ccdots%2C%5Cmathbf%7Ba%7D_q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{a}_1,&#92;cdots,&#92;mathbf{a}_q' title='&#92;mathbf{a}_1,&#92;cdots,&#92;mathbf{a}_q' class='latex' /> and the corresponding projection matrix <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7BA%7D%5Cmathbf%7BA%7D%5ET&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{A}&#92;mathbf{A}^T' title='&#92;mathbf{A}&#92;mathbf{A}^T' class='latex' /> with <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7BA%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{A}' title='&#92;mathbf{A}' class='latex' /> defined as <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7BA%7D+%3D+%5B%5Cmathbf%7Ba%7D_1%5C+%5Cmathbf%7Ba%7D_2+%5C+%5Ccdots+%5C+%5Cmathbf%7Ba%7D_q%5D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{A} = [&#92;mathbf{a}_1&#92; &#92;mathbf{a}_2 &#92; &#92;cdots &#92; &#92;mathbf{a}_q]' title='&#92;mathbf{A} = [&#92;mathbf{a}_1&#92; &#92;mathbf{a}_2 &#92; &#92;cdots &#92; &#92;mathbf{a}_q]' class='latex' />, then expanding the objective function <img src='http://s0.wp.com/latex.php?latex=L_%7B%5Ctext%7Bplane%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='L_{&#92;text{plane}}' title='L_{&#92;text{plane}}' class='latex' /> with the condition <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7BA%7D%5ET%5Cmathbf%7BA%7D+%3D+%5Cmathbf%7BI%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{A}^T&#92;mathbf{A} = &#92;mathbf{I}' title='&#92;mathbf{A}^T&#92;mathbf{A} = &#92;mathbf{I}' class='latex' />.)</p>
<p>The <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bx%7D_i+%3D+%5Cmathbf%7BV%7D_q%5Cmathbf%7BV%7D_q%5ET%5Cmathbf%7By%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{x}_i = &#92;mathbf{V}_q&#92;mathbf{V}_q^T&#92;mathbf{y}_i' title='&#92;mathbf{x}_i = &#92;mathbf{V}_q&#92;mathbf{V}_q^T&#92;mathbf{y}_i' class='latex' /> are <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' />-dimensional vectors, but are completely confined to the <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' />-dimensional subspace spanned by <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1%2C%5Ccdots%2C%5Cmathbf%7Bv%7D_q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1,&#92;cdots,&#92;mathbf{v}_q' title='&#92;mathbf{v}_1,&#92;cdots,&#92;mathbf{v}_q' class='latex' />. In other words, the &#8220;effective&#8221; dimension of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bx%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{x}' title='&#92;mathbf{x}' class='latex' /> in this case is only <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' />. For this reason, in a typical Principal Component Analysis, one is not interested in the <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bx%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{x}_i' title='&#92;mathbf{x}_i' class='latex' /> themselves, but in the coordinates of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bx%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{x}_i' title='&#92;mathbf{x}_i' class='latex' /> in the new coordinate system defined by the orthonormal basis <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1%2C%5Ccdots%2C+%5Cmathbf%7Bv%7D_q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1,&#92;cdots, &#92;mathbf{v}_q' title='&#92;mathbf{v}_1,&#92;cdots, &#92;mathbf{v}_q' class='latex' />. The coordinates of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bx%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{x}_i' title='&#92;mathbf{x}_i' class='latex' /> in this system are <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bv%7D_1%5ET%5Cmathbf%7By%7D_i%2C+%5Ccdots%2C+%5Cmathbf%7Bv%7D_q%5ET%5Cmathbf%7By%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{v}_1^T&#92;mathbf{y}_i, &#92;cdots, &#92;mathbf{v}_q^T&#92;mathbf{y}_i' title='&#92;mathbf{v}_1^T&#92;mathbf{y}_i, &#92;cdots, &#92;mathbf{v}_q^T&#92;mathbf{y}_i' class='latex' />. Therefore, further analyses after PCA often work directly with the <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' />-dimensional column vector <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bz%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{z}_i' title='&#92;mathbf{z}_i' class='latex' /> defined as</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bz%7D_i+%3D+%28%5Cmathbf%7Bv%7D_1%5ET%5Cmathbf%7By%7D_i%2C+%5Ccdots%2C%5Cmathbf%7Bv%7D_q%5ET%5Cmathbf%7By%7D_i%29%5ET&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{z}_i = (&#92;mathbf{v}_1^T&#92;mathbf{y}_i, &#92;cdots,&#92;mathbf{v}_q^T&#92;mathbf{y}_i)^T' title='&#92;mathbf{z}_i = (&#92;mathbf{v}_1^T&#92;mathbf{y}_i, &#92;cdots,&#92;mathbf{v}_q^T&#92;mathbf{y}_i)^T' class='latex' /></p>
<p>On a side note, the <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bz%7D_i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{z}_i' title='&#92;mathbf{z}_i' class='latex' /> have a diagonal sample covariance matrix (i.e. the dimensions of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bz%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{z}' title='&#92;mathbf{z}' class='latex' /> are uncorrelated). Since <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7By%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{y}' title='&#92;mathbf{y}' class='latex' /> has (sample) covariance matrix <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7B%5COmega%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{&#92;Omega}' title='&#92;mathbf{&#92;Omega}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bz%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{z}' title='&#92;mathbf{z}' class='latex' /> is a transformation of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7By%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{y}' title='&#92;mathbf{y}' class='latex' /> with <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bz%7D+%3D+%5Cmathbf%7BV%7D_q%5ET%5Cmathbf%7By%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{z} = &#92;mathbf{V}_q^T&#92;mathbf{y}' title='&#92;mathbf{z} = &#92;mathbf{V}_q^T&#92;mathbf{y}' class='latex' />, the covariance matrix of <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7Bz%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{z}' title='&#92;mathbf{z}' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=%5Cmathbf%7BV%7D_q%5ET%5Cmathbf%7B%5COmega%7D%5Cmathbf%7BV%7D_q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;mathbf{V}_q^T&#92;mathbf{&#92;Omega}&#92;mathbf{V}_q' title='&#92;mathbf{V}_q^T&#92;mathbf{&#92;Omega}&#92;mathbf{V}_q' class='latex' />, which is a diagonal matrix.</p>
<p>2. PCA as a solution of a ridge regression problem:</p>
<p>[the need for sparse PCA here]</p>
<p>The above formulation of PCA based on orthogonal projection onto optimal hyperplanes is actually a ridge regression formulation, and the final reformulation for SparsePCA will be very close to this.</p>
<p>[sparse PCA ]</p>
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		<title>RJMCMC in clustering</title>
		<link>http://thethong.wordpress.com/2011/04/24/rjmcmc-in-clustering/</link>
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		<pubDate>Sun, 24 Apr 2011 02:18:18 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Markov Chains]]></category>
		<category><![CDATA[Monte Carlo]]></category>
		<category><![CDATA[Probability]]></category>
		<category><![CDATA[RJMCMC]]></category>
		<category><![CDATA[Bayesian Variable Selection]]></category>
		<category><![CDATA[Clustering]]></category>
		<category><![CDATA[Green]]></category>
		<category><![CDATA[High-dimensional data]]></category>
		<category><![CDATA[MCMC]]></category>
		<category><![CDATA[Mixture model]]></category>
		<category><![CDATA[model selection]]></category>
		<category><![CDATA[Reversible Jump MCMC]]></category>
		<category><![CDATA[Richardson&Green]]></category>
		<category><![CDATA[Trans-dimensional MC]]></category>

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			<content:encoded><![CDATA[<p>Slide from a 30-minute presentation. There are some mistakes in the slide.</p>
<p><span id="more-737"></span></p>
<iframe src='http://www.slideshare.net/slideshow/embed_code/7718190' width='570' height='467'></iframe>
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		<title>Nonparametric Bayesian Seminar 1 : Notes</title>
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		<pubDate>Thu, 11 Nov 2010 10:43:48 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Bayesian]]></category>
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		<description><![CDATA[(mục đích chính là viết ra để khỏi quên nên sẽ  lộn xộn. Không có hình vẽ) Paper: Introduction to Nonparametric Bayesian Models (Naonori Ueda, Takeshi Yamada) 1. Generative model: gán cho quá trình tạo ra data một mô hình xác suất. Latent variable model: cho thêm các latent variable vào, tạo mô hình có [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=706&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>(mục đích chính là viết ra để khỏi quên nên sẽ  lộn xộn. Không có hình vẽ)</p>
<p>Paper: Introduction to Nonparametric Bayesian Models (Naonori Ueda, Takeshi Yamada)</p>
<p><span id="more-706"></span></p>
<p>1. Generative model: gán cho quá trình tạo ra data một mô hình xác suất.<br />
Latent variable model: cho thêm các latent variable vào, tạo mô hình có degree of freedom cao.<br />
Xét bài toán clustering thì có mô hình tiêu biểu Mixture model <img src='http://s0.wp.com/latex.php?latex=p%28x+%7C+%5Ctheta%29%3D%5Csum_%7Bi%7D+%5Cpi_%7Bi%7D+p_%7Bi%7D%28x+%7C+%5Ctheta_%7Bi%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p(x | &#92;theta)=&#92;sum_{i} &#92;pi_{i} p_{i}(x | &#92;theta_{i})' title='p(x | &#92;theta)=&#92;sum_{i} &#92;pi_{i} p_{i}(x | &#92;theta_{i})' class='latex' /> với <img src='http://s0.wp.com/latex.php?latex=%5Csum_%7Bi%7D%5Cpi_%7Bi%7D+%3D+1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sum_{i}&#92;pi_{i} = 1' title='&#92;sum_{i}&#92;pi_{i} = 1' class='latex' /> . latent variable <img src='http://s0.wp.com/latex.php?latex=z_%7Bi%7D+%3D+k&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z_{i} = k' title='z_{i} = k' class='latex' /> nếu <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> được tạo ra từ class <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k' title='k' class='latex' /> , <img src='http://s0.wp.com/latex.php?latex=k+%3D+1%2C...%2CK&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k = 1,...,K' title='k = 1,...,K' class='latex' />. <img src='http://s0.wp.com/latex.php?latex=p_%7Bi%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p_{i}' title='p_{i}' class='latex' /> thường dùng là Normal.</p>
<ul>
<li>Frequentist: Định sẵn <img src='http://s0.wp.com/latex.php?latex=K%28+%3C+%5Cinfty%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='K( &lt; &#92;infty)' title='K( &lt; &#92;infty)' class='latex' />, parameter là <img src='http://s0.wp.com/latex.php?latex=%5Ctheta+%3D+%5B%5Ctheta_%7B1%7D%2C...%2C%5Ctheta_%7BK%7D%5D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta = [&#92;theta_{1},...,&#92;theta_{K}]' title='&#92;theta = [&#92;theta_{1},...,&#92;theta_{K}]' class='latex' />. rồi dùng EM tìm maximum của <img src='http://s0.wp.com/latex.php?latex=P%28X+%7C+%5Ctheta%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P(X | &#92;theta)' title='P(X | &#92;theta)' class='latex' /></li>
</ul>
<ul>
<li>Parametric Bayesian :  Vẫn định sẵn <img src='http://s0.wp.com/latex.php?latex=K&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='K' title='K' class='latex' />.</li>
</ul>
<p>a. Định 1 prior cho <img src='http://s0.wp.com/latex.php?latex=%5Ctheta+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta ' title='&#92;theta ' class='latex' />: <img src='http://s0.wp.com/latex.php?latex=%5Ctheta_%7B%28k%29%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta_{(k)}' title='&#92;theta_{(k)}' class='latex' /> sinh ra theo phân bố <img src='http://s0.wp.com/latex.php?latex=G_%7B0%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G_{0}' title='G_{0}' class='latex' /> với<img src='http://s0.wp.com/latex.php?latex=k+%3D+1%2C..%2CK+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k = 1,..,K ' title='k = 1,..,K ' class='latex' /></p>
<p>b. Định prior cho <img src='http://s0.wp.com/latex.php?latex=%5Cpi&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi' title='&#92;pi' class='latex' /> với Dirichlet(<img src='http://s0.wp.com/latex.php?latex=%5Cpi%2C+%5Cgamma_%7B1%7D%2C...%2C%5Cgamma_%7BK%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi, &#92;gamma_{1},...,&#92;gamma_{K}' title='&#92;pi, &#92;gamma_{1},...,&#92;gamma_{K}' class='latex' />) (conjugate with multinomial). Đến phiên <img src='http://s0.wp.com/latex.php?latex=%5Cgamma&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;gamma' title='&#92;gamma' class='latex' /> thì muốn truy lên nữa cũng có thể nhưng thường thì vậy là đủ.<br />
c.Đến đây rồi thì tìm <img src='http://s0.wp.com/latex.php?latex=%5Ctheta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta' title='&#92;theta' class='latex' /> để Max <img src='http://s0.wp.com/latex.php?latex=P%28%5Ctheta+%7C+X%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P(&#92;theta | X)' title='P(&#92;theta | X)' class='latex' /> (cũng bằng EM?)<br />
2.<strong> Nonparametric  Bayesian model</strong>: mô hình <em>Dirichlet Process Mixture</em> (DPM) làm việc với <img src='http://s0.wp.com/latex.php?latex=K+%3D+%5Cinfty&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='K = &#92;infty' title='K = &#92;infty' class='latex' /> class. Nói chung thì cũng giống như ở parametric bayesian model: sản sinh vô hạn <img src='http://s0.wp.com/latex.php?latex=%5Ctheta_%7B%281%29%7D%2C...%2C%5Ctheta_%7B%28k%29%7D%2C...&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta_{(1)},...,&#92;theta_{(k)},...' title='&#92;theta_{(1)},...,&#92;theta_{(k)},...' class='latex' /> , <img src='http://s0.wp.com/latex.php?latex=k+%3D+1%2C...%2C%5Cinfty&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k = 1,...,&#92;infty' title='k = 1,...,&#92;infty' class='latex' /> theo phân bố nền tảng <img src='http://s0.wp.com/latex.php?latex=G_%7B0%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G_{0}' title='G_{0}' class='latex' />, sản sinh <img src='http://s0.wp.com/latex.php?latex=%5Cpi_%7B1%7D%2C...%2C%5Cpi_%7Bk%7D%2C...&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi_{1},...,&#92;pi_{k},...' title='&#92;pi_{1},...,&#92;pi_{k},...' class='latex' /> sao cho <img src='http://s0.wp.com/latex.php?latex=%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+%5Cpi_%7Bk%7D%3D+1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sum_{k=1}^{&#92;infty} &#92;pi_{k}= 1' title='&#92;sum_{k=1}^{&#92;infty} &#92;pi_{k}= 1' class='latex' />.  Khi đó <img src='http://s0.wp.com/latex.php?latex=%5Ctheta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta' title='&#92;theta' class='latex' /> sẽ có phân bố rời rạc <img src='http://s0.wp.com/latex.php?latex=G%28%5Ctheta%29+%3D+%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cpi_%7Bk%7D%5Cdelta_%7B%5Ctheta%28k%29%7D%28%5Ctheta%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G(&#92;theta) = &#92;sum_{k=1}^{&#92;infty}&#92;pi_{k}&#92;delta_{&#92;theta(k)}(&#92;theta)' title='G(&#92;theta) = &#92;sum_{k=1}^{&#92;infty}&#92;pi_{k}&#92;delta_{&#92;theta(k)}(&#92;theta)' class='latex' /> với <img src='http://s0.wp.com/latex.php?latex=%5Cdelta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;delta' title='&#92;delta' class='latex' /> là Krocneker Delta.<br />
Khoan nói đến sản sinh <img src='http://s0.wp.com/latex.php?latex=%5Ctheta_%7B%28k%29%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta_{(k)}' title='&#92;theta_{(k)}' class='latex' /> từ <img src='http://s0.wp.com/latex.php?latex=G_%7B0%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G_{0}' title='G_{0}' class='latex' /> , xem trước cách chọn(tạo ra) <img src='http://s0.wp.com/latex.php?latex=%5Cpi_%7Bk%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi_{k}' title='&#92;pi_{k}' class='latex' />, là xác suất mà một data <img src='http://s0.wp.com/latex.php?latex=x_%7Bi%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x_{i}' title='x_{i}' class='latex' /> sẽ thuộc class <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k' title='k' class='latex' /></p>
<p>a.<strong> Stick-Breaking Process(SBP)</strong>: <img src='http://s0.wp.com/latex.php?latex=%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+%3D1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sum_{k=1}^{&#92;infty} =1' title='&#92;sum_{k=1}^{&#92;infty} =1' class='latex' /> nên có nhiều cách chia, 1 trong những cách đó là SBP. Tưởng tượng ban đầu có thanh gỗ có độ dài 1. SBP sẽ bẻ dần dần <img src='http://s0.wp.com/latex.php?latex=%5Cpi_%7B1%7D+%2C+%5Cpi_%7B2%7D+%2C+...+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi_{1} , &#92;pi_{2} , ... ' title='&#92;pi_{1} , &#92;pi_{2} , ... ' class='latex' /> từ thanh gỗ này.</p>
<ul>
<li>generate <img src='http://s0.wp.com/latex.php?latex=ratio&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='ratio' title='ratio' class='latex' /> theo Beta<img src='http://s0.wp.com/latex.php?latex=%281%2C%5Cgamma%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(1,&#92;gamma)' title='(1,&#92;gamma)' class='latex' />. <img src='http://s0.wp.com/latex.php?latex=ratio&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='ratio' title='ratio' class='latex' /> sẽ thuộc <img src='http://s0.wp.com/latex.php?latex=%5B0%2C1%5D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='[0,1]' title='[0,1]' class='latex' /></li>
<li>Bẻ <img src='http://s0.wp.com/latex.php?latex=%5Cpi_%7Bi%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi_{i}' title='&#92;pi_{i}' class='latex' /> có độ dài bằng <img src='http://s0.wp.com/latex.php?latex=ratio&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='ratio' title='ratio' class='latex' /> * Độ dài còn lại của thanh gỗ sau lần bẻ <img src='http://s0.wp.com/latex.php?latex=i+-+1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i - 1' title='i - 1' class='latex' />.</li>
</ul>
<p><span style="text-decoration:underline;">Tính chất</span>: <img src='http://s0.wp.com/latex.php?latex=%5Cgamma&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;gamma' title='&#92;gamma' class='latex' /> nhỏ <img src='http://s0.wp.com/latex.php?latex=ratio&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='ratio' title='ratio' class='latex' /> lớn và ngược lại<br />
<img src='http://s0.wp.com/latex.php?latex=E%5B%5Cpi_%7Bk%7D%5D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='E[&#92;pi_{k}]' title='E[&#92;pi_{k}]' class='latex' /> giảm exponentially theo <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k' title='k' class='latex' /> : lúc đầu thanh gỗ dài còn dễ bẻ, càng về sau càng ngắn càng khó bẻ.<br />
Dùng SBP thì  thấy rõ normalizing condition được thỏa,  nhưng phải tạo đủ <img src='http://s0.wp.com/latex.php?latex=%5Ctheta_%7B1%7D%2C%5Ctheta_%7B2%7D%2C%5Ctheta_%7Bk%7D%2C...%2C%5Ctheta_%7B%5Cinfty%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta_{1},&#92;theta_{2},&#92;theta_{k},...,&#92;theta_{&#92;infty}' title='&#92;theta_{1},&#92;theta_{2},&#92;theta_{k},...,&#92;theta_{&#92;infty}' class='latex' /> từ đầu (?).</p>
<p>b.  <strong>Chinese Restaurant Process(CRP)</strong>:</p>
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		<title>Flows: the psychology of optimal experience</title>
		<link>http://thethong.wordpress.com/2010/11/07/flows-the-psychology-of-optimal-experience/</link>
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		<pubDate>Sun, 07 Nov 2010 12:00:55 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Books]]></category>

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		<description><![CDATA[There is no inherent problem in our desire to escalate our goals, as long as we enjoy the struggle along the way. The problem arises when people are so fixated on what they want to achieve that they cease to derive pleasure from the present.<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=703&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><span id="more-703"></span></p>
<blockquote><p>There is no inherent problem in our desire to escalate our goals, as long as we enjoy the struggle along the way. The problem arises when people are so fixated on what they want to achieve that they cease to derive pleasure from the present.</p></blockquote>
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		<title>Ising model</title>
		<link>http://thethong.wordpress.com/2010/08/17/ising-model/</link>
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		<pubDate>Tue, 17 Aug 2010 12:22:44 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Monte Carlo]]></category>

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		<description><![CDATA[Boltzman distribution: Partition function (also called normalizing constant) Potential energy The symbol means that they are neighboring pair, is the interaction strength, is the external magnetic field. In zero field we can also define the potential energy as We define the expectation of with respect to as internal energy: The free energy of the system [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=666&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Boltzman distribution: <img src='http://s0.wp.com/latex.php?latex=%5Cpi+%28%5Ctext%7Bx%7D%29+%3D+%5Cfrac%7B1%7D%7BZ%7De%5E%7B-U%28%5Ctext%7Bx%7D%29%2FkT%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi (&#92;text{x}) = &#92;frac{1}{Z}e^{-U(&#92;text{x})/kT}' title='&#92;pi (&#92;text{x}) = &#92;frac{1}{Z}e^{-U(&#92;text{x})/kT}' class='latex' /></p>
<p><span id="more-666"></span></p>
<p><em>Partition function</em> <img src='http://s0.wp.com/latex.php?latex=Z%3DZ%28T%29+%3D+%5Cint_D+e%5E%7B-U%28%5Ctext%7Bx%7D%29%2FkT%7Dd%5Ctext%7Bx%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='Z=Z(T) = &#92;int_D e^{-U(&#92;text{x})/kT}d&#92;text{x}' title='Z=Z(T) = &#92;int_D e^{-U(&#92;text{x})/kT}d&#92;text{x}' class='latex' /> (also called normalizing constant)</p>
<p>Potential energy <img src='http://s0.wp.com/latex.php?latex=U%28%5Ctext%7Bx%7D%29+%3D-J%5Csum_%7B%5Csigma+%5Cthicksim+%5Csigma%5E%7B%27%7D%7D+x_%5Csigma+x_%7B%5Csigma%5E%7B%27%7D%7D+%2B+%5Csum_%5Csigma+h_%5Csigma+x_%5Csigma&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='U(&#92;text{x}) =-J&#92;sum_{&#92;sigma &#92;thicksim &#92;sigma^{&#039;}} x_&#92;sigma x_{&#92;sigma^{&#039;}} + &#92;sum_&#92;sigma h_&#92;sigma x_&#92;sigma' title='U(&#92;text{x}) =-J&#92;sum_{&#92;sigma &#92;thicksim &#92;sigma^{&#039;}} x_&#92;sigma x_{&#92;sigma^{&#039;}} + &#92;sum_&#92;sigma h_&#92;sigma x_&#92;sigma' class='latex' /></p>
<p>The symbol <img src='http://s0.wp.com/latex.php?latex=%5Csigma+%5Cthicksim+%5Csigma%5E%7B%27%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sigma &#92;thicksim &#92;sigma^{&#039;}' title='&#92;sigma &#92;thicksim &#92;sigma^{&#039;}' class='latex' /> means that they are neighboring pair, <img src='http://s0.wp.com/latex.php?latex=J&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='J' title='J' class='latex' /> is the interaction strength, <img src='http://s0.wp.com/latex.php?latex=h_%5Csigma&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='h_&#92;sigma' title='h_&#92;sigma' class='latex' /> is the external magnetic field.</p>
<p>In zero field we can also define the potential energy as</p>
<p><img src='http://s0.wp.com/latex.php?latex=U%28%5Ctext%7Bx%7D%29%3D%5Csum_%7B%5Csigma+%5Cthicksim+%5Csigma%5E%7B%27%7D%7D+1_%7Bx_%5Csigma+%5Cne+x_%7B%5Csigma%5E%7B%27%7D%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='U(&#92;text{x})=&#92;sum_{&#92;sigma &#92;thicksim &#92;sigma^{&#039;}} 1_{x_&#92;sigma &#92;ne x_{&#92;sigma^{&#039;}}}' title='U(&#92;text{x})=&#92;sum_{&#92;sigma &#92;thicksim &#92;sigma^{&#039;}} 1_{x_&#92;sigma &#92;ne x_{&#92;sigma^{&#039;}}}' class='latex' /></p>
<p>We define the expectation of <img src='http://s0.wp.com/latex.php?latex=U%28%5Ctext%7Bx%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='U(&#92;text{x})' title='U(&#92;text{x})' class='latex' /> with respect to <img src='http://s0.wp.com/latex.php?latex=%5Cpi&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi' title='&#92;pi' class='latex' /> as <em>internal energy</em>:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%3CU%3E%3D+E_%5Cpi+%5C%7BU%28%5Ctext%7Bx%7D%29%5C%7D+%3D+%5Cint_D+U%28%5Ctext%7Bx%7D%29%5Cpi%28%5Ctext%7Bx%7D%29d%5Ctext%7Bx%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&lt;U&gt;= E_&#92;pi &#92;{U(&#92;text{x})&#92;} = &#92;int_D U(&#92;text{x})&#92;pi(&#92;text{x})d&#92;text{x}' title='&lt;U&gt;= E_&#92;pi &#92;{U(&#92;text{x})&#92;} = &#92;int_D U(&#92;text{x})&#92;pi(&#92;text{x})d&#92;text{x}' class='latex' /></p>
<p>The<em> free energy</em> of the system is <img src='http://s0.wp.com/latex.php?latex=F+%3D+-kT%5Ctext%7Blog%7DZ&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='F = -kT&#92;text{log}Z' title='F = -kT&#92;text{log}Z' class='latex' /></p>
<p>The <em>specific heat</em> of the system is <img src='http://s0.wp.com/latex.php?latex=C+%3D+%5Cfrac%7B%5Cpartial+%3CU%3E%7D%7B%5Cpartial+T%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='C = &#92;frac{&#92;partial &lt;U&gt;}{&#92;partial T}' title='C = &#92;frac{&#92;partial &lt;U&gt;}{&#92;partial T}' class='latex' /></p>
<p>The <em>mean magnetization per spin</em> is <img src='http://s0.wp.com/latex.php?latex=%3Cm%3E+%3D+E_%5Cpi+%5C%7B%5Cfrac%7B1%7D%7BN%5E2%7D+%7C%5Csum_%7B%5Csigma+%5Cin+S%7Dx_%5Csigma%7C%5C%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&lt;m&gt; = E_&#92;pi &#92;{&#92;frac{1}{N^2} |&#92;sum_{&#92;sigma &#92;in S}x_&#92;sigma|&#92;}' title='&lt;m&gt; = E_&#92;pi &#92;{&#92;frac{1}{N^2} |&#92;sum_{&#92;sigma &#92;in S}x_&#92;sigma|&#92;}' class='latex' /></p>
<p>1.</p>
<p>Let <img src='http://s0.wp.com/latex.php?latex=%5Cbeta+%3D+1%2FkT&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;beta = 1/kT' title='&#92;beta = 1/kT' class='latex' />, we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cpartial+%5Ctext%7Blog%7DZ+%2F%5Cpartial+%5Cbeta+%3D+%5Cfrac%7B1%7D%7BZ%7D%5Cpartial+Z%2F%5Cpartial+%5Cbeta+%3D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;partial &#92;text{log}Z /&#92;partial &#92;beta = &#92;frac{1}{Z}&#92;partial Z/&#92;partial &#92;beta =' title='&#92;partial &#92;text{log}Z /&#92;partial &#92;beta = &#92;frac{1}{Z}&#92;partial Z/&#92;partial &#92;beta =' class='latex' /> <img src='http://s0.wp.com/latex.php?latex=%5Cfrac%7B1%7D%7BZ%7D%5Cpartial%5B%5Cint_D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;frac{1}{Z}&#92;partial[&#92;int_D' title='&#92;frac{1}{Z}&#92;partial[&#92;int_D' class='latex' /> <img src='http://s0.wp.com/latex.php?latex=e%5E%7B-%5Cbeta+U%28%5Ctext%7Bx%7D%29%7Dd%5Ctext%7Bx%7D%5D%2F+%5Cpartial%5Cbeta+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='e^{-&#92;beta U(&#92;text{x})}d&#92;text{x}]/ &#92;partial&#92;beta ' title='e^{-&#92;beta U(&#92;text{x})}d&#92;text{x}]/ &#92;partial&#92;beta ' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D%5Cfrac%7B1%7D%7BZ%7D%5Cint_D+%5C%7B%5Cpartial%5Be%5E%7B-%5Cbeta+U%28%5Ctext%7Bx%7D%29+%7D%5D+%2F+%5Cpartial+%5Cbeta%5C%7Dd%7B%5Ctext%7Bx%7D%7D+%3D%5Cfrac%7B1%7D%7BZ%7D%5Cint_D-U%28%5Ctext%7Bx%7D%29+e%5E%7B-%5Cbeta+U%28%5Ctext%7Bx%7D%29+%7D+d%7B%5Ctext%7Bx%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='=&#92;frac{1}{Z}&#92;int_D &#92;{&#92;partial[e^{-&#92;beta U(&#92;text{x}) }] / &#92;partial &#92;beta&#92;}d{&#92;text{x}} =&#92;frac{1}{Z}&#92;int_D-U(&#92;text{x}) e^{-&#92;beta U(&#92;text{x}) } d{&#92;text{x}}' title='=&#92;frac{1}{Z}&#92;int_D &#92;{&#92;partial[e^{-&#92;beta U(&#92;text{x}) }] / &#92;partial &#92;beta&#92;}d{&#92;text{x}} =&#92;frac{1}{Z}&#92;int_D-U(&#92;text{x}) e^{-&#92;beta U(&#92;text{x}) } d{&#92;text{x}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D%5Cfrac%7B1%7D%7BZ%7D%5Cint_D+%5Cfrac%7B%5Cpartial+e%5E%7B-%5Cbeta+U%28%5Ctext%7Bx%7D%29+%7D%7D+%7B%5Cpartial+%5Cbeta%7D+d%7B%5Ctext%7Bx%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='=&#92;frac{1}{Z}&#92;int_D &#92;frac{&#92;partial e^{-&#92;beta U(&#92;text{x}) }} {&#92;partial &#92;beta} d{&#92;text{x}}' title='=&#92;frac{1}{Z}&#92;int_D &#92;frac{&#92;partial e^{-&#92;beta U(&#92;text{x}) }} {&#92;partial &#92;beta} d{&#92;text{x}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+-%3CU%3E&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='= -&lt;U&gt;' title='= -&lt;U&gt;' class='latex' /></p>
<p>Thus <img src='http://s0.wp.com/latex.php?latex=%5Cfrac%7B%5Cpartial+%5Ctext%7Blog%7DZ%7D%7B%5Cpartial+%5Cbeta%7D+%3D+-%3CU%3E&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;frac{&#92;partial &#92;text{log}Z}{&#92;partial &#92;beta} = -&lt;U&gt;' title='&#92;frac{&#92;partial &#92;text{log}Z}{&#92;partial &#92;beta} = -&lt;U&gt;' class='latex' /></p>
<p>2. <img src='http://s0.wp.com/latex.php?latex=C+%3D+%5Cfrac%7B%5Cpartial+%3CU%3E%7D%7B%5Cpartial+T%7D+%3D+%5Cfrac%7B1%7D%7BkT%5E2%7DVar_%5Cpi%5C%7BU%28%5Ctext%7Bx%7D%29%5C%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='C = &#92;frac{&#92;partial &lt;U&gt;}{&#92;partial T} = &#92;frac{1}{kT^2}Var_&#92;pi&#92;{U(&#92;text{x})&#92;}' title='C = &#92;frac{&#92;partial &lt;U&gt;}{&#92;partial T} = &#92;frac{1}{kT^2}Var_&#92;pi&#92;{U(&#92;text{x})&#92;}' class='latex' /></p>
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		<title>Q</title>
		<link>http://thethong.wordpress.com/2010/05/05/critical-questions/</link>
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		<pubDate>Wed, 05 May 2010 06:42:45 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Books]]></category>
		<category><![CDATA[Write up]]></category>
		<category><![CDATA[Avoid boring people and other lessons from a life in science]]></category>

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		<description><![CDATA[(These are some thoughts I got while reading the inspiring book by James D.Watson &#8220;Avoid boring people and other lessons from a life in science.&#8221; Maybe I will give a full summary on the lessons from the book later.) Before performing an experiment, ask yourself (or your instructor, if you dare) these questions: + What [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=659&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>(These are some thoughts I got while reading the inspiring book by James D.Watson &#8220;<a href="http://www.amazon.co.jp/Avoid-Boring-People-Lessons-Science/dp/0192802739/">Avoid boring people and other lessons from a life in science</a>.&#8221; Maybe I will give a full summary on the lessons from the book later.)</p>
<p><span id="more-659"></span></p>
<p>Before performing an experiment, ask yourself (or your instructor, if you dare) these questions:</p>
<p>+ What is this experiment all about? What are its objectives?<br />
+ Are these objectives really worth achieving? Why do you think so?<br />
+ By what theory/reason that we should believe the experiment will really fulfill the objectives? Are there any reasons to fear it will <strong>not</strong> fulfill the objectives?<br />
+ Can the objectives be achieved by other methods?<br />
If so, what is the best method (in the point of highest probability to achieve the full objectives), and why are we choosing <strong>this</strong> method?<br />
If not, what is that made the objectives so hard to be achieved, and why is <strong>this</strong> method can overcome the obstacles while others can&#8217;t?</p>
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		<title>Mathematics and the Unexpected</title>
		<link>http://thethong.wordpress.com/2010/04/24/mathematics-and-the-unexpected/</link>
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		<pubDate>Fri, 23 Apr 2010 15:09:23 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Books]]></category>
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		<description><![CDATA[(Just some thoughts while I read Mathematics and the Unexpected and Innumeracy: Mathematical Illiteracy and its consequence. Maybe some more detailed post later.) Nature is too complicated for us to model with finite equations. All mathematical models are wrong. Do not expect to find (new) simple mathematical models which are useful in real world problems. [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=652&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>(Just some thoughts while I read <a href="http://www.amazon.co.jp/Mathematics-Unexpected-Ivar-Ekeland/dp/0226199908/">Mathematics and the Unexpected</a> and <a href="Mathematical Illiteracy and Its Consequences">Innumeracy: Mathematical Illiteracy and its consequence</a>. Maybe some more detailed post later.)</p>
<p><span id="more-652"></span></p>
<p>Nature is too complicated for us to model with finite equations. All mathematical models are wrong.<br />
Do not expect to find (new) simple mathematical models which are useful in real world problems. If they do exist, the chances are that they had been discovered, or they are not so useful.<br />
Do not expect to find relations (simple or complicated) in the small scale , or in a quantitative manner. The only choice is to find relations in the whole picture: a large scale relation in a qualitative manner is more feasible.<br />
Do not cook up mathematical models for the problem without testing it with a large amount of data, or having a good theory to explain why it must be like that. Finding out meaning among seemingly meaningless data is surely a big deal, but to resist the temptation of cooking up an artificial meaning is no less important.</p>
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		<title>Probability and Computing: Chapter 7 Exercises</title>
		<link>http://thethong.wordpress.com/2009/11/28/probability-and-computing-chapter-7-exercises/</link>
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		<pubDate>Sat, 28 Nov 2009 09:12:22 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Markov Chains]]></category>
		<category><![CDATA[Probability]]></category>
		<category><![CDATA[Computer Science]]></category>
		<category><![CDATA[Maths]]></category>

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		<description><![CDATA[Exercise 7.12: Let be the sum of independent rolls of a fair dice. Show that, for any , . Solution: Let be a Markov chain on state space consist of states: , where the chain reaches state if and only if . The transition matrix is for and . The claim is equivalent to We [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=591&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><strong>Exercise 7.12: </strong>Let <img src='http://s0.wp.com/latex.php?latex=X_%7Bn%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X_{n}' title='X_{n}' class='latex' /> be the sum of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> independent rolls of a fair dice. Show that, for any <img src='http://s0.wp.com/latex.php?latex=k+%3E+2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k &gt; 2' title='k &gt; 2' class='latex' />, <img src='http://s0.wp.com/latex.php?latex=%5Clim_%7Bn+%5Crightarrow+%5Cinfty%7D%28X_%7Bn%7D+%5Ctext%7Bis+divisible+by+k%7D%29+%3D+%5Cfrac%7B1%7D%7Bk%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;lim_{n &#92;rightarrow &#92;infty}(X_{n} &#92;text{is divisible by k}) = &#92;frac{1}{k}' title='&#92;lim_{n &#92;rightarrow &#92;infty}(X_{n} &#92;text{is divisible by k}) = &#92;frac{1}{k}' class='latex' />.</p>
<p><span id="more-591"></span></p>
<p><strong>Solution: </strong>Let <img src='http://s0.wp.com/latex.php?latex=%28Y_%7Bn%7D%29_%7Bn+%5Cge+0%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(Y_{n})_{n &#92;ge 0}' title='(Y_{n})_{n &#92;ge 0}' class='latex' /> be a Markov chain on state space <img src='http://s0.wp.com/latex.php?latex=I&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I' title='I' class='latex' /> consist of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k' title='k' class='latex' /> states: <img src='http://s0.wp.com/latex.php?latex=0%2C1%2C...%2C+k+-1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='0,1,..., k -1' title='0,1,..., k -1' class='latex' />, where the chain reaches state <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i' title='i' class='latex' /> if and only if <img src='http://s0.wp.com/latex.php?latex=X_%7Bn%7D%5Cequiv+i+%5Cpmod%7Bk%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X_{n}&#92;equiv i &#92;pmod{k}' title='X_{n}&#92;equiv i &#92;pmod{k}' class='latex' />.<br />
The transition matrix <img src='http://s0.wp.com/latex.php?latex=P&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P' title='P' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=p_%7Bi%2C%28i+%2B+j%29+%5Cbmod+k%7D+%3D+%5Cfrac%7B1%7D%7B6%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p_{i,(i + j) &#92;bmod k} = &#92;frac{1}{6}' title='p_{i,(i + j) &#92;bmod k} = &#92;frac{1}{6}' class='latex' /> for <img src='http://s0.wp.com/latex.php?latex=0+%5Cle+i+%5Cle+k-1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='0 &#92;le i &#92;le k-1' title='0 &#92;le i &#92;le k-1' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=1+%5Cle+j+%5Cle+6&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1 &#92;le j &#92;le 6' title='1 &#92;le j &#92;le 6' class='latex' />. The claim is equivalent to</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Clim_%7Bn+%5Crightarrow+%5Cinfty%7D%28Y_%7Bn%7D+%3D+0%29+%3D+%5Cfrac%7B1%7D%7Bk%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;lim_{n &#92;rightarrow &#92;infty}(Y_{n} = 0) = &#92;frac{1}{k}' title='&#92;lim_{n &#92;rightarrow &#92;infty}(Y_{n} = 0) = &#92;frac{1}{k}' class='latex' /></p>
<p>We know that if a Markov chain <img src='http://s0.wp.com/latex.php?latex=%28Y_%7Bn%7D%29_%7Bn+%5Cge+0%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(Y_{n})_{n &#92;ge 0}' title='(Y_{n})_{n &#92;ge 0}' class='latex' /> has an irreducible, aperiodic transition matrix and an invariant distribution <img src='http://s0.wp.com/latex.php?latex=%5Cpi&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi' title='&#92;pi' class='latex' />, then <img src='http://s0.wp.com/latex.php?latex=%5Clim_%7Bn+%5Crightarrow+%5Cinfty%7D%28Y_%7Bn%7D+%3D+j%29+%3D+%5Cpi_%7Bj%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;lim_{n &#92;rightarrow &#92;infty}(Y_{n} = j) = &#92;pi_{j}' title='&#92;lim_{n &#92;rightarrow &#92;infty}(Y_{n} = j) = &#92;pi_{j}' class='latex' /> for all <img src='http://s0.wp.com/latex.php?latex=j&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='j' title='j' class='latex' /> of the state space.</p>
<p>We will prove the defined above <img src='http://s0.wp.com/latex.php?latex=P&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P' title='P' class='latex' /> is irreducible, aperiodic and then find the invariant distribution <img src='http://s0.wp.com/latex.php?latex=%5Cpi&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi' title='&#92;pi' class='latex' /> of <img src='http://s0.wp.com/latex.php?latex=P&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P' title='P' class='latex' />(it will turn out that <img src='http://s0.wp.com/latex.php?latex=%5Cpi_%7B0%7D+%3D+%5Cfrac%7B1%7D%7Bk%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi_{0} = &#92;frac{1}{k}' title='&#92;pi_{0} = &#92;frac{1}{k}' class='latex' /> as we need).</p>
<p><strong>Exercise 7.21:</strong> Consider a Markov chain on the states <img src='http://s0.wp.com/latex.php?latex=%5C%7B+0%2C1%2C%5Cldots+%2C+n%5C%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;{ 0,1,&#92;ldots , n&#92;}' title='&#92;{ 0,1,&#92;ldots , n&#92;}' class='latex' />, where for <img src='http://s0.wp.com/latex.php?latex=i+%3C+n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i &lt; n' title='i &lt; n' class='latex' /> we have <img src='http://s0.wp.com/latex.php?latex=P_%7Bi%2Ci%2B1%7D+%3D+%5Cfrac%7B1%7D%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P_{i,i+1} = &#92;frac{1}{2}' title='P_{i,i+1} = &#92;frac{1}{2}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=P_%7Bi%2C0%7D+%3D+%5Cfrac%7B1%7D%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P_{i,0} = &#92;frac{1}{2}' title='P_{i,0} = &#92;frac{1}{2}' class='latex' />. Also, <img src='http://s0.wp.com/latex.php?latex=P_%7Bn%2Cn%7D+%3D+%5Cfrac%7B1%7D%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P_{n,n} = &#92;frac{1}{2}' title='P_{n,n} = &#92;frac{1}{2}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=P_%7Bn%2C0%7D+%3D+%5Cfrac%7B1%7D%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='P_{n,0} = &#92;frac{1}{2}' title='P_{n,0} = &#92;frac{1}{2}' class='latex' />. This process can be viewed as a random walk on a directed graph with vertices <img src='http://s0.wp.com/latex.php?latex=%5C%7B+0%2C1%2C+%5Cldots+%2Cn%5C%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;{ 0,1, &#92;ldots ,n&#92;}' title='&#92;{ 0,1, &#92;ldots ,n&#92;}' class='latex' />, where each vertex has two directed edges: one that returns to <img src='http://s0.wp.com/latex.php?latex=0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='0' title='0' class='latex' /> and one that moves to the vertex with the next higher number(with a self-loop at vertex <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' />). Find the stationary distribution of this chain.(This example shows that random walks on directed graphs are very different than random walks on undirected graph).</p>
<p><strong>Solution: </strong></p>
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		<title>An exercise about Markov Chains</title>
		<link>http://thethong.wordpress.com/2009/11/05/an-exercise-about-markov-chains/</link>
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		<pubDate>Thu, 05 Nov 2009 13:33:28 +0000</pubDate>
		<dc:creator>thethong</dc:creator>
				<category><![CDATA[Markov Chains]]></category>
		<category><![CDATA[Probability]]></category>
		<category><![CDATA[Maths]]></category>

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		<description><![CDATA[Lately  I have stopped reading &#8220;Probability and computing&#8221;, since I found some gaps in the exposition of the text, especially at chapter 7 &#8220;Markov Chains and Random Walks&#8221;-the authors left undefined some terminologies such as absorption. Certainly this is not the text for anybody who has little background of probability and want to learn it [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=thethong.wordpress.com&amp;blog=9253470&amp;post=482&amp;subd=thethong&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Lately  I have stopped reading &#8220;Probability and computing&#8221;, since I found some gaps in the exposition of the text, especially at chapter 7 &#8220;Markov Chains and Random Walks&#8221;-the authors left undefined some terminologies such as absorption. Certainly this is not the text for anybody who has little background of probability and want to learn it rigorously (though it is a good introductory text for randomized algorithm). So I bought &#8220;Markov Chains&#8221; of James Norris.<br />
Exercises in &#8220;Markov Chains&#8221; are easy (at least for the first chapter ), though there are some problems that I am not quite sure. Here is one of them:</p>
<p><span id="more-482"></span></p>
<p>A particle moves on the eight vertices of a cube in the following way: at each step the particle is equally likely to move to each of the three adjacent vertices, independently of it past motion. Let <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i' title='i' class='latex' /> be the initial vertex occupied by the particle, <img src='http://s0.wp.com/latex.php?latex=o&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='o' title='o' class='latex' /> the vertex oppostite <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i' title='i' class='latex' />. Calculate each of the following quantities:<strong><br />
(i) </strong>the expected number of steps until the particle returns to <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i' title='i' class='latex' />;<br />
<strong>(ii) </strong>the expected number of visits to <img src='http://s0.wp.com/latex.php?latex=o&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='o' title='o' class='latex' /> until the first return to <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i' title='i' class='latex' />;<br />
<strong>(iii) </strong>the expected number of steps until the first visit to <img src='http://s0.wp.com/latex.php?latex=o&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='o' title='o' class='latex' />.</p>
<p>This exercise is placed under &#8220;Invariant distribution&#8221; section. My answer is <strong>(i)</strong> 8, <strong>(ii) </strong>1, <strong>(iii)</strong> 16 (?). I did not use anything from this section to solve <strong>(iii) </strong>and the answer also is suspicious, at least counter-intuitive to me. So I would like to know if there is any connection between the expected number of steps until the first visit to <img src='http://s0.wp.com/latex.php?latex=o&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='o' title='o' class='latex' />, <img src='http://s0.wp.com/latex.php?latex=E_%7Bi%7D%5BT_%7Bo%7D%5D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='E_{i}[T_{o}]' title='E_{i}[T_{o}]' class='latex' />, and the invariant distribution <img src='http://s0.wp.com/latex.php?latex=%5Cpi&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi' title='&#92;pi' class='latex' />.</p>
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