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218 lines
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HTML
218 lines
7.6 KiB
HTML
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<!DOCTYPE html>
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<html>
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<meta name="author" content="Fredrik Danielsson, http://lostkeys.se">
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</aside>
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<main class="main-content">
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<article class="article">
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<h1>Principal Component Analysis Part 2: Formal Theory</h1>
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<h2>Properties of PCA</h2>
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<p>There are a number of ways to maximize the variance of a principal component. To create an unique solution we should impose a constraint. Let us say that the sum of the square of the coefficients must equal 1. In vector notation this is the same as
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$$
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a_i^Ta_i = 1
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$$
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Every future principal component is said to be orthogonal to all the principal components previous to it.
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$$
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a_j^Ta<em>i = 0, i < j
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$$
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The total variance of the $q$ principal components will equal the total variance of the original variables
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$$
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\sum</em>{i = 1}^q {\lambda_i} = trace(S)
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$$
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Where $S$ is the sample covariance matrix.</p>
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<p>The proportion of accounted variation in each principle component is
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$$
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P_j = \frac{\lambda<em>j}{trace(S)}
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$$
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From this, we can generalize to the first $m$ principal components where $m < q$ and find the proportion $P^{(m)}$ of variation accounted for
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$$
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P^{(m)} = \frac{\sum</em>{i = 1}^m{\lambda_i}}{trace(S)}
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$$
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You can think of the first principal component as the line of best fit that minimizes the residuals orthogonal to it.</p>
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<h3>What to watch out for</h3>
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<p>As a reminder to the last lecture, <em>PCA is not scale-invariant</em>. Therefore, transformations done to the dataset before PCA and after PCA often lead to different results and possibly conclusions.</p>
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<p>Additionally, if there are large differences between the variances of the original variables, then those whose variances are largest will tend to dominate the early components.</p>
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<p>Therefore, principal components should only be extracted from the sample covariance matrix when all of the original variables have roughly the <strong>same scale</strong>.</p>
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<h3>Alternatives to using the Covariance Matrix</h3>
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<p>But it is rare in practice to have a scenario when all of the variables are of the same scale. Therefore, principal components are typically extracted from the <strong>correlation matrix</strong> $R$</p>
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<p>Choosing to work with the correlation matrix rather than the covariance matrix treats the variables as all equally important when performing PCA.</p>
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<h2>Example Derivation: Bivariate Data</h2>
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<p>Let $R$ be the correlation matrix
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$$
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R = \begin{pmatrix}
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1 & r \
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r & 1
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\end{pmatrix}
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$$
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Let us find the eigenvectors and eigenvalues of the correlation matrix
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$$
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det(R - \lambda I) = 0
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$$</p>
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<p>$$
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(1-\lambda)^2 - r^2 = 0
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$$</p>
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<p>$$
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\lambda_1 = 1 + r, \lambda_2 = 1 - r
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$$</p>
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<p>Let us remember to check the condition "sum of the principal components equals the trace of the correlation matrix":
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$$
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\lambda_1 + \lambda_2 = 1+r + (1 - r) = 2 = trace(R)
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$$</p>
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<h3>Finding the First Eigenvector</h3>
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<p>Looking back at the characteristic equation
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$$
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Ra_1 = \lambda a<em>1
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$$
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We can get the following two formulas
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$$
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a</em>{11} + ra<em>{12} = (1+r)a</em>{11} \tag{1}
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$$</p>
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<p>$$
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ra<em>{11} + a</em>{12} = (1 + r)a_{12} \tag{2}
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$$</p>
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<p>Now let us find out what $a<em>{11}$ and $a</em>{12}$ equal. First let us solve for $a<em>{11}$ using equation $(1)$
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$$
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ra</em>{12} = (1+r)a<em>{11} - a</em>{11}
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$$</p>
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<p>$$
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ra<em>{12} = a</em>{11}(1 + r - 1)
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$$</p>
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<p>$$
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ra<em>{12} = ra</em>{11}
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$$</p>
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<p>$$
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a<em>{12} = a</em>{11}
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$$</p>
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<p>Where $r$ does not equal $0$.</p>
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<p>Now we must apply the condition of sum squares
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$$
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a_1^Ta_1 = 1
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$$</p>
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<p>$$
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a<em>{11}^2 + a</em>{12}^2 = 1
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$$</p>
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<p>Recall that $a<em>{12} = a</em>{11}$
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$$
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2a_{11}^2 = 1
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$$</p>
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<p>$$
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a_{11}^2 = \frac{1}{2}
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$$</p>
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<p>$$
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a_{11} =\pm \frac{1}{\sqrt{2}}
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$$</p>
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<p>For sake of choosing a value, let us take the principal root and say $a_{11} = \frac{1}{\sqrt{2}}$</p>
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<h3>Finding the Second Eigenvector</h3>
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<p>Recall the fact that each subsequent eigenvector is orthogonal to the first. This means
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$$
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a<em>{11}a</em>{21} + a<em>{12}a</em>{22} = 0
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$$
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Substituting the values for $a<em>{11}$ and $a</em>{12}$ calculated in the previous section
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$$
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\frac{1}{\sqrt{2}}a<em>{21} + \frac{1}{\sqrt{2}}a</em>{22} = 0
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$$</p>
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<p>$$
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a<em>{21} + a</em>{22} = 0
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$$</p>
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<p>$$
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a<em>{21} = -a</em>{22}
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$$</p>
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<p>Since this eigenvector also needs to satisfy the first condition, we get the following values
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$$
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a<em>{21} = \frac{1}{\sqrt{2}} , a</em>{22} = \frac{-1}{\sqrt{2}}
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$$</p>
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<h3>Conclusion of Example</h3>
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<p>From this, we can say that the first principal components are given by
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$$
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y_1 = \frac{1}{\sqrt{2}}(x_1 + x_2), y_2 = \frac{1}{\sqrt{2}}(x_1-x_2)
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$$
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With the variance of the first principal component being given by $(1+r)$ and the second by $(1-r)$</p>
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<p>Due to this, as $r$ increases, so does the variance explained in the first principal component. This in turn, lowers the variance explained in the second principal component.</p>
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<h2>Choosing a Number of Principal Components</h2>
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<p>Principal Component Analysis is typically used in dimensionality reduction efforts. Therefore, there are several strategies for picking the right number of principal components to keep. Here are a few:</p>
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<ul>
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<li>Retain enough principal components to account for 70%-90% of the variation</li>
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<li>Exclude principal components where eigenvalues are less than the average eigenvalue</li>
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<li>Exclude principal components where eigenvalues are less than one.</li>
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<li>Generate a Scree Plot
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<ul>
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<li>Stop when the plot goes from "steep" to "shallow"</li>
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<li>Stop when it essentially becomes a straight line.</li>
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</ul></li>
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</ul>
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</article>
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