Difference between revisions of "Derivation of Fisher's Linear Discriminant Old Kiwi" - Rhea

Revision as of 17:43, 25 March 2008

See also: Fisher Linear Discriminant_Old Kiwi

In this derivation, we assume the dimension of the space being projected to, k, is k=1.

We attempt to find a linear projection (line that passes through the origin) s.t. the projected data can be "best" separated. The following animation shows how we can achieve the "best" separation between classes on the projected data by changing the projection plane $ \omega $.

sorry, I did not realize that this image would change while I was working on Lecture 11_Old Kiwi. I changed the name of the other image so whoever has the original animation can put it back

Assume $ y_1,..., y_{d_1} $ belongs to Class 1 and $ y_{d_1+1} $,..., $ y_d $ belongs to Class 2.

Consider the sample mean of the projected data for each class

$ \tilde{m_1} = \frac{1}{d_1} \sum_{i=1}^{d_1} \pi(y_i) = \frac{1}{d_1}\sum_{i=1}^{d_1} \vec{w} \cdot y_i $, if $ ||\omega || =1 $

Observation 1

$ m1 = \vec{w} \frac{1}{d_1} \sum_{i=1}^{d_1} y_i $, which is the projection of the sample mean for Class 1.

Also, $ \tilde{m_2} = \frac{1}{d-d_1}\sum_{i=d_1+1}^{d} \vec{w} \cdot y_i = \vec{w} \cdot m_2 $ .

Consider the scatter projections:

$ \tilde{S_1}^2 = \sum_{i=1}^{d_1} (\vec{w} \cdot y_i - \tilde{m_1})^2 $

$ \tilde{S_2}^2 = \sum_{i=d_1+1}^{d} (\vec{w} \cdot y_i - \tilde{m_2})^2 $

Define the following cost function, $ J(\vec{w}) = \frac{|\tilde{m1} - \tilde{m2}|^2}{\tilde{S_1}^2+ \tilde{S_2}^2} $ .

The above equation is the "Fisher linear discriminant_Old Kiwi".

Observation 2

Demanding $ ||\tilde{m_1} - \tilde{m_2}|| $ to be maximized would not do, since $ ||\tilde{m_1} - \tilde{m_2}|| = |\vec{w} \cdot m_1 - \vec{w} \cdot m_2 | = | \vec{w} \cdot (m_1 - m_2)| \rightarrow \infty $

Need to fix $ || \omega || = 1 $ , resulting in a Constrained Optimization Problem.

By dividing by $ \tilde{S_1}^2 + \tilde{S_2}^2 $ we demand that $ \vec{w} \cdot (m_1 - m_2) $ be large in comparison with the scatter. This allows you to bypass the coordinate scale problem because $ J(\vec{w}) $ is independent of $ \omega $.

$ |\tilde{m_1} - \tilde{m_2}|^2 = | w(m_1 - m_2)|^2 = |w|^2 | \frac{w}{|w|}(m_1-m_2)|^2 $

Similarly,

$ \tilde{S_1}^2 = \sum (\vec{w}(y_i - m_1))^2 = |w|^2 \sum (\frac{w}{|W|} (y_i - m_1))^2 $

and,

$ \tilde{S_2}^2 = \sum (\vec{w}(y_i - m_1))^2 = |w|^2 \sum (\frac{w}{|W|} (y_i - m_2))^2 $

and,

$ J(\vec{w}) = \frac{|m_1-m_2|^2}{S_1^2+S_2^2} =\frac{|w|^2 | \frac{w}{|w|}(m_1-m_2)|^2}{|w|^2(\sum (\frac{w}{|W|} (y_i - m_1))^2 + \sum (\frac{w}{|W|} (y_i - m_2))^2 )} $ is independent of $ ||\omega|| $.

Importance of scatter:

If we find $ \omega $ which makes $ Jw $ large, we are guaranteed that the classes are well-separated.

We can write $ Jw $ using matrices:

$ J(\vec{w}) = \frac{\vec{w}^{\top}S_B \vec{w}}{\vec{w}^{\top}S_w \vec{w}} $, which is called the "[Generalized Rayleigh Quotient]".

(Lecture 11_Old Kiwi begins here)

Last time we considered $ J(\vec{w})=\frac {{\vec{w}}^{T}{S}_{B}\vec{w}} {{\vec{w}}^{T}{S}_{W}\vec{w}} $, explicit function of $ \vec{w} $

one can do this because

(numerator of J) = $ \| \tilde{m_1}-\tilde{m_2} \|^2 = \| w(m_1-m_2) \|^2 = \| (m_1-m_2)^T w \|^2 $

$ =w^T (m_1-m_2)(m_1^T-m_2^T)w $

where, $ (m_1-m_2)(m_1^T-m_2^T) =S_B $ is 'between class scatter matrix'.

(denominator of J) = $ \tilde{s_1}^2 + \tilde{s_2}^2 $

$ \tilde{s_1}^2 =\displaystyle \sum_{\vec y \; in \; class} (w y_i- \tilde{m_i} )^2 $

$ = \displaystyle \sum_{\vec{y} \; in \; class} w^T (y_i-m_i)(y_i^T-m_i^T)w $

$ w^T \displaystyle \sum_{\vec{y} \; in \; class} (y_i-m_i)(y_i^T-m_i^T)w $

where, $ \displaystyle \sum_{\vec y \; in \; class} (y_i-m_i)(y_i^T-m_i^T) =S_w $ is 'within class scattter matrix'.