Fisher Discriminant Analysis

November 16, 2020
Table of Content

One challenge is in classification is the curse dimensionality in which the amount of finite samples required to properly estimate the underlying distribution of the data goes exponentially with the dimension of the data. Therefore, it might be a reasonable and efficient way to first perform dimensionality reduction and operate on this subspace.

However, one issue might arise is that we might lose some information that is meaningful for classification. To illustrate, the figure shows the effect of projection on distinguishing two groups of data.

Fig. 1: Projection direction that does not separate the two distribution well.

How could we find such a direction?

Let wRd\mathbf w \in \Reals^d be the direction that we aim to find; We also assume that w=1\| \mathbf w \| = 1. Consider two groups of data N1={xRd}\mathcal{N}_1 = \{ \mathbf x \in \Reals^d \} and N2={xRd}\mathcal{N}_2 = \{ \mathbf x \in \Reals^d \}; The projection of these samples are : wTx\mathbf w^T\mathbf x. Noting here that we implicitly represent the bias term in the inner production, i.e. the last feature of these samples xN1N2\mathbf x \in N_1 \cup N_2 is always 1.

From Fig 1, we can observe that the direction that would allow us to separate the two classes best is the one that maximizes the distance between the means (μ^1,μ^2R)\hat \mu_1, \hat \mu_2 \in \Reals ) of these data after projection. Moreover, the projection would be more discriminate if the two distributions after projection are narrow; in other word, it implies that the variances of these distributions should be small. With this, we can construct the objective function:

J(w)=(μ^1μ^2)2σ12+σ22,J(\mathbf w) = \frac{(\hat \mu_1 - \hat \mu_2)^2 }{\sigma_1^2+\sigma_2^2},

where our goal is arg maxwRd,w=1J(w)\argmax_{\mathbf w \in \Reals^d, \|\mathbf w \| = 1} J(\mathbf w).

where our goal is arg maxwRd,w=1J(w)\argmax_{\mathbf w \in \Reals^d, \|\mathbf w \| = 1} J(\mathbf w). For the nominator, we can rewrite it as

(μ^1μ^2)2=(1N1xN1wTx1N2xN2wTx)2=(1N1xN1wTx1N2xN2wTx)(1N1xN1xTw1N2xN2xTw)=wT(1N1xN1x1N2xN2x)(1N1xN1xT1N2xN2xT)SBw,\begin{aligned} (\hat \mu_1 - \hat \mu_2)^2 &= \Bigg( \frac{1}{|\mathcal{N}_1|} \sum_{\mathbf x \in \mathcal{N}_1}\mathbf w^T\mathbf x - \frac{1}{|\mathcal{N}_2|} \sum_{\mathbf x \in \mathcal{N}_2}\mathbf w^T\mathbf x \Bigg)^2 \\ &= \Bigg( \frac{1}{|\mathcal{N}_1|} \sum_{\mathbf x \in \mathcal{N}_1}\mathbf w^T\mathbf x - \frac{1}{|\mathcal{N}_2|} \sum_{\mathbf x \in \mathcal{N}_2}\mathbf w^T\mathbf x \Bigg)\Bigg( \frac{1}{|\mathcal{N}_1|} \sum_{\mathbf x \in \mathcal{N}_1} \mathbf x^T \mathbf w - \frac{1}{|\mathcal{N}_2|} \sum_{\mathbf x \in \mathcal{N}_2}\mathbf x^T \mathbf w\Bigg) \\ &= \mathbf w^T \underbrace{\Bigg( \frac{1}{|\mathcal{N}_1|} \sum_{\mathbf x \in \mathcal{N}_1}\mathbf x - \frac{1}{|\mathcal{N}_2|} \sum_{\mathbf x \in \mathcal{N}_2}\mathbf x \Bigg) \Bigg( \frac{1}{|\mathcal{N}_1|} \sum_{\mathbf x \in \mathcal{N}_1}\mathbf x^T - \frac{1}{|\mathcal{N}_2|} \sum_{\mathbf x \in \mathcal{N}_2}\mathbf x^T \Bigg)}_{\triangleq S_B} \mathbf w, \end{aligned}

where SBS_B is called between-class covariate matrix. Similarly, we can expand the denominator term:

σ12+σ22=(1N1xN1(wTxμ^1)21N2xN2(wTxμ^2)2).\begin{aligned} \sigma_1^2 + \sigma_2^2 &= \Bigg( \frac{1}{\mathcal |\mathcal{N}_1|} \sum_{\mathbf x \in \mathcal N_1} (\mathbf w^T\mathbf x - \hat \mu_1)^2 - \frac{1}{\mathcal |\mathcal N_2|} \sum_{\mathbf x \in \mathcal N_2} (\mathbf w^T\mathbf x - \hat \mu_2)^2 \Bigg). \end{aligned}

Let look at x(wTxμ^1)2\sum_{\mathbf x} (\mathbf w^T\mathbf x - \hat \mu_1)^2 : we can complete the square and get the outer product:

(wTxμ^i)2=[wTx1NiwTx][xTw1NixTw]=wT[x1Nix][xT1NixT]Siw.\begin{aligned} \sum (\mathbf w^T\mathbf x - \hat \mu_i)^2 &= \sum \Bigg[\mathbf w^T\mathbf x - \frac{1}{|\mathcal N_i|} \sum \mathbf w^T\mathbf x \Bigg] \Bigg[\mathbf x^T\mathbf w - \frac{1}{\mathcal |\mathcal N_i|} \sum \mathbf x^T\mathbf w \Bigg] \\ &=\mathbf{w}^T \underbrace{\Bigg[\mathbf x - \frac{1}{|\mathcal N_i|} \sum \mathbf x \Bigg] \Bigg[\mathbf x^T - \frac{1}{|\mathcal N_i|} \sum \mathbf x^T \Bigg]}_{\triangleq S_i} \mathbf w. \end{aligned}

With this derivation, we have σ12+σ22=wT(S1+S2)w\sigma_1^2 + \sigma_2^2 = \mathbf w^T(S_1 + S_2) \mathbf w. Let define SWS1+S2S_W \triangleq S_1 + S_2 called within-class covariate matrix. Therefore, the objective function becomes

J(w)=wTSBwwTSWw.J(\mathbf w) = \frac{\mathbf w^T S_B \mathbf w}{\mathbf w^TS_W\mathbf w}.

Solving the optimization problem

With the assumption that the underlying distribution of x\mathbf x is a multivariate Gaussian distribution. SWS_W is positive definite; therefore, we can use Cholesky decomposition: SW=RTR,S_W = R^TR, where RRd×dR \in \Reals^{d\times d} is an upper-triangular matrix. Let define vRw\mathbf v \triangleq R\mathbf w and rewrite wTSWw\mathbf w^TS_W \mathbf w:

wTSWw=wTRTRw=(Rw)TRw=vTv.\begin{aligned} \mathbf w^T S_W \mathbf w &= \mathbf w^TR^TR\mathbf w \\ &= (R\mathbf w)^T R\mathbf w \\ &= \mathbf v^T \mathbf v. \end{aligned}

Because w=R1v\mathbf w = R^{-1}\mathbf v, we have

wTSBw=(R1v)TSBR1v=vT(R1)TSBR1Av.\begin{aligned} \mathbf w^TS_B\mathbf w &= \mathbf (R^{-1}\mathbf v)^T S_B R^{-1}\mathbf v \\ &= \mathbf v^T \underbrace{(R^{-1})^T S_B R^{-1}}_{\triangleq A}\mathbf v. \end{aligned}

With this reparameterization, the optimization problem is simplified to

v=arg maxvRd;v=1vTAvvTv.\mathbf v^* = \argmax_{\mathbf v \in \Reals^d; \|\mathbf v \| = 1} \frac{\mathbf v^T A \mathbf v}{\mathbf v^T\mathbf v}.

This is the Rayleigh quotient, and it can be shown that the maximum is attained when v\mathbf v^* is the eigenvalue corresponds to the largest eigenvalue λ\lambda^* (See ... for the proof) . Therefore, we have

Av=λv((R1)TSBR1)Rw=λRwRT(R1)TSBIw=λRTRwSBw=λSWw(SW)1SBw=λw\begin{aligned} A \mathbf v^* &= \lambda^* \mathbf v^* \\ ((R^{-1})^T S_B R^{-1}) R \mathbf w^* &= \lambda^* R \mathbf w^* \\ R^T(R^{-1})^T S_B I \mathbf w^* &= \lambda^* R^T R \mathbf w^* \\ S_B \mathbf w^* &= \lambda^* S_W \mathbf w^* \\ (S_W)^{-1}S_B \mathbf w^* &= \lambda^* \mathbf w^* \end{aligned}

Noting here that SBS_B is the outer product of the vector μ1μ2\mathbf \mu_1 - \mathbf \mu_2; hence rank(SB)=1\text{rank}(S_B) = 1. Therefore, SBwS_B \mathbf w^* is in the direction of μ1μ2\mathbf \mu_1 - \mathbf \mu_2. With this observation, we do not need to solve the eigenvalue problem above and rather compute

w(SW)1(μ1μ2).\mathbf w^* \propto (S_W)^{-1} (\mathbf \mu_1 - \mathbf \mu_2).
Fig. 2: Projection direction that maximizes the class separation and minimize the variance of the distributions along the projection direction.

Decision Boundary

So far, what we have now is a direction in space that maximally separates the two classes once they are projected onto this direction. To prediction whether a sample belongs to which class, we can derive the discriminant function of this projection. Let ω1\omega_1 and ω2\omega_2 be the two classes we consider. We know that

P(ωix)p(xωi)P(ωi).P(\omega_i|\mathbf x) \propto p(\mathbf x | \omega_i) P(\omega_i).

Using Bayes' decision theory, we decide x\mathbf x to belong to ω1\omega_1 if P(ω1x)>P(ω2x)P(\omega_1| \mathbf x ) > P(\omega_2| \mathbf x ) and vice versa; hence the decision boundary lies at P(ω1x)P(ω2x)=0P(\omega_1| \mathbf x ) - P(\omega_2| \mathbf x ) = 0. Applying logarithm, we have

logp(xω1)+logP(ω1)logp(xω2)logP(ω2)=0.\log p(\mathbf x| \omega_1) + \log P(\omega_1) - \log p(\mathbf x| \omega_2) - \log P(\omega_2) = 0.

Let assume that after the projection the data of each class follows a Gaussian distribution (in this case one dimension). The equation is reduced to

(xμ1)2σ12+logP(ω1)+(xμ2)2σ22logP(ω2)=0.- \frac{(\mathbf x - \mathbf \mu_1)^2}{\sigma_1^2} + \log P(\omega_1) + \frac{(\mathbf x - \mathbf \mu_2)^2}{\sigma_2^2} - \log P(\omega_2) = 0.

If we assume that P(ω1)=P(ω2)P(\omega_1) = P(\omega_2) , the decision boundary then depends only on μ1,μ2,σ1,σ2\mathbf \mu_1, \mathbf \mu_2, \sigma_1, \sigma_2, which can be estimated using maximum likelihood. Furthermore, one might further assume that σ1=σ2\sigma_1 = \sigma_2, hence they can be dropped from the equation. In this case, the decision boundary becomes the point between μi\mu_i's, which yields a linear decision boundary, called Linear Discriminant Analysis (LDA). On the other hand, if we estimate σi\sigma_i's and use them, we have a quadratic decision boundary, called Quadratic Discriminant Analysis (QDA), i.e. it can curve following the data distributions.

Below are decision boundaries derived from these two cases on three different datasets.

Fig. 3: Decision boundaries from LDA and QDA on three different datasets. Numbers on the bottom right of each plot are test set accuracies.


Google Colab used for making Figure 3.

Some references for this blog:

Proof: Maximization Rayleigh Quotient

Consider a symmetric and positive definite ARd×dA \in \Reals^{d\times d}. Denote vRd\mathbf v \in \Reals^d the vector that we seek under the following objective:

arg maxvRdvTAvvTvJ(v).\argmax_{\mathbf v \in \Reals^d} \underbrace{\frac{\mathbf v^T A \mathbf v}{\mathbf v^T\mathbf v}}_{\triangleq J(\mathbf v)}.

Because v\mathbf v would be scale invariant, we only need to find the solution in a unit ball, i.e. v=1\| \mathbf v \| = 1. Because A is symmetric and positive definite, we know that it can be diagonalizable. In particular, the diagonalization is A=QΣQTA = Q\Sigma Q^T, where QQ is an orthogonal matrix and Σ\Sigma is an diagonal matrix. We can write the nominator as follows:

vTAv=vTQΣQTv=(QTv)TΣQTvu=uTΣu.\begin{aligned} \mathbf v^T A \mathbf v &= \mathbf v^T Q\Sigma Q^T \mathbf v \\ &= ( Q^T \mathbf v )^T \Sigma \underbrace{Q^T \mathbf v}_\mathbf u \\ &= \mathbf u^T \Sigma \mathbf u. \end{aligned}

We can see that

u=uTu=vTQTQv=vTv=1\begin{aligned} \|\mathbf u \| &= \mathbf u ^T \mathbf u \\ &= \mathbf v^T Q^T Q \mathbf v \\ &= \mathbf v^T \mathbf v \\ &= 1 \end{aligned}

Without loss of generality, let's first assume that the eigenvalues of AA is λ1λ2λn\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n. Because Σ\Sigma is a diagonal matrix, we know that

uTΣu=i=1dλiui2.\begin{aligned} \mathbf u^T \Sigma \mathbf u = \sum_{i=1}^d \lambda_i u_i^2. \end{aligned}

With the assumption that the eigenvalues are in descending order, it implies that u12=1u_1^2=1 and ui2=0,i1u_i^2=0, \forall i\ne1. Hence, we have maxuTΣu=λ1\max \mathbf u^{*T} \Sigma \mathbf {u}^* = \lambda_1 and u\mathbf u^* is the basis vector (after the rotation with the eigen basis) corresponding to λ1\lambda_1, i.e. u=[1,0,,0]T\mathbf u^* = [1, 0, \dots, 0]^T. Therefore, we have v=Qu\mathbf v^* = Q\mathbf u^*, which is the eigenvector corresponding to λ1\lambda_1. ◾️

Conversely, if the problem is minimization, the minimum is attained at λd\lambda_d.