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Discrimination and Classification

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Discrimination Situation: We have two or more populations 1, 2, etc (possibly p-variate normal). The populations are known (or we have data from each population) We have data for a new case (population unknown) and we want to identify the which population for which the new case is a member.

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The Basic Problem Suppose that the data from a new case x 1, …, x p has joint density function either : 1 : g(x 1, …, x n ) or 2 : h(x 1, …, x n ) We want to make the decision to D 1 : Classify the case in 1 (g is the correct distribution) or D 2 : Classify the case in 2 (h is the correct distribution)

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The Two Types of Errors 1.Misclassifying the case in 1 when it actually lies in 2. Let P[1|2] = P[D 1 | 2 ] = probability of this type of error 2.Misclassifying the case in 2 when it actually lies in 1. Let P[2|1] = P[D 2 | 1 ] = probability of this type of error This is similar Type I and Type II errors in hypothesis testing.

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Note: 1. C 1 = the region were we make the decision D 1. (the decision to classify the case in 1 ) A discrimination scheme is defined by splitting p – dimensional space into two regions. 2. C 2 = the region were we make the decision D 2. (the decision to classify the case in 2 )

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1.Set up the regions C 1 and C 2 so that one of the probabilities of misclassification, P[2|1] say, is at some low acceptable value . Accept the level of the other probability of misclassification P[1|2] = . There can be several approaches to determining the regions C 1 and C 2. All concerned with taking into account the probabilities of misclassification P[2|1] and P[1|2]

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2.Set up the regions C 1 and C 2 so that the total probability of misclassification: P[Misclassification] = P[1] P[2|1] + P[2]P[1|2] is minimized P[1] = P[the case belongs to 1 ] P[2] = P[the case belongs to 2 ]

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3.Set up the regions C 1 and C 2 so that the total expected cost of misclassification: E[Cost of Misclassification] = ECM = c 2|1 P[1] P[2|1] + c 1|2 P[2]P[1|2] is minimized P[1] = P[the case belongs to 1 ] P[2] = P[the case belongs to 2 ] c 2|1 = the cost of misclassifying the case in 2 when the case belongs to 1. c 1|2 = the cost of misclassifying the case in 1 when the case belongs to 2.

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The Optimal Classification Rule Suppose that the data x 1, …, x p has joint density function f(x 1, …, x p ; ) where is either 1 or 2. Let g(x 1, …, x p ) = f(x 1, …, x n ; 1 ) and h(x 1, …, x p ) = f(x 1, …, x n ; 2 ) We want to make the decision D 1 : = 1 (g is the correct distribution) against D 2 : = 2 (h is the correct distribution)

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and where then the optimal regions (minimizing ECM, expected cost of misclassification) for making the decisions D 1 and D 2 respectively are C 1 and C 2

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Proof: ECM = E[Cost of Misclassification] = c 2|1 P[1] P[2|1] + c 1|2 P[2]P[1|2]

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Therefore Thus ECM is minimized if C 1 contains all of the points (x 1, …, x p ) such that the integrand is negative

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Fishers Linear Discriminant Function. Suppose that x 1, …, x p is either data from a p-variate Normal distribution with mean vector: The covariance matrix is the same for both populations 1 and 2.

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The Neymann-Pearson Lemma states that we should classify into populations 1 and 2 using: That is make the decision D 1 : population is 1 if > k

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or and

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Finally we make the decision D 1 : population is 1 if where and Note: k = 1 and ln k = 0 if c 1|2 = c 2|1 and P[1] = P[2].

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The function Is called Fisher’s linear discriminant function

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In the case where the populations are unknown but estimated from data Fisher’s linear discriminant function

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Example 2 Annual financial data are collected for firms approximately 2 years prior to bankruptcy and for financially sound firms at about the same point in time. The data on the four variables x 1 = CF/TD = (cash flow)/(total debt), x 2 = NI/TA = (net income)/(Total assets), x 3 = CA/CL = (current assets)/(current liabilties, and x 4 = CA/NS = (current assets)/(net sales) are given in the following table.

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The data are given in the following table:

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Examples using SPSS

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Classification or Cluster Analysis Have data from one or several populations

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Situation Have multivariate (or univariate) data from one or several populations (the number of populations is unknown) Want to determine the number of populations and identify the populations

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Example

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Hierarchical Clustering Methods The following are the steps in the agglomerative Hierarchical clustering algorithm for grouping N objects (items or variables). 1.Start with N clusters, each consisting of a single entity and an N X N symmetric matrix (table) of distances (or similarities) D = (d ij ). 2.Search the distance matrix for the nearest (most similar) pair of clusters. Let the distance between the "most similar" clusters U and V be d UV. 3.Merge clusters U and V. Label the newly formed cluster (UV). Update the entries in the distance matrix by a)deleting the rows and columns corresponding to clusters U and V and b)adding a row and column giving the distances between cluster (UV) and the remaining clusters.

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4.Repeat steps 2 and 3 a total of N-1 times. (All objects will be a single cluster a termination of this algorithm.) Record the identity of clusters that are merged and the levels (distances or similarities) at which the mergers take place.

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Different methods of computing inter-cluster distance

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Example To illustrate the single linkage algorithm, we consider the hypothetical distance matrix between pairs of five objects given below:

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Treating each object as a cluster, the clustering begins by merging the two closest items (3 & 5). To implement the next level of clustering we need to compute the distances between cluster (35) and the remaining objects: d (35)1 = min{3,11} = 3 d (35)2 = min{7,10} = 7 d (35)4 = min{9,8} = 8 The new distance matrix becomes:

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The next two closest clusters ((35) & 1) are merged to form cluster (135). Distances between this cluster and the remaining clusters become:

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Distances between this cluster and the remaining clusters become: d (135)2 = min{7,9} = 7 d (135)4 = min{8,6} = 6 The distance matrix now becomes: Continuing the next two closest clusters (2 & 4) are merged to form cluster (24).

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Distances between this cluster and the remaining clusters become: d (135)(24) = min{d (135)2,d (135)4 )= min{7,6} = 6 The final distance matrix now becomes: At the final step clusters (135) and (24) are merged to form the single cluster (12345) of all five items.

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The results of this algorithm can be summarized graphically on the following "dendogram"

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Dendograms for clustering the 11 languages on the basis of the ten numerals

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Dendogram Cluster Analysis of N=22 Utility companies Euclidean distance, Average Linkage

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Dendogram Cluster Analysis of N=22 Utility companies Euclidean distance, Single Linkage

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Classification Course web page: vision.cis.udel.edu/~cv May 14, 2003 Lecture 34.

Classification Course web page: vision.cis.udel.edu/~cv May 14, 2003 Lecture 34.

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