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Applied Statistics Using SPSS, STATISTICA, MATLAB and R

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244 6 Statistical Classification<br />

Figure 6.14. Classification results obtained with <strong>STATISTICA</strong>, of two classes of<br />

cork stoppers using: (a) Ten features; (b) Four features.<br />

Let us denote:<br />

Pe – Probability of error of a given classifier;<br />

Pe *<br />

– Probability of error of the optimum Bayesian classifier;<br />

Ped(n) – Training (design) set estimate of Pe based on a classifier<br />

designed on n cases;<br />

Pet(n) – Test set estimate of Pe based on a set of n test cases.<br />

The quantity Ped(n) represents an estimate of Pe influenced only by the finite<br />

size of the design set, i.e., the classifier error is measured exactly, <strong>and</strong> its deviation<br />

from Pe is due solely to the finiteness of the design set. The quantity Pet(n)<br />

represents an estimate of Pe influenced only by the finite size of the test set, i.e., it<br />

is the expected error of the classifier when evaluated using n-sized test sets. These<br />

quantities verify Ped(∞) = Pe <strong>and</strong> Pet(∞) = Pe, i.e., they converge to the theoretical<br />

value Pe with increasing values of n. If the classifier happens to be designed as an<br />

optimum Bayesian classifier Ped <strong>and</strong> Pet converge to Pe * .<br />

In normal practice, these error probabilities are not known exactly. Instead, we<br />

compute estimates of these probabilities, Pˆ<br />

ed<br />

<strong>and</strong> Pˆ<br />

et<br />

, as percentages of<br />

misclassified cases, in exactly the same way as we have done in the classification<br />

matrices presented so far. The probability of obtaining k misclassified cases out of<br />

n for a classifier with a theoretical error Pe, is given by the binomial law:<br />

⎛ n⎞<br />

k n−k<br />

P(<br />

k)<br />

= ⎜ Pe − Pe<br />

k ⎟ ( 1 ) . 6.26<br />

⎝ ⎠<br />

The maximum likelihood estimation of Pe under this binomial law is precisely<br />

(see Appendix C):<br />

P ˆ e = k / n , 6.27<br />

with st<strong>and</strong>ard deviation:<br />

Pe ( 1−<br />

Pe)<br />

σ =<br />

. 6.28<br />

n

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