Let us consider the problem of nearest-mean classifier. Suppose we are given N training samples (x₁, y₁),..., (XN, YN) from two classes with yn € {+1, -1}. We saw in lecture 2 that we can decide a label for a test vector x as g(x) = sign (w²x + b), where w = 2(µ+ − µ-) and b = ||µ–||²3 − ||µ+||³2. - μ+ is a mean vector for samples in the +ve class and p_ is a mean vector for samples the -ve class. Show that w²x + b = 1 an(xn, x) + b and calculate the values of the an.
Let us consider the problem of nearest-mean classifier. Suppose we are given N training samples (x₁, y₁),..., (XN, YN) from two classes with yn € {+1, -1}. We saw in lecture 2 that we can decide a label for a test vector x as g(x) = sign (w²x + b), where w = 2(µ+ − µ-) and b = ||µ–||²3 − ||µ+||³2. - μ+ is a mean vector for samples in the +ve class and p_ is a mean vector for samples the -ve class. Show that w²x + b = 1 an(xn, x) + b and calculate the values of the an.
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Transcribed Image Text:Let us consider the problem of nearest-mean classifier. Suppose we are given N training samples
(x1,Y1), ..., (XN,YN) from two classes with yn E {+1, –1}. We saw in lecture 2 that we can decide a
label for a test vector x as g(x) = sign(w'x+b), where w =
2(u+ - µ-) and b = ||µ- ||3 – ||4+|3.
H+ is a mean vector for samples in the +ve class and u_ is a mean vector for samples the -ve class.
Show that w'x+b = E=1 an(Xn, x) + b and calculate the values of the an.
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