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15-781 Machine Learning (Recitation 1)

15-781 Machine Learning (Recitation 1). By Jimeng Sun 9/15/05. Entropy. Entropy, Information Gain Decision Tree Probability. Suppose X can have one of m values… V 1, V 2, … V m

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15-781 Machine Learning (Recitation 1)

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  1. 15-781 Machine Learning(Recitation 1) By Jimeng Sun 9/15/05

  2. Entropy • Entropy, Information Gain • Decision Tree • Probability

  3. Suppose X can have one of m values… V1, V2, … Vm What’s the smallest possible number of bits, on average, per symbol, needed to transmit a stream of symbols drawn from X’s distribution? It’s H(X) = The entropy of X “High Entropy” means X is from a uniform (boring) distribution “Low Entropy” means X is from varied (peaks and valleys) distribution Entropy

  4. Entropy H(*) X = College Major Y = Likes “XBOX” H(X) = 1.5 H(Y) = 1

  5. Specific Conditional Entropy H(Y|X=v) X = College Major Y = Likes “XBOX” Definition of Specific Conditional Entropy: H(Y |X=v) = The entropy of Y among only those records in which X has value v

  6. Specific Conditional Entropy H(Y|X=v) X = College Major Y = Likes “XBOX” • Definition of Specific Conditional Entropy: • H(Y |X=v) = The entropy of Y among only those records in which X has value v • Example: • H(Y|X=Math) = 1 • H(Y|X=History) = 0 • H(Y|X=CS) = 0

  7. Conditional Entropy H(Y|X) X = College Major Y = Likes “XBOX” Definition of Conditional Entropy: H(Y |X) = The average specific conditional entropy of Y = if you choose a record at random what will be the conditional entropy of Y, conditioned on that row’s value of X = Expected number of bits to transmit Y if both sides will know the value of X = Σj Prob(X=vj) H(Y | X = vj)

  8. Conditional Entropy X = College Major Y = Likes “XBOX” • Definition of Conditional Entropy: • H(Y|X) = The average conditional entropy of Y • = ΣjProb(X=vj) H(Y | X = vj) Example: H(Y|X) = 0.5 * 1 + 0.25 * 0 + 0.25 * 0 = 0.5

  9. Information Gain X = College Major Y = Likes “XBOX” • Definition of Information Gain: • IG(Y|X) = I must transmit Y. How many bits on average would it save me if both ends of the line knew X? • IG(Y|X) = H(Y) - H(Y | X) • Example: • H(Y) = 1 • H(Y|X) = 0.5 • Thus IG(Y|X) = 1 – 0.5 = 0.5

  10. Decision Tree

  11. Tree pruning

  12. Probability

  13. Test your understanding • All the time? • Only when X and Y are independent? • It can fail even if X and Y are independent?

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