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# lecture15 - CSE 6740 Lecture 15 What Error Guarantees Can...

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CSE 6740 Lecture 15 What Error Guarantees Can We Make? (Learning Theory and Generalization) Alexander Gray [email protected] Georgia Institute of Technology CSE 6740 Lecture 15 – p. 1/2

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Today 1. Statistical inequalities (How can we bound values that can appear in the future?) 2. Confidence intervals (How good is the estimation/learning?) CSE 6740 Lecture 15 – p. 2/2
Statistical inequalities How can we bound values that can appear in the future? CSE 6740 Lecture 15 – p. 3/2

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Markov’s Inequality Theorem ( Markov’s inequality ): Suppose X is a non-negative random variable and E ( X ) exists. Then for any t > 0 , P ( X > t ) E ( X ) t . (1) CSE 6740 Lecture 15 – p. 4/2
Markov’s Inequality: Proof Since X > 0 , E ( X ) = integraldisplay 0 xf ( x ) dx (2) = integraldisplay t 0 xf ( x ) dx + integraldisplay t xf ( x ) dx (3) integraldisplay t xf ( x ) dx (4) t integraldisplay t f ( x ) dx (5) = t P ( X > t ) . (6) CSE 6740 Lecture 15 – p. 5/2

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Chebyshev’s Inequality Theorem ( Chebyshev’s inequality ): If μ = E ( X ) and σ 2 = V ( X ) , then P ( | X - μ | ≥ t ) σ 2 t 2 (7) and P parenleftbigg vextendsingle vextendsingle vextendsingle vextendsingle X - μ σ vextendsingle vextendsingle vextendsingle vextendsingle u parenrightbigg 1 u 2 (8) (or P ( | Z | ≥ u ) 1 u 2 if Z = ( X - μ ) ). For example, P ( | Z | > 2) 1 / 4 and P ( | Z | > 3) 1 / 9 . CSE 6740 Lecture 15 – p. 6/2
Chebyshev’s Inequality: Proof Using Markov’s inequality, P ( | X - μ | ≥ t ) = P ( | X - μ | 2 t 2 ) (9) E ( X - μ ) 2 t 2 (10) = σ 2 t 2 . (11) The second part follows by setting t = . CSE 6740 Lecture 15 – p. 7/2

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Chebyshev’s Inequality: Example Suppose we test a classifier on a set of N new examples. Let X i = 1 if the prediction is wrong and X i = 0 if it is right; then X N = 1 N N i =1 X i is the observed error rate. Each X i may be regarded as a Bernoulli with unknown mean p ; we would like to estimate this. How likely is X N to not be within ǫ of p ? CSE 6740 Lecture 15 – p. 8/2
Chebyshev’s Inequality: Example We have that V ( X N ) = V ( X ) /N = p (1 - p ) /N and P ( | X N - p | > ǫ ) V ( X N ) ǫ 2 (12) = p (1 - p ) 2 (13) 1 4 2 (14) since p (1 - p ) 1 / 4 for all p .

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