MIT15_097S12_lec15

10 11 i1 the ml estimate is equivalent to ordinary

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Unformatted text preview: assume that the noise is distributed normally with zero mean and some known variance σ 2 : E ∼ N (0, σ 2 ). This, together with our assumption of independence, allows us to express the likelihood function for the observations y = {y1 , . . . , ym }: p(y |x, θ) = = m m i=1 m m i=1 p(yi |xi , θ) = m m N (yi ; θT xi , σ 2 ) i=1 1 1 √ exp − 2 (yi − θT xi )2 , 2σ 2πσ 2 where the first line follows from independence. As usual, it is more convenient to work with the log-likelihood: �m � m1 1 √ exp − 2 (yi − θT xi )2 R(θ) = log 2σ 2πσ 2 i=1 m m 1m = − log(2πσ 2 ) − 2 (yi − θT xi )2 . 2 2σ i=1 The ML estimate is m 1m ˆ θML ∈ arg max R(θ) = arg max − 2 (yi − θT xi )2 θ θ 2σ i=1 = arg min θ m m (yi − θT xi )2 . (10) (11) i=1 The ML estimate is equivalent to ordinary least squares! Now let us try the MAP estimate. We saw in our coin toss example that the MAP estimate acts as a regularized ML estimate. For probabilistic regression, we will use a multiva...
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