10-701/ Machine Learning, Fall 2005 Homework 3
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1 10-701/ Machne Learnng, Fall 2005 Homework 3 Out: 10/20/05 Due: begnnng of the class 11/01/05 Instructons Contact questons-10701@autonlaborg for queston Problem 1 Regresson and Cross-valdaton [40 ponts] Part 1: Multple regresson [15 ponts] The multple regresson model s Y = Xβ + ɛ where Y 1 ɛ 1 β 0 1 x 11 x 12 x 1n Y 2 ɛ 2 β 1 1 x 21 x 22 x 2n Y =, ɛ =, β = and X = Y r ɛ r β n 1 x r1 x r2 x rn Assume Y N(Xβ, σ 2 I) and ɛ N(0, σ 2 I) where I s the n n dentty matrx From the class we know that the least square estmator β = SY where S = (X T X) 1 X T (a) prove that β s unbased, e, E( β) = β Soluton: E( β) = E(SY ) = SE(Y ) = SXβ = (X T X) 1 X T Xβ = β (b) fnd the covarance matrx of β: V( β) (hnt: V(Cx) = CV(x)C T matrx) Soluton: f C s a constant V(β) = V(SY ) = SV(Y )S T = S(σ 2 I)S T = (X T X) 1 X T (σ 2 I)((X T X) 1 X T ) T = σ 2 (X T X) 1 X T ((X T X) 1 X T ) T = σ 2 (X T X) 1 X T X(X T X) 1 = σ 2 (X T X) 1 The estmator Ŷ = X β = HY where H = X(X T X) 1 X T (H s called the hat matrx) 1
2 (c) prove H s symmetrc (H = H T ) and dempotent (H 2 = H) Soluton: H = X(X T X) 1 X T H T = (X(X T X) 1 X T ) T = X(X T X) 1 X T H = H T (d) prove the trace of H equals the rank of X, e, tr(h) = n + 1 (hnt: what s the relatonshp between tr(ab) and tr(ba) f AB and BA are defned?) Soluton: Frst, t s easy to show tr(ab) = tr(ba) tr(h) = tr(x(x T X) 1 X T ) = tr((x T X) 1 X T X) = tr(i) = n + 1 Part 2: Leave-one-out cross-valdaton [25 ponts] The least square estmator mnmzes the sums of squared errors: r SSE = (Y Ŷ) 2 Recall the defnton of leave-one-out cross-valdaton score r LOOCV = (Y Ŷ ( ) ) 2 =1 =1 where Ŷ ( ) s the estmator of Y after removng -th observaton (e, t mnmzes j (Y j Ŷ ( ) j ) 2 ) In partcular, Ŷ ( ) s the estmated value of Y after removng -th observaton (a) wrte Ŷ n terms of H and Y Soluton: Recall Ŷ = HY, then the -th element of Ŷ s Ŷ = j H jy j { (b) prove that Ŷ ( ) Yj, j s also the estmator that mnmzes SSE for Z where Z j = Ŷ ( ), j = Soluton: By defnton, Ŷ ( ) = argmn j (Y j Ŷ ( ) j ) 2 Note that j (Y j ) 2 s equvalent to j (Z j Ŷ ( ) ) 2 snce Z = Ŷ ( ) Ŷ ( ) j Therefore, Ŷ ( ) = argmn j (Z j Ŷ ( ) j ) 2 (c) prove that Ŷ ( ) = Ŷ H Y + H Ŷ ( ) Soluton: From (a), we know Ŷ = j H jy j and Ŷ ( ) j H j(y j Z j ) = H (Y Ŷ ( ) ) Therefore Ŷ ( ) = Ŷ H Y + H Ŷ ( ) j 2 = j H jz j And Ŷ Ŷ ( ) =
3 (d) prove that LOOCV = r =1 ( Y Ŷ 1 H ) 2 Soluton: Just plug n (c) nto the defnton LOOCV, you get the result 3
4 Problem 2 Kernelzaton [40 ponts] In the lecture on SVM, we learned a trck called kernelzaton for classfcaton The dea s to map a feature vector n low dmensonal space X nto a hgher dmensonal space Z Ths can yeld a more flexble classfer whle retanng computatonal smplcty In other words: a lnear classfer n a hgher dmensonal space corresponds to a non-lnear classfer n the orgnal space In general, kernelzaton nvolves fndng a mappng φ : X Z such that 1 Z has a hgher dmenson than X ; 2 the computaton n Z only uses nner product; 3 there s a functon K called kernel such that the nner product of φ(x ) and φ(x j ) s K(x, x j ) 1 The standard logstc regresson has the followng form: n P (Y = 1 X) = g(ω 0 + ω X ) where g(a) = 1/(1 + e a ) =1 P (Y = 0 X) = 1 P (Y = 1 X) (a) Consder a functon φ maps X from a low dmensonal space X (dmensonalty=n) nto a hgh dmensonal space Z (dmensonalty s m, m > n) The logstc regresson becomes m P (Y = 1 φ(x)) = g(ω 0 + ω φ(x) ) where m s the dmenson of Z 2 Assume the weght vector ω s the lnear combnaton of all nput feature vector φ(x ); more formally, (ω 1,, ω m ) T = R =1 α φ(x () ) and ω 0 = α 0 where R s the number of data ponts and X () s the -th data pont Use kernelzaton trck to compute P (Y = 1 φ(x)) (e, to avod explctly computng n Z ) Soluton: P (Y = 1 φ(x)) = g(α 0 + =1 α K(X, X)) 1 And K has to be postve defnte, eg gaussan kernel s one of such kernel And you don t have to worry t for ths queston 2 X s a n-dmensonal feature vector; φ(x) s the correspondng m-dmensonal vector; φ(x) s the -th element of φ(x) 4 =1
5 (b) Wrte down the gradent descent update rule for kernel logstc regresson Soluton: The lkelhood of α s l(α) = Y l (α 0 + α j (X j, X))) ln(1 + exp(α 0 + l=1 The dervatve of l(α) s (l(α)) α = The update rule s j=1 α j (X j, X))) j=1 (Y l exp(α 0 + R j=1 α j(x j, X)) 1 + exp(α 0 + R j=1 α j(x j, X)) )K(X, X) l=1 α (t+1) = α (t) + η (l(α)) α (c) Implement the kernel logstc regresson usng the gaussan kernel K σ (x, x ) = exp ( x x ) 2 2σ 2 And run your program on ds2txt (frst two columns are X, last column s Y) wth σ = 1 Report the tranng error Set stepsze to be 001 and maxmum number of teratons 100 (Please use ths settng and don t try alternatve settngs) The scatterplot of the ds2txt s the follows: Soluton: 53 msclassfcatons (d) Use the 10-folds cross-valdaton to fnd the best σ and plot the total number of mstakes for σ = {05, 1, 2, 3, 4, 5, 6} Soluton:The best σ = 2 5
6 Problem 3 Computatonal Learnng Theory [20 ponts] Part1:VC-dmenson [12 ponts] Consder the space of nstances X correspondng to all ponts n the 2D plane Gve the VC-dmenson of the followng hypothess spaces: (a) H r : the set of all axs-parallel rectangles n the 2D plane Ponts nsde the rectangle are postve examples Soluton: VC-dmenson = 4 (b) H c : crcles n the 2D plane Ponts nsde the crcle are classfed as postve examples Soluton: VC-dmenson = 3 (c) How many tranng examples suffce to assure wth probablty 9 that a consstent learner usng H c wll learn the target functon wth accuracy of at least 095? Soluton: The bound s m 1 ɛ (4log 2(2/δ)+8VC(H)log 2 (13/ɛ)) Then just by pluggng n the numbers (VC(H)=4 or 3, δ = 1 and ɛ = 05), we have m 5480 for (a) and m 4197 for (b) (d) What exactly does t mean n part (c) when we say the learner wll succeed wth probablty 09? Answer ths queston by descrbng a smple experment whch you could run repeatedly, for whch the success rate s expected to be at least 09 Soluton: Choose a dstrbuton P(X) and target functon f Now repeatedly draw a tranng set {< x, y >} of sze 4197, based on P(X) and f For each tranng set, fnd a hypothess h n H c that perfectly classfes all 4197 tranng examples, and measure the true accuracy of h (e, the expected accuracy of h relatve to f and P(X)) In at least 09 of these experments (e, wth probablty 09) the true accuracy of h wll be at least 095 Part2: Mstake bounds [8 ponts] Consder learnng a boolean valued functon f : X Y, where X = X 1 X N, where Y and the X are all boolean valued varables You decde to consder a hypothess space H where each hypothess s of the form f [(X = a) (X j = b)] then Y = 1 else Y = 0 where j, and where a and b can be ether 0 or 1 Notce each hypothess constrans exactly two of the features of X Please answer the followng questons: 6
7 (a) How many dstnct hypotheses are there n H? Soluton: N 2 (N 1) (b) Consder the followng Weghted Majorty algorthm, appled to the entre space of hypotheses H: You begn wth all hypotheses n H assgned an ntal weght equal to 1 Every tme you see a new example, you predct based on a weghted majorty vote of the hypotheses n H After each predcton, any hypothess that made an ncorrect predcton has ts weght dvded by two How many mstakes wll ths Weghted Majorty algorthm make when shown a sequence of tranng examples, as a functon of the number of mstakes made by the most accurate hypothess n H? Soluton: Based on the theorem from p224 of Tom Mtchell s book, the number of mstake s at most 24 log 2 (k + 2N(N 1)) where k s the mnmal number of errors (c) Suppose X has N=1024 features, the tranng sequence contans 1000 examples, and the best hypothess n H has a true error of 005 What bound can you gven on the expected number of mstakes made by the Weghted Majorty algorthm n ths case? Soluton: Usng the same theorem as (b), the number of mstakes s at most 24 (05 N + log 2 (2N(N 1)) where N=1024 7
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