Solution Manual of a first course in machine learning 1st -2nd edition by Simon

1. cl i ck h ere t o d ow nl oad solutions MAnuAl FoR A First Course in Machine learning (2nd Edition) Exercise solutions by Simon Rogers and Mark Girolami @solutionmanual1 K26591_SM_Cover.indd 1 05/04/16 3:37 pm

2. cl i ck h ere t o d ow nl oad @solutionmanual1 K26591_SM_Cover.indd 2 05/04/16 3:37 pm

3. cl i ck h ere t o d ow nl oad solutions MAnuAl FoR A First Course in Machine learning (2nd Edition) Exercise solutions by Simon Rogers and Mark Girolami Boca Raton London New York CRC Press is an imprint of the Taylor & Francis Group, an informa business @solutionmanual1 K26591_SM_Cover.indd 3 05/04/16 3:37 pm

4. cl i ck h ere t o d ow nl oad CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2017 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20160404 International Standard Book Number-13: 978-1-4987-3859-0 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com @solutionmanual1 K26591_SM_Cover.indd 4 05/04/16 3:37 pm

5. cl i ck h ere t o d ow nl oad Chapter 1 EX 1.1. A high positive value of w0 and a small negative value for w1. These reflect the high intercept on the t axis (corresponding to the theoretical time winning time at x = 0 and the small decrese in winning time over the years. EX 1.2. The following would do the job: % Attributes are stored in Nx1 vector x % Targets are stored in Nx1 vector t xb = mean(x); tb = mean(t); x2b = mean(x.*x); xtb = mean(x.*t); w1 = (xtb − xt*xb)/(x2b−xbˆ2); w0 = tb−w1*xb; % Plot the data plot(x,t,'b.','markersize',25); % Plot the model hold on; plot(x,w0+w1*x,'r','linewidth',2); 1 2 3 4 5 6 7 8 9 10 11 12 13 EX 1.3. We need to find wTXTXw. We’ll start with XTX. Multiplying XTby X gives: ? ? ?N ?N n=1x2 n=1xn2xn1 n=1xn1xn2 ?N XTX = n1 ?N n=1x2 n2 Multiplying this by w gives: ? ? ?N ?N n=1x2 ?N w0 w0 n1+ w1 n=1xn1xn2 ?N XTXw = n=1x2 n=1xn2xn1+ w1 n2 1 @solutionmanual1 K26591_SM_Cover.indd 5 05/04/16 3:37 pm

6. cl i ck h ere t o d ow nl oad 2 CHAPTER 1. Finally, pre-multiplying this by wTgives: ? ? ? ? N N ? N ? ? wTXTXw x2 = w0 w0 n1+ w1 xn1xn2 + n=1 n=1 N ? x2 w1 w0 xn2xn1+ w1 n2 n=1 n=1 N N N ? ? ? w2 x2 xn1xn2+ w2 x2 = n1+ 2w0w1 0 1 n2 n=1 n=1 n=1 as required. EX 1.4. Let’s first work out Xw:     w0x11+ w1x12 w0x21+ w1x22 ... w0xN1+ w1xN2   Xw = Therefore (Xw)T= [w0x11+ w1x12, w0x21+ w1x22, ... , w0xN1+ w1xN2] Finally, work out wTXT: wTXT= [w0x11+ w1x12, w0x21+ w1x22, ... , w0xN1+ w1xN2] as required. EX 1.5. Starting with? nxntn. The result of this is a column vector of the same size as x (2 × 1). Now, using the definition of X, ? (which is a 2×N vector). Multiplying this by t gives a 2×1 vector that looks like this: ??N which is? ? w0 ? x11,x21,...,xN1 x12,x22,...,xN2 XT= ? n=1xn1tn ?N XTt = n=1xn2tn nxntnas required. The second example, XTXw. We already know what XTXw is (Exercise 1.3) ? ?N ?N n=1x2 ?N w0 n1+ w1 n=1xn1xn2 ?N XTXw = n=1x2 n=1xn2xn1+ w1 n2 Now, xnxT nis the following matrix: ? ? x2 xn1xn2 x2 n2 xnxT n1 n= xn2xn1 @solutionmanual1 K26591_SM_Cover.indd 6 05/04/16 3:37 pm