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## Mathematical Methods of Systems Analysis Stochastic

by: Buck Ankunding

72

0

7

# Mathematical Methods of Systems Analysis Stochastic CMPE 107

Marketplace > University of California - Santa Cruz > Computer Engineering > CMPE 107 > Mathematical Methods of Systems Analysis Stochastic
Buck Ankunding
UCSC
GPA 3.56

Staff

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COURSE
PROF.
Staff
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Class Notes
PAGES
7
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KARMA
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## Popular in Computer Engineering

This 7 page Class Notes was uploaded by Buck Ankunding on Monday September 7, 2015. The Class Notes belongs to CMPE 107 at University of California - Santa Cruz taught by Staff in Fall. Since its upload, it has received 72 views. For similar materials see /class/182220/cmpe-107-university-of-california-santa-cruz in Computer Engineering at University of California - Santa Cruz.

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Date Created: 09/07/15
Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers Roy D Yates and David J Goodman Problem Solutions Yates and Goodman312 332 342 352 362 363 and 373 Problem 312 On the X Y plane the joint PMF is y PXY y a To nd 0 we sum the PMF over all possible values ofX and Y We choose 0 so the sum equals one ZZPnyxy 2 2 cxy6c2c6cl4c x y x727072y717071 Thus 0 114 b PY ltX Pny0ilgtPny2ilgtPny20gtPny2lgt cc2c3c7c 12 C PY gtX Pny2ilgtPny20gtPny2lgtPny0lgt 3c2ccc7c12 d From the sketch of PXVY xy given above PX Y 0 6 PW lt1iPXY 27 1gtPX7Y 270gtPXY 271gtPX7Y07 1gtPX7Y071gt Sc 814 Problem 332 On the X 7Y plane the joint PMF is a To nd 0 we sum the PMF over all possible values ofX and Y We choose 0 so the sum equals one ZZPnyxy 2 2 cxy6c206cl4c x y x727072y717071 Thus 0 114 b PY ltX Pny01gtPny21gtPny20gtPny21gt cc203c7c 12 C PY gtX Pny21gtPny20gtPny21gtPny01gt 3c20cc 70 12 1 From the sketch of PXVY xy given above PX Y 0 9 PW lt1PX7Y 27 1gtPX7Y 270gtPX7Y 271gtPX7Y07 1gtPX7Y071gt Sc 814 Problem 342 In Problem 322 we found that the joint PMF of X and Y was y The expected values and variances were found to be EX0 VarX247 EY0 VarY57 We will need these results in the solution to this problem a Random variable W 2XY has expected value E 2 Z Z 2WPXYxygt x27072y17071 3 2 l l l awn 72ltogti imp own on 2 142 142 142 142 14 l 2 3 21i 20i 21i 2 14 2 14 2 14 6128 b The correlation ofX and Y is 7XY ElXYl 2 Z WPXJ x01 x27072y17071 72ltilgtlt3gt 72lt0gtlt2gt 72lt1gtlt1gt 2ltilgtlt1gt 2lt0gtlt2gt 2lt1gtlt3gt TT17T 14 14 47 c The covariance ofX and Y is GXYEPYl EleElYl 47 d The correlation coef cient is Cov XY 2 pX Y Var Var Y m Problem 352 The event B occurs iff X S 5 and Y S 5 and has probability 5 5 PB PlXS 5Y s 5 Z Z 001 025 x1y1 From Theorem 311 I mXV x A y EA PX Y B Jay 7 0 NB otherwise 004 x15y15 0 otherwise Problem 362 We can make a table of the possible outcomes and the corresponding values of W and Y In the following table we write the joint PMF PW7Y wy along with the marginal PMFs PY and PW Using the de nition PW Y PW7Y wy PY y we can nd the conditional PMFs of W given Y 1 w0 12 w 711 FWDWm 0 otherwise FWDWu 0 otherwise 1 w 0 FWDM2 0 otherwise Similarly the conditional PMFs of Y given W are lt17pgt2 y 0 1y1 PY WOl71gt7 0 otherwise PY WOlmi 172p2p2 y2 0 otherwise 0 otherwise Pyrww1gt1 F1 Problem 363 a First we observe thatA takes on the values SA 711 while B takes on values from SE 01 To construct a table describing PAJ 5117 we build a table for all possible values of pairs AB The general form of the entries is PAJab 120 121 6171 iPBlA071gtPA71gtiPBlA171gtPA 1gti a1 PBiltoI1gtPAlt1gt PBMlt1I1gtPAltIgt Now we ll in the entries using the conditional PMFs PB bla and the marginal PMF PA a This yields b 1 m 13gt13gt 23gt13gt a 12gt23gt 12gt23gt which simpli es to m wwlanmau IEEHHMIIMI IEEEIIEIIEI b If A 1 then the conditional expectation of B is 1 EBA 1 1 bPBVb1 PBM11 12 0 c Before nding the conditional PMF PA 3 al 1 we rst sum the columns of the joint PMF table to nd 49 b 0 PBbgt 59 121 The conditional PMF of A given B l is P 611 517 awwval32a d Now that we have the conditional PMF PAM al 1 calculating conditional expectations is easy we 11 21 1an am 712sgtlt3sgt 15 a 7 EAZB 1 2 azPAw all 25 35 1 2171 The conditional variance is then VarAB 1 EAZB 1 7EAB 12 14152 2425 e To calculate the covariance we need EA 2 aPAltagt 7113gt1lt23gt 13 117171 EB i bPB b 049159 59 b0 EAB 2 i abPAJg my a7171b0 710gt19gt11gt29gt10gt13gt11gt13gt 19 The covariance is just CovAB EAB7EAEB 1971359 7227 Set Theory Venn Diagram Basic Properties of Sets A U j A A 1 gt Logarithms Basic Properties of Logarithms logbm logam logba logamy logam 10ng logaE logam 7 logay y logaml7 blogam Combinatorics Formula for Combination Formula for Binomial Expansion M byv i lt I gt n0 Differential Calculus Power Rule Product Rule Chain Rule Partial Derivatives Integral Calculus Substitution Method Double Integrals Exercise 1 Let A7 B7 and C be sets such that A O B C Complete the following identities a A C 10 AUG c AUB C Exercise 2 a A simple pocket calculator can only compute natural logarithms Suppose you want to compute the logarithm base 2 of a number m using such a calculator How do you do it 1 9272 b Compute 1093 Exercise 3 40 a Compute lt 38 gt b Using the binomial expansion formula7 compute 84 by expanding 10 7 24 Exercise 4 Compute the following derivatives or partial derivatives a loglom 1 iv 11 c m bm24 5 5 7 d E as my 8 6e257ltwygt2 Exercise 5 Compute the following de nite integrals 1 a f amz bz3dz 71 2 b1fzdz

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