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Unformatted text preview: /. ' Assignment 2: Bayesian Networks and Classifiers Name: Siyu Liu
Student Number.“
CDF Email: 1. Bayesian Network Inference (4 points). In the Bayesian Network shown below, all variables are
binary with values Dom(A)={A,~A}, Dom(B)={B,~B}, etc. @ﬁ
90 The probability table values are as follows: P(A)=0.9 P(B)=0.7 P(G)=0.75 P(D  C) = 0.9 , P(D  ~C) = 0.06 P(C I AB) = 0.95 , P(C I A,~B) = 0.01 , P(C I ~A,B) = 0.6 , P(C I ~A,~B) = 0.03
P(E I C,G) = 0.95 , P(E I C,~G) = 0.01 , P(E I ~C,G) # 0.9 , P(E  ~C,~G) = 0.3
P(F l (3,6) = 0.8 , P(F I C,~G) = 1.0 , P(F I ~C,G) = 0.1 , P(F  ~C,~G) = 0.5  Compute the following probabilities (show your work): P(EI A) (1/2 point)
P(D I A,B) (1/2 point)
P(F I ~E) (1.5 points)
P(G I B,F) (1.5 points) Solution: 8) P(E  A) = '3‘“) P(A) _ 23,13,054; P(A,B,C.D.E.F.G)
HA) 11111 P(A) = 23 1303) EC P03151025 P03 I (02.; 13mm: P(F I c, G) P(E  c. G) '
= 2:3 13002.; P0:  A1025 90) I a) 26 mm Pa: I c.9130: I 99) Define term(G) = 2C 9(0) 2:, pa: I c, G) 1205 I c, G)
term(~G) = EC P(~G) ZF P(F I C, ~G) PCE I C, ~G) thEILZB P03) 2c PU:  A: B) 20 PC}  C) 20 PH?) 25 HF I C. @1303 I C; G)
= 213 P0321: P(C  A; 1920 P03 I C) termm) +
23 P (13) 2c FCC  A, B) 20 PG) I C) term(~G)
= 23 13(3) 2c PCC  A: 3) termCG) Zn PCB I C) +
213 PCB) 2c PCC  A 3) term(~G) 20 PG) I C)
— ‘23 13(3) (P (C IA B) (termm C) 20 PCD I CD +
' P(~C I A, B)(term(G, ~C) ED P(D I ~C))) +
21; 13(3) (FCC I A B) (term(~G C) 20 P (D  CD +
_ P(~C I A B)(term(~G 9'0 20 PC) I ~CD) 0") Notice that ED P(D I C)‘ = 2D P(D I ~C) = 2F P(F I C, G) = 1, and term(G, C) = P(G)P(E I C, G) = c1 ' term(G,~C) = P(G)P(E  ~c, G) = c2 term(~G, C) = P(~G)P(E I C,~G) = c3 term(~G, ~C) = P(~G)P(E I ~C, ~G) = c4 (:5) = 23 P(B) (PG: IA, B)c1 + P(~C  A, B)c2) + 23 PCB) (P(C I A, B)c3 + P(~C IA, B)c4) : P(B)(P(C I A, B)c1 + P(~C I A, B)c2) +
P(~B)(P(C I A, ~B)C1 + P(~C I A, ~B)c2) +
P(B)(P(C I A, B)c3 + P(~C I A, B)c4) +
P(~B)(P(C I A, ~B)c3 + P(~C I A,~B)c4) = 0.7(0.95c1 + 0.05c2) + 0.3 (0.01c1 + 0.99c2) + 0.7(0.95c3 + 0.05c4) + 0.3(0.01c3 + 0.99c4)
=‘0.7(0.95 I. 0.75 1 0.95 + 0.05 =5 0.75 1. 0.9) + 0.3(0.01 * 0.75 1. 0.95 + 0.99 * 0.75 a: 0.9) +
0.7(0.95 1 0.25 I: 0.01 + 0.05 * 0.25 1. 0.3) + 0.3(0.01 * 0.25 I: 0.01 + 0.99 * 0.25 a: 0.3)
= 0.7 * 0.710625 + 0.3 1. 0.575375 + 0.7 1 0.005125 + 0.3 ... 0.074275 = 0.72662
\I I, b) P(D I A, B) = P03“) P (AB) ﬂ W
— P(A,B) : P(A)P(B) 2C,E,F,G P(F IC,G)P{E IC,G)P(D I C)P(C IA,B)P (G) since A J_ B
PCA)F(B) = 2c P(C I A. 13) P03 I C) 25 PG) 213 PCE I C. G) 21: PG“ I C: G) (*)
Deﬁne term(G) = 2C 13(0) 2E P(E  c, G) 2F P(F  c, G) '
term(~G) = 2c P(~G) 2E P03 I C. ~G) 2F P(F I C, 4G)
then, (=9 _ ‘_
=zc P(C A,B) P(D I C)term(G) + 2C P(C  A, B) P03  C)term(~G)
= (P(C  A, B)P(D I C)term(G) + P(~C I A, B)P(D I ~C)term(G)) +
(P(C I A, B)P(D I C)term(~G) + P(~C I A, B)P(D I ~C)term(~G))
Notice if we have C, G in the evidence set, then 25 P(E I C, G) 2 SF P(F I C, G) = 1,
regardiess what C and G are, which means term(G) = P(G) and term(~G) = P(~G)
then, (* ’) '
= (P(C  A,B)P(D  C)P(G) + P(~C I A, B)P(D I ~C)P(G)) +
(PCC  A,B)P(D I C)PC~G) + P(~C  A.B)P(D I ~C)P(~G))
'='(P(C I'A,‘B)P(D  C) + P(~C I A,B)P(D  ~C))(P(G) + P(~G))   
= POE I A, B)P(D I C) + P(~C  A,B)P(D I ~C) —
= 0.95 * 0.9 + 0.05 1: 0.06 = 0.858 \
I ’I/ C) P(F  ~E) = POM” P (~E) __ EA,B,C,D,G P(A,B.C.D.~E.F,G)
ZA.B,C.D,F,G P (A,B,C,D,~E,F,G) 2 EA P0923 1’03) 2c PCC IAB) Zn PCG)P(F I C.G)P(~E I (LG) 21) P03 I C)
2A PUDZB 13(3)::[2 NC I AB) 2:: P(G)P(~E  CrG)ZF P(F I (LG) 21) PCD I C) The numerator 2A,B,C,D,G P(F I C, G)P(~E I C, G)P(D I C)P(C I A, B)P(A)P(B)P(G)
=ZA P(A) EB P03) 2:: P“: I A,.B) 2G P(G) PU“ I C. G)P(~E I C; (3)21) PU) I C) (1)
Define term(G) 2 2c P(G)P(F I C, G)P(~E I C, G) term(~G) = 2c P(~G)P(F I C,~G)P(~E I C,~G)
then, (1) _ '
= EA PTA) EB PCB) 2c P“: I A: B) term(G) Zn P(D I C) + EA PC1023 1’03) 2c PU:  A. 13) term(~G) Zn PU) I C) WWW“ ' = EA POD 21; P03) (P(C I A. B)term(G) En P(D I C) + P(~C  A. B)term(G) ED P(D' I ~C)) +
2A 130023 P(B) (P(C I A. B)term(~G) in P0) I C) + P(~C I A. B)term(~G) 25 Pa) I ~C))
= 2A P(A) EB P(B) (P(C  A, B)term(G)  1 + P(~C I A, B)term(G)  1) 4—
2A P(A) EB P(B) (P(C  A, B)term(~G)  1 + P(~C I A, B)term(~G)' 1)
= 2A 1’09 213 P03) (P(C I A, BDPCG)P(F I C. G)P(~E  C) G) +
P(~C I A, B)P(G)P(F I ~C, G)P(~E I ~C, 8)) 4—
2A 130023 P(B) (P(C  A, B)P(~G)P(F I C. ~G)P(~E  C,~G) +
P(~C I A, B)P(~G)P(F I ~C,~G)P(~E I ~C,~G)) (2)
Notice that if C is in evidence set, then
P(G)P(F  C, G)P(~E I C, G) = 0.75 4 0.8 r: 0.05 = 0.03 = c1,
P(G)P(F  ~C,G)P(~E I ~C, G) = 0.75 =I= 0.1 * 0.1 = 0.0075 = (:2,
P(~G)P(FI C,~G)P(~E  C,~G) = 0.25 * 1.0 4 0.99 = 0.2475 2 c3,
P(~G)P(F ~C, ~G)P(~E I ~C,~G) = 0.25 * 0.5 =I= 0.7 = 0.0875 = c4 are constant
thus, (2)
22A P(A) EB P(B) (P(C I A, B)c1 + P(~C IA, B)c2) +
2A P(A) 2:3 P(B) (P(C I A, B)c3 + P(~C  A, B)c4)
= P(A)P(B)(P(C  A, B)c1 + P(~C I A, B)c2 + P(C I A, B)c3 + P(~C I A, B)c4) +
P(A)P(~B)(P(C I A, ~B)c1 + P(~C  A, ~B)cZ + P(C I A, ~B)c3 + P(~C I A, ~B)c4) +
P(~A)P(B)(P(C  ~A,B)cl + P(~C  ~A, B)c2 + P(C I ~A,B)c3 + P(~C I ~A, B)c4) +
P(~A)P(~B)(P(C I ~A, ~B)cl + P(~C  ~A, ~B)c2 + P(C  ~A, ~B)c3 + P(~C I ~A,~B)c4)
= 0.9 * 0.7(0.95c1 + 0.05c2 + 0.95c3 + 0.05c4) + 0.9 * 0.3 (0.01c1 + 0.99C2 + 0.01c3 + 0.99c4) +
0.1 * 0.7(0.6c1 + 0.4c2 + 0.6c3 —I— 0.484) + 0.1 * 0.3(0.03£1 + 0.97c2 + 0.03c3 + 0.97c4)
= 0.63(0.95 * (0.03 + 0.2475) + 0.05(0.0075 + 00875)) +
0.27(0.01 * (0.03 + 0.2475) + 0.99(0.0075 + 00875)) +
0.07(0.6 * (0.03 + 0.2475) + 0.4 4: (0.0075 + 00875)) +
0.03(0.03 =I= (0.03 + 0.2475) + 0.97 * (0.0075 + 00875))
= 0.63(0.95 * 0.2775 + 0.05 * 0.095) + 0.27(0.01 =I= 0.2775 + 0.99 * 0.095) +
0.07(0.6 4 0.2775 + 0.4 >I= 0.095) + 0.03(0.03 * 0.2775 + 0.97 =1= 0.095)
= 0.63 * 0.268375 + 0.27 * 0.096825 + 0.07 a: 0.2045 + 0.03 * 0.100475 4 0.212548 = P(F,~E) On the other hand, the denominator P(~E) = EA,B,C,D,F,G P(A, B, C, D, ~E, F, G)
= 2A 130023 P(B) 2.; P(C  A. 3m. P(G) P(~E I c. G) 21: P(F I c. 6) ED P0) I C) (3)
Similarly, define term’(G) = 2c P(G) 2F P(F I C, G)P(~E I C, G)
term’(~G) — 2,; 13(4):, P(F I c ~G)P(~E  c, ~G)
Then, by follow the same steps as the numerator, (3)
— —ZA P(A) EB P(B) (P(C I A, B)term (G) + P(~C IA, B)term' (8)) —I—
2,1300}:B P(B) (P(C I A, B)term’(~G) + P(~C  A, B)term'(~G)) (4)
When C and G are known, then 21: P(F I C, G) = 1, then (4) = 2A P(A) 23 P(B) (0(0 1 A,B)P(G)P(~E 1 c, G) + P(~C 1 A, B)P(G)P(~E 1 ~12, 0)) +
21. P00 23 P(B) (PU: l A,B)P(~G)P(~E  C, ~G) + P(~C 1A.B)P(~G)P(~E l ~C, ~G)) (5)
Similarly, we have
P(G)P(~E  c, G) = 0.75 4 0.05 = 0.0375 = c5,
P(G)P(~E 1 ~c, G) = 0.75 4 0.1 = 0.075 = c6,
P(~G)P(~E 1 C,~G) = 0.25 4 0.99 = 0.2475 = c7.
P(~G)P(~E _~C,~G) = 0.25 * 0.7 = 0.175 = (:8 are constant, then (5)
: 2A MA):B P(B) (P(c 1 A, B)c1 + P(~C 1 A,B)c2) +
211 13010213 P(8) (P(0 1 A, B)c3 + P(~C 1 A, B)c4)
= P(A)P(B)(P(c 1A, B)cS + P(~C 1 A,B)c6 + P(C 1 A, B)c7 + P(~c 1 A,B)c8) +
P(A)P(~B)(P(C 1 A, ~B)CS + P(~C  A,~B)c6 + P(c 1 A,~B)c7 + P(~C 1 A,~B)c8) +
P(~A)P(B)(P(C 1 ~A,B)c5 + P(~C1~A,B)c6 + P(C1~A,B)c7 + P(~C 1 ~A,B)c8) +
P(~A)P(~B)(P(C 1 ~A,.~B)c5 + P(~C 1 ~A, ~B)c6 + P(C 1 ~A, ~B)c7 + P(~C 1 ~A,.~B)c8) _ = 0.9 =5 0.7(0.95C5 + 0.0506 + 0.95C7 + 0.05128) + 0.9 * 0.3(0.01C5 + 0.99c6 + 0.01c7 + 0.99118) + . 0.1 * 0.7(0.6C5 + 0.4c6 + 0.6137 + 0.4c8) + 0.1 * 0.3(0.03C5 + 0.97c6 + 0.03117 + 0.97c8) = 063(095 * (0.0375 + 0.2475) + 0.05(0.075 + 0.175)) +
027(001 4 (0.0375 + 0.2475) + 0.99(0.075 + 0.175)) +
0.07(0.6 4 (0.0375 + 0.2475) + 0.4 * (0.075 + 0.175)) +
003(003 * (0.0375 + 0.2475)+ 0.97 4 (0.075 + 0.175))
= 0.63(0.95 * 0.285 + 0.05 ,* 0.25) + 027(001 .. 0.285 + 0.99 .. 0.25) + 0.07(0.6'* 0.285 + 0.4 * 0.25) + 0.03(0.03 =1: 0.285 +0.97 * 0.25) 
= 0.63 * 0.28325 + 0.27 * 0.25035 + 0.07 * 0.271 + 0.03 * 0.25105 = 0.2725435 = P(~E) Therefore, P(F l ~E) = 119251;? 2 g% z 0.779868 \ 5 d) P(G 1 3,17) = 70.3.7) 703,7) _ 2.1.0.012 P(A,B,C,D,E,F,G)
ZA,C,D,E,G P(A.B.C.D.E.F.G) = P030512. POD 21: Pm I ArB) ED PCD I C)P(G)P(F  C.G)2£ 1903  C.G)
P03) EA P00 20 PU? I A.B)20 PU) E (3)20 P(G)P(F  (LG) 28 P03 IC.G) : EA PU!) 2c P(C 13:3)20 PG) l C)P(G)P(F  (LG) Z13 P(E I (LG)
2A P(A) 2c PCC 1AJB)ED P(D I C) 20 PCG)P(F I (LG) 28 PCE  CIG) The numerator 2A P(A) 2c P(C 1 A, 13) Zn 13(1) 1 C) 13(0) P(F  (1sz P03  c, 0) mem m ———————__...__..._. _ _. v f.
‘ ' i: z, P(A) (pa: I A, B) 25 PG) I C) P(G)P(F I 11.9)26 P(E I QG) + P(~C  A, B) Zn 1’0) I "*0 P(G)P(F  ~C. G) 22 1’03  ~C, G)) (1)
Notice that when C, G both known. . 26 Fail c, G) = 26 P03 I C) = 2.; P13  ~c. G) = 26 MD I ~C) '= 1, then (I) '= 2A P—(A) (P(G  A, B)  1  P(G)P(F I c, G)  1 + P(~C I A, B)  1 'P(G)P(F  ~C, G)  1)
= P(A)(P(C I A, B)P(G)P(F I c, G) + P(~C I A, B)P(G)P(F I ~C, G)) +
P(~A)(P(C I ~A,B)P(G)P(F I C, G) + P(~C I ~A, B)P(G)_P(F I ~c, G))
'2 0.9(095 * 0.75 4 0.8 + 0.05 a: 0.75 4 0.1) + O.1(0.6 A 0.75 A 0.8 + 0.4 4 0.75 A 0.1)
= 0.9 4 0.57375 + 0.1 =1 0.39 : 0.555375 = P(G, B, F)/P(B) Similarly, define term(G) = 2c P(G) P(F I C, G) XE P(E  C, G)
term(~G) = 2c P(~G) P(F I C, ~G) 2E P(E  C,~G) The denominatorZA P(A) EC P(C  A, B) 2D P(D I C) ZG P(G) P(F I C, G) 2E P(E I C, G)
= 211 NM 2c P(C  A. 3) 20 PO)  C) term(G) + EA P(A) 2c P(C  A) B) 20 ND  C) term(~G)
= 2A P(A) (190: I A, B) Z], P(D I G) term(G) + P(~C I A, B) 2,, P(D I ~c) term(G)) + 2,, P(A) (P(G I A, 8) ED P(D I C) term(~G) + P(~C I A, B) 2,, P(D I ~c) term(~G)) (2)
Notice that when G, G both known, ED PC) I C) = 2013(1) I NC) = 25 P03 I ~C,G) = 213 P(E I C’G) = 2}}: P(E I "CH”CD = Es P(EII ~C,~G) = 1, then (2) = 2A P(A) (P(C I A,B)P(G)P(F I c, G) + P(~C I A, B)P(G)P(F I ~C,G)) +
2A P(A) (P(C I A, B)P(~G)P(F I C,~G) + P(~C I A, B)P(~G)P(F I ~c, ~G)))
= P(A)(P(C IA, B)P(G)P(F I c, G) + P(~C I A, B)P(G)P(F ~C,G)) +.
P(~A)(P(C I ~A,B)P(G)P(F I G, G) + P(~C I ~A,B)P(G)P(F I ~c, G)) +
P(A)(P(C I A, B)P(~G)P(F I C, ~G) + P(~C I A, B)P(~G)P(F I ~c, ~G))) +
P(~A)(P(C I ~A,B)P(~G)P(F I C,~G) + P(~C I ~A,B)P(~G)P(F I ~C,~G)))
= 0.9(095 4 0.75 4 0.8 + 0.05 4 0.75 4 0.1) + 0.1 * (0.6 4 0.75 a: 0.8 + 0.4 A 0.75 a: 0.1) +
0.9(0.95 4 0.25 4 1.0 + 0.05 4 0.25 4 0.5) + 0.1 4 (0.6 4 0.25 4 1.0 + 0.4 .1. 0.254 0.5)
= 0.9 4 0.57375 + 0.1 4 0.39 + 0.9 4 0.24375 + 0.1 4 0.2 = 0.79475 = P(B,F)/P(B) Therefore,P(G I B, F) = P(G’B’F) = ”555375 =9 0.6988047 \ ,5 P(B,F) 0.79475 /\ r»va—v—v——v—«—«n——..__________._____.
I 2. ‘(5 points, 1 point each) For the Bayesian Network below, answer the following questions. Justify
your answers using the dseparation criterion. # QQKMI List all nodes dependent on 4.
Dependent set of 4 = {6, 8, 9, ll} ’\ 6 is dependent on 4 since 64 forms a path with no blocks in between;
8 is dependent on 4 since 864 forms a path with no blocks in'between given 6 not in evidence set; 9 is dependent on 4 since 96—4 forms a path with no blocks in between given 6 not in evidence set;
11 is dependent on 4 since 11864 forms a path with no blocks in between given 6, 8 not in evidence set. 1 is independent from 4, since 152—8—64 is the only path but blocked by 5 given 5 not in evidence set;
2 is independent from 4, since 2864 is the only path but blocked by 8 given 8 not in evidence set; 3 is independent from 4, since 3—64 is the only path but blocked by 6 given 6 not in evidence set; 5 is independent from 4, since, 5—2864 is the only path but blocked by 8 given 8 not in evidence set; 7 is independent from 4, since 4697 is the only path but blocked by 9 given 9 not in evidence set; 10 is independent from 4, since 9 depends on 4 but 9 and 10 are independent given 7 not in evidence set. List all nodes dependent on 4 given 11. Dependent set of4 given 11 i {2, 3, 5, 6, 8, 9} \ 2 is dependent on” 4 since 2—864 is a path with no blocks in between given 11 in the evidence set, i.e.
path 286 no longer blocked because 11 is a decedent of 8; ..._.._~.__..._.._....._._....mummwmwwwwwacuwwmmmhmmmmm 3 is dependent on 4 since 364 is a path with no blocks in between given I] in the evidence set, i.e. path
364 no longer bidcked because 1 1 is a decedent of 6; ' 5 is dependent on 4 since 5—2864 is a path with no blocks in between given 11 in the evidence set,
same reason with 2; 6 is dependent on 4 since 64 forms a path with no‘ blocks in between; 8 is dependent on 4 since 864 forms a path with no blocks in between given 6 not in evidence set; 9 is dependent on 4 since 964 forms a path with no blocks in between given 6 not in evidence set; 7 is independent from 4, since 469—7 is the only path but blocked by 9 given 9 not in evidence set;
10' is independent from 4, since 9 depends on 4 but 9 and 10 are independent given 7 not in evidence set. List all nodes dependent on 1 given 5.
Dependent set of 1 given 5 = {2, 8, 11} \ 2 is dependent on 1 given 5, since 251 fonns a path withno blocks in between given 5 in evidence set;
8 is dependent on 1 given 5, since 8251 forms a path with no blocks in between given 5 in evidence set but 2 not;
11 is dependent on 1 given 5, since 11—82—51 forms a path with no blocks in between given 5 in evidence 'set but 2, 8 not. 3 is independent from 1 given 5, since 3682a5—l is the only path but blocked by 8 given 8 not in
evidence set; ‘ ' ' ' , '
4 is independent from 1 given 5, since 468251 is the only path but blocked by 8 given 8 not in evidence set; 
6 is independent from 1 given 5, since 682—51 is the only path but blocked by 8 given 8 not in evidence set;
7 is independent from 1 given 5, since 79—6—8—2~5~1 is the only path but blocked by 8 given 8 not in evidence set;
9 is independent from 1 given 5, since 968251 is the only path but blocked by 8 given 8 not in evidence set; .
10 is independent from 1 given 5, since 10796825~l is the only path but blocked by 8 given 8 not in evidence set. ' List all nodes dependent on 7 given 9. Dependent set of? given 9 = {3, 4, 6, 3, 10, 11} « i 3 is dependent on 7 given 9, since 3697 is a path with no blocks in between given 9 in evidence set but 6 not;
4 is dependent on 7 given 9, since 46—9—7 is a path with no blocks in between given 9 in evidence set but 6 not;
6 is dependent on 7 given 9, since 6937 is a path with no blocks in between given 9 in evidence set;
8 is dependent on 7 given 9, since 8—6~97 is a path with no blocks in between given 9 in evidence set but 6 is not;
10 is dependent on 7 given 9, since 107 is a path with no blocks in between;
11 is dependent on 7 given 9, since 1186—97 is a path with no blocks in between given 9 in evidence set but 6, 8 is not. 1 is independent from 1 given 9, since 152—8697 forms the only path but blocked by 5 given 5 not in evidence set;
2 is independent from 7 given 9, since 2869—7 forms the only path but blocked by 8 given 8 not in evidence set;
5 is independent from 7 given 9, since 528697 1°0th the only path but blocked by 8 given 8 not in evidence set.
List all nodes dependent on 2 given 8.
Dependent set of 2 given 8 =,{3, 4, 5, 6, 9}  N 3 is dependent on 2 since 3682 is a path with no blocks in between given 8 in evidence set but 6 not;
4 is dependent on 2 since 4682 is a path with no blocks in between given 8 in evidence set but 6 not;
5 is dependent on 2 since 52 is a path with no blocks in between; 6 is dependent on 2 since 682 is a path with no blocks, in between given 8 in evidence; 9 is dependent on 2 since 9682 is a path with no blocks in between given 8 in evidence set but 6 not. 1 is independent from 2 since 152 forms the only path but blocked by 5 given 5 not in evidence set;
7 is independent from 2 since 7—96—82 forms the only path but blocked by 9 given 9 not in evidence set;
10 is independent from 2 since 10—79682 forms the only path but blocked by 9 given 9 not in evidence set;
11 is independent from 2 since 1182 forms the only path but blocked by 8 given 8 in evidence set. ...
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 Fall '05
 Sheila
 Artificial Intelligence

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