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COMP9418 – Advanced Topics in Statistical Machine Learning
创建日期：2024-06-05 16:22:08
标签： COMP9418
Advanced Topics in Statistical Machine Learning

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COMP9418 – Advanced Topics in Statistical Machine Learning

Lecture: Propositional Logic and Probability Calculus
Topic: Questions from lecture topics
Question 1
For each of the following pairs of sentences, decide whether the first sentence implies the second. If the
implication does not hold, identify a world in which the first sentence is true, but the second is not.
a. (A⇒ B) ∧ ¬B and A.
b. (A ∨ ¬B) ∧B and A.
c. (A ∨B) ∧ (A ∨ ¬B) and A.
Answer
Let us build a table with all possible worlds for each item.
a. (A⇒ B) ∧ ¬B and A
A B ¬B A⇒ B (A⇒ B) ∧ ¬B (A⇒ B) ∧ ¬B |= A?
1 1 0 1 0 Yes
1 0 1 0 0 Yes
0 1 0 1 0 Yes
0 0 1 1 1 No
b. (A ∨ ¬B) ∧B and A
A B ¬B A ∨ ¬B (A ∨ ¬B) ∧B (A ∨ ¬B) ∧ ¬B |= A?
1 1 0 1 1 Yes
1 0 1 1 0 Yes
0 1 0 0 0 Yes
0 0 1 1 0 Yes
c. (A ∨B) ∧ (A ∨ ¬B) and A
A B ¬B A ∨B A ∨ ¬B (A ∨B) ∧ (A ∨ ¬B) (A ∨ ¬B) ∧ ¬B |= A?
1 1 0 1 1 1 Yes
1 0 1 1 1 1 Yes
1
A B ¬B A ∨B A ∨ ¬B (A ∨B) ∧ (A ∨ ¬B) (A ∨ ¬B) ∧ ¬B |= A?
0 1 0 1 0 0 Yes
0 0 1 0 1 0 Yes
Question 2
Which of the following pairs of sentences are mutually exclusive? Which are exhaustive? If a pair of sentences
are not mutually exclusive, identify a world in which they both hold. If a pair of sentences are not exhaustive,
identify a world neither holds.
a. A ∨B and ¬A ∨ ¬B.
b. A ∨B and ¬A ∧ ¬B.
c. A and (¬A ∨B) ∧ (¬A ∨ ¬B).
Answer
a. A∨B and ¬A∨¬B are not mutually exclusive. The worlds both hold are marked with “yes” in column
“Both”. These sentences are exhaustive.
A B ¬A ¬B A ∨B ¬A ∨ ¬B Both
1 1 0 0 1 0 No
1 0 0 1 1 1 Yes
0 1 1 0 1 1 Yes
0 0 1 1 0 1 No
b. A ∨B and ¬A ∧ ¬B are mutually exclusive and exhaustive.
A B ¬A ¬B A ∨B ¬A ∧ ¬B
1 1 0 0 1 0
1 0 0 1 1 0
0 1 1 0 1 0
0 0 1 1 0 1
c. A and (¬A ∨B) ∧ (¬A ∨ ¬B) are mutually exclusive and exhaustive.
A B ¬A ¬B ¬A ∨B ¬A ∨ ¬B (¬A ∨B) ∧ (¬A ∨ ¬B)
1 1 0 0 1 0 0
1 0 0 1 0 1 0
0 1 1 0 1 1 1
0 0 1 1 1 1 1
Question 3
Suppose that 24% of the population are smokers and that 5% of the population has cancer. Suppose further
that 86% of the population with cancer are also smokers. What is the probability that a smoker will also
have cancer?
2
Answer
Let us take note of the given probabilities. We will use variable S to denote smokers and C for people with
cancer.
P (s) = .24
P (c) = .05
P (s|c) = .86
The exercise asks for P (c|s). Therefore, we will need to invert the conditional probabilities given. It is a
direct application of the Bayes rule.
P (c|s) = P (s|c)p(c)p(s) = .86×.05.24 = .1792
Therefore, it is expected that 17.91% of the smokers also have cancer.
Question 4
(After Koller & Friedman) Suppose a tuberculosis (TB) skin test is 95% accurate. If the patient is TB-infected,
the test will be positive with a probability of 0.95; if the patient is not infected, then the test will be negative
with a probability of 0.95. Now, suppose that a person gets a positive test result. What is the probability
that he is infected? Suppose that 1 in 1000 of the subjects who get tested is infected.
To answer this question, provide the following intermediate quantities:
P (TB = +) =
P (Test = +|TB = +) =
P (Test = +|TB = −) =
P (Test = +) =
Which equation provides a direct answer to the question posed in this problem?
Answer
From the problem statement, we get:
P (TB = +) = 0.001
P (Test = +|TB = +) = 0.95
P (Test = +|TB = −) = 0.05
For the next one, we can first use the sum rule in the following way:
P (Test = +) = P (Test = +, TB = +) + P (Test = +, TB = −)
As we do not have the joint probabilities, we need to calculate them from the conditional probabilities using
the product rule:
P (Test = +, TB = +) = P (Test = +|TB = +)P (TB = +) = 0.95× 0.001 = 0.00095
P (Test = +, TB = −) = P (Test = +|TB = −)P (TB = −) = 0.05× 0.999 = 0.04995
Therefore,
P (Test = +) = 0.00095 + 0.04995 = 0.0509
The problem asks for P (TB = +|Test = +). Therefore, we will need the Bayes Rule to invert P (Test =
+|TB = +)
P (TB = +|Test = +) = P (Test=+|TB=+)P (TB=+)P (Test=+) = 0.95×0.0010.0509 ≈ 0.0187
3
Thus, although a subject with a positive test is much more likely to be TB-infected than a random subject,
fewer than 2 per cent of these subjects are TB-infected.
Question 5
Consider the following distribution over three variables:
A B C P (A,B,C)
1 1 1 .27
1 1 0 .18
1 0 1 .03
1 0 0 .02
0 1 1 .02
0 1 0 .03
0 0 1 .18
0 0 0 .27
For each pair of variables, state whether they are independent. State also whether they are independent given
the third variable. Justify your answer.
Answer
Let us compute the marginals P (A,B), P (A,C) and P (B,C).
A B P (A,B)
1 1 .45
1 0 .05
0 1 .05
0 0 .45
A C P (A,C)
1 1 .30
1 0 .20
0 1 .20
0 0 .30
B C P (B,C)
1 1 .29
1 0 .21
0 1 .21
0 0 .29
Also, let us compute the priors P (A), P (B) and P (C).
4
A P (A)
1 .50
0 .50
B P (B)
1 .50
0 .50
C P (C)
1 .50
0 .50
From these probability distributions, we observe that no pair of variables is independent since P (X,Y ) ̸=
P (X)P (Y ). Now, let us use these tables to compute the conditional probabilities using the equation
P (X,Y |Z) = P (X,Y,Z)P (Z) .
C A B P (A,B|C)
1 1 1 .54
1 1 0 .06
1 0 1 .04
1 0 0 .36
0 1 1 .36
0 1 0 .04
0 0 1 .06
0 0 0 .54
Also, use the equation P (X|Y ) = P (X,Y )P (Y ) to compute the following conditional probabilities.
A C P (A|C)
1 1 .60
1 0 .40
0 1 .40
0 0 .60
C B P (B|C)
1 1 .58
1 0 .42
0 1 .42
0 0 .58
From these probability tables, we observe that A is not independent of B given C since P (A,B|C) ̸=
P (A|C)P (B|C).
We can use the same approach to conditional independence of B and C given A. First, compute P (B,C|A).
5
A B C P (B,C|A)
1 1 1 .54
1 1 0 .36
1 0 1 .06
1 0 0 .04
0 1 1 .04
0 1 0 .06
0 0 1 .36
0 0 0 .54
Also, we need to compute P (B|A) and P (C|A).
B A P (B|A)
1 1 .90
1 0 .10
0 1 .10
0 0 .90
C A P (C|A)
1 1 .60
1 0 .40
0 1 .40
0 0 .60
From these probability tables, we observe that B |= C|A since P (B,C|A) = P (B|A)P (C|A).
Finally, we test if A |= B|C. We start computing P (A,C|B).
B A C P (A,C|B)
1 1 1 .54
1 1 0 .36
1 0 1 .04
1 0 0 .06
0 1 1 .06
0 1 0 .04
0 0 1 .36
0 0 0 .54
Also, we need to compute P (A|B) and P (C|B).
B A P (A|B)
1 1 .90
1 0 .10
0 1 .10
0 0 .90
6
B C P (C|B)
1 1 .58
1 0 .42
0 1 .42
0 0 .58
From these probability tables, we observe that A is not independent of B given C since P (A,C|B) =
P (A|B)P (C|B).
Question 6
We have three binary random variables: season (S), temperature (T ) and weather (W ). Let us suppose you
are given the following conditional probability distributions (CPDs):
S P (S)
summer 0.5
winter 0.5
S T P (T |S)
summer hot 0.7
summer cold 0.3
winter hot 0.3
winter cold 0.7
S T W P (W |S, T )
summer hot sun 0.86
summer hot rain 0.14
summer cold sun 0.67
summer cold rain 0.33
winter hot sun 0.67
winter hot rain 0.33
winter cold sun 0.43
winter cold rain 0.57
Calculate the joint probability distribution P (S, T,W ) using the chain rule.
S T W P (S, T,W )
summer hot sun
summer hot rain
summer cold sun
summer cold rain
winter hot sun
winter hot rain
winter cold sun
7
S T W P (S, T,W )
winter cold rain
Answer
We need to remember that the chain rule has n! possibilities. We need to choose one that matches the
information at hand. In this case, we need to obtain P (S), P (T |S) and P (W |S, T ) from P (S, T,W ). Therefore,
we choose the following one:
P (S, T,W ) = P (S)P (T |S)P (W |T, S)
Now, filling in the table is similar to a database join operation. We multiply the matching rows for values of
T , S and W .
S T W P (S, T,W )
summer hot sun 0.5× 0.7× 0.86 = 0.301
summer hot rain 0.5× 0.7× 0.14 = 0.049
summer cold sun 0.5× 0.3× 0.67 = 0.1005
summer cold rain 0.5× 0.3× 0.33 = 0.0495
winter hot sun 0.5× 0.3× 0.67 = 0.1005
winter hot rain 0.5× 0.3× 0.33 = 0.0495
winter cold sun 0.5× 0.7× 0.43 = 0.1505
winter cold rain 0.5× 0.7× 0.57 = 0.1995
Question 7
(From Ben Lambert’s book “A Student’s Guide to Bayesian Statistics”) Suppose that, in an idealised world,
the ultimate fate of a thrown coin - head or tails - is deterministically given by the angle at which you throw
the coin and its height above the table. Also, in this ideal world, the heights and angles are discrete. However,
the system is chaotic1 (highly sensitive to initial conditions), and the results of throwing a coin at a given
angle (in degrees) and height (in meters) are shown in the following table.
Angle (degree) 0.2 0.4 0.6 0.8 1
0 T H T T H
45 H T T T T
90 H H T T H
135 H H T H T
180 H H T H H
225 H T H T T
270 H T T T H
315 T H H T T
a. Suppose all combinations of angles and heights are equally likely to be chosen. What is the probability
that the coin lands heads up?
b. Now suppose that some combinations of angles and heights are more likely to be chosen than others,
with the probabilities shown in the following table. What are the new probabilities that the coin lands
heads up?
1The authors of the following paper experimentally tested this and found it to be the case, "The three-dimensional dynamics
of the die throw", Chaos, Kapitaniak et al. (2012).
8
Angle (degree) 0.2 0.4 0.6 0.8 1
0 0.05 0.03 0.02 0.04 0.04
45 0.03 0.02 0.01 0.05 0.02
90 0.05 0.03 0.01 0.03 0.02
135 0.02 0.03 0.04 0.00 0.04
180 0.03 0.02 0.02 0.00 0.03
225 0.00 0.01 0.04 0.03 0.02
270 0.03 0.00 0.03 0.01 0.04
315 0.02 0.03 0.03 0.02 0.01
c. We force the coin thrower to throw the coin at an angle of 45 degrees. What is the probability that the
coin lands heads up?
d. We force the coin-thrower to throw the coin at a height of 0.2m. What is the probability that the coin
lands heads up?
e. If we constrained the angle and height to be fixed, what would happen in repetitions of the same
experiment?
Answer
a. A count of heads in the first table results in P (H) = 1940 .
b. A weighted average where the second Table provides the weights results in P (H) = 0.5.
c. A weighted average given an angle of 45 degrees results in P (H|angle = 45) ≈ 0.23.
d. Similar to the previous question but constrained to a height of 0.2, we have P (H|height = 0.2) ≈ 0.70.
e. If both height and angle are fixed, the outcome becomes deterministic.
Question 8
(After Koller & Friedman) An often useful rule in dealing with probabilities is known as reasoning by cases.
Let X, Y , and Z be random variables, then P (X|Y ) =∑z P (X, z|Y ).
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MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A PHIL2617 PHYS3116 MIE250 BCPM0008 ECN 620 QBUS6830 ECSE-415 ECON5102 MGMT20005 MAT 653 ECO 480 COMP 5/600 CSSE 5600 ME 571 7CCMMS61 CS 550 COMP2865 MATH3075 STATS 506 QTM 220 ISYS2120 FIT5106 STAT 153 MKTG3112 ECON2102 ELE2018 COMP3031 MSIN0010 SCC461 MATH237 CS246 CSC3065 CSE 101 ICTWEB411 COMP9313 ECOS 2025 FN620 MATH11111 MATH 3018 KE1068 FINA 6112 C++ MISM 6202 MSIN0105 CSC8631 MECO6935 COMP4121 ACCT2011 DPBS1150 EEL 3701C F19PL OPM 661 CSE 310 CS918 BIOL30001 ACCT3563 BANK3600 MET CS622 MATH367 ACCT5907 Continuing COMP5048 COMP61021 CS3334 GSOE9210 CSCI-570 COMP 5322 HW 2 ALY-6020 IEOR 4706 DATA 200 EBUS504 CGE12412 COMP4108 7SSMM800 ECMM171P MATE50002 CS225 RS 6003 BH2274 ELEC3111 MAT6669 ECON-UA 227 PHIL 1012 7SSMM603 TB028A P6 ECS708 CSC 427 CS6476 CSE 250A A1A2 ITS61504 EECS 376 model HW6 UV0010 ECON 6074 MG4F7 CMP-7030Y INFO3008 BFIP013 ECON 5068 MATH6002 CS3483 PSTAT 131 C31FM CSOR W4246 AAEC 5307 DATA ELEC362 COMSM0047 ECN6006 MA20224 BST351 MATH 323 MATH323 AA2021 COMP3811 COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 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ACCOUNTING BUSS1040 CHEM 181 MAT136H1 GY 6183 COMP3170 AGRI90089 PHYSICS 1002 ECON10003 IEOR E4004 ACS61012 MATH 40550 MKTG2113 TELE4653 ENSC1004 ECE 232E MK726 MTRX2700 SHEET 3 CSCA08 COMP30027 Math 2XX3 statistic MMAN3200 BISM7213 ECMM109 ECON 2022 FNCE 20005 GR5067 CSCB09 ME 3017 CS 5100 PHYS 2903 COMS 4771 COMP222 COMP 1046 CSC148H1 ISYS2038 FINS2615 OT5206 COMP2048 A6 MECH5770M MA826/20 CMPT 726 STA457 INFS6071 INFS588 integral ECON 216 IRDR 0017 MA323 PA 15213 PSYC2001 ENV222 Statistics ACCT5930 ACCT12 LAND2121 ME 5405 COMM1180 TELE3113 A7 ECE 568 ACCT2019 INFS1200 ENGG1340 CS 1400 ASST3 FINM7409 SWEN30006 INFO2222 FIN20150 SOSS1000 CSSE7231 FINS5513 COMS3200 CECN 702 ECON4060H MN3515 HW2 PSYA02H3 ECON 2112 COMP90051 CS 1652 MA30059 AF1605 MKTG7501 ECO-6003B COM4521 CTM ECON3350 COMP 3804 ELEC9762 BISM 7255 ST22 MATH20014 MNE3122 COMP343 MATH5165 ELE7029 ENGR30002 SEEM3500 ECON7350 RISK5002 FINM7405 COMP90059 MOS 3320B STAT 431 BISM7221 MATH3014 MTH783P EBA 35302 ACS133 BUS2206 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FN1024 MATH0094 MANG 6554 ECOS3003 ELEC6252 C2511C PHY328 EC1060 ECON32202 COMP4002-E1 MATH3063 EFIMM0124 QBUS2310 ECON4411 FUNDAMENTALS ACCT20001 18YX RPM COMP 2711 QBUS6600 BUSS5221 ST3189 FIT1047 MAT344 EGB375 MATH60013 MATH96011 COMP3557 MAST20029 ECM159 EAE1011 ENSCEN 105 SECTION A STA 141C SOSE 2022 PAM006 COMP0048 M40007 ISOM2500 CMPE 142 A06472 MCC150 ACST 8083 GEOG6087 ECON920 FINM7008 ENVENG 4110 COMPSCI 361 ENGRCEE 21 PHYS 1112 ACF2100 ECMM108 COMPS267F MAT00050M MAST30025 CSIT970 MSCI 224 INT202 PUBL0054 MSDM5004 BMAN 21020 BMAN21020 MGMT1002 MATH11150 LAWS7023 ACTL30002 CIVL3340 MATH1002 ECON 8069 COMPSCI 351 IBUS7302 STAT4528 CHEM1101 STAT3925 BUSN 7008 Comp 3120 ECOS2040 SWEN90010 ENGN6528 ECM158 ECON20022 COMP6452 Calculation CITS1401 S1889112 BUS 351 ACTL30008 FIT2091 ECON6002 IBUS6020 PHI 001 ECON30290 MATH7501 ECON10004 MFIN7017 ISYS90043 LAWS7012 ECON30025 COMP30023 COMM1190 ECOM20001 GSBS6009 STAT 337 EMET8002 ECOM90001 ECOM30001 FINC 2012 BUSINESS ENGCV512 ECO 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