Chapter 2: CONDITIONAL PROBABILITY

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Charles Boncelet, “Probability, Statistics, and Random Signals,"Oxford University Press, 2016. ISBN: 978-0-19-020051-0Chapter 2: CONDITIONAL PROBABILITYSections2.1 Definitions of Conditional Probability2.2 Law of Total Probability and Bayes Theorem2.3 Example: Urn Models2.4 Example: A Binary Channel2.5 Example: Drug Testing2.6 Example: A Diamond NetworkSummaryProblemsNotes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20201 of 25ECE 3800

Joint Probability – Compound ExperimentsDefining probability based on multiple events two classes for considerations. Independent experiments: The outcome of one experiment is not affected by pastor future experiments.o flipping coinso repeating an experiment after initial conditions have been restoredo Note: these problems are typically easier to solve Dependent experiments: The result of each subsequent experiment is affected bythe results of previous experiments.o drawing cards from a deck of cardso drawing strawso selecting names from a hato for each subsequent experiment, the previous results change the possibleoutcomes for the next event.o Note: these problems can be very difficult to solve (the “next experiment”changes based on previous outcomes!)Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20202 of 25ECE 3800

Conditional ProbabilityDefining the conditional probability of event A given that event B has occurred.Using a Venn diagram, we know that B has occurred then the probability that A hasoccurred given B must relate to the area of the intersection of A and B Pr A B Pr A B Pr B , for Pr B 0ThereforePr A B Pr A B , for Pr B 0Pr B For elementary events,Pr A B Pr A B Pr A, B , for Pr B 0Pr B Pr B Special cases for A B , B A , and B A .Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20203 of 25ECE 3800

Special cases for A B , B A , and B A . If A is a subset of B, then the conditional probability must bePr A B Pr A B Pr A , for A BPr B Pr B Therefore, it can be said thatPr A B Pr A B Pr A Pr A , for A BPr B Pr B If B is a subset of A, then the conditional probability becomesPr A B Pr A B Pr B 1 , for B APr B Pr B If A and B are mutually exclusive,Pr A B Pr A B 0 0 , for B A Pr B Pr B Conditional probabilities are generally not symmetric!Pr A B Pr A B Pr B , for Pr B 0Pr A B Pr B A Pr A , for Pr A 0andPr A B , for Pr B 0Pr B Pr A B Pr B A , for Pr A 0Pr A Pr A B Pr B A Pr A B Pr B Pr A B Pr B Pr B Pr A B Pr A Pr B Pr B Pr A Pr A Therefore, we would expect (unless Pr(A) and Pr(B) are equal)Pr B A Pr A B Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20204 of 25ECE 3800

Total ProbabilityFor a space, S, that consists of multiple mutually exclusive events, the probability of arandom event, B, occurring in space S, can be described based on the conditionalprobabilities associated with each of the possible events.Proof:S A1 A2 A3 AnandAi A j , for i jB B S B A1 A2 A3 An B A1 B A2 B A3 B An Pr B Pr B A1 Pr B A2 Pr B A3 Pr B An ButPr B Ai Pr B Ai Pr Ai , for Pr Ai 0ThereforePr B Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An Remember your math properties: distributive, associative, commutative etc. applied to settheory.Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20205 of 25ECE 3800

Continuing ConceptsConditional Probability When the probability of an event depends upon prior events.If trials are performed without replacement and/or the initial conditions arenot restored, you expect trial outcomes to be dependent on prior results orconditions.Pr A B Pr A when A follows B The joint probability is.Pr A, B Pr B, A Pr A B Pr B Pr B A Pr A Applicable for objects that have multiple attributes and/or for trialsperformed without replacement.Experiment 3: A bag of marbles, draw 2 without replacement Experiment: Draw two marbles, without replacement Sample Space: {BB, BR, BY, RR, RB, RY, YB, YR}Therefore1st-rows \2nd-col1st-Blue1st-Red1st-YellowTotal 2ndMarble2nd-Blue2nd-Red2nd-Yellow 3 2 6 6 5 30 2 3 6 6 5 30 1 3 3 6 5 3036 3 2 6 6 5 30 2 1 2 6 5 30 1 2 2 6 5 3026 3 1 3 6 5 30 2 1 2 6 5 30 1 0 0 6 5 3016Total1stMarble36261666Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20206 of 25ECE 3800

Resistor Example: Joint and Conditional ProbabilityAssume we have a bunch of resistors (150) of various impedances and powers Similar to old textbook problems (more realistic resistor values)50 ohms100 ohms200 ohmsWatt Subtotal¼ watt40201070½ watt30205551 watt1010525ohm Subtotal805020150Each object has two attributes: impedance (ohms) and power rating (watts)Better Be Right Or Your Great Big Plan Goes Wrong. (p purple for violet)Bat Brained Resistor Order You Gotta Be Very Good WithMarginal Probabilities: (uses subtotals)Pr(¼ watt) 70/150Pr(½ watt) 55/150Pr(1 watt) 25/150Pr(50 ohms) 80/150Pr(100 ohms) 50/150Pr(200 ohms) 20/150These are called the marginal probabilities when fewer than all the attributes areconsidered (or don’t matter).Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20207 of 25ECE 3800

Joint Probabilities: divided each member of the table by 150!50 ohms100 ohms200 ohmsSubtotal¼ watt40/150 0.26620/150 0.13310/150 0.06670/150 0.466½ watt30/150 0.2020/150 0.1335/150 0.03355/150 0.3661 watt10/150 0.06610/150 0.0665/150 0.03325/150 0.166Subtotal80/150 0.53350/150 0.33320/150 0.133150/150 1.0These are called the joint probabilities when all unique attributes must be considered.(Concept of total probability things that sum to 1.0)Conditional Probabilities:When one attributes probability is determined based on the existence (or non-existence)of another attribute. Therefore,The probability of a ¼ watt resistor given that the impedance is 50 ohm.Pr(¼ watt given that the impedance is 50 ohms) Pr(¼ watt 50 ohms) 40/80 0.5050 ohms¼ watt40/80 0.50½ watt30/80 0.3751 watt10/80 0.125Total80/80 1.0Simple math that does not work to find the solution: (they are not independent)Pr(¼ watt) 70/150 andPr(50 ohms) 80/150Pr(¼ watt) x Pr(50 ohms) 70/150 x 80/150 56/225 0.249 NO!!! Not independent!!Math that does workPr A B 40Pr A B Pr A, B 150 40 0.508080Pr B Pr B 150Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20208 of 25ECE 3800

What about Pr(50 ohms given the power is ¼ watt)¼ watt50 ohms100 ohms200 ohmsTotal40/70 0.57120/70 0.28610/70 0.14370/70 1.0Pr(50 ohms ¼ watt) Pr(50 ¼) 40/70 0.571Pr A B 40Pr A B Pr A, B 150 40 0.5717070Pr B Pr B 150Can you determine?Pr(100, ½) Pr(100) Pr(50, ½) Pr(½ 50) Pr(50 ½) Pr( 1 ) Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 20209 of 25ECE 3800

Using the “table” it is rather straight forward 50 ohms100 ohms¼ watt4020½ watt30201 watt1010Subtotal8050200 ohms105520Subtotal705525150Joint Probabilities Pr A B Pr A, B Pr(100, ½) Conditional Probabilities Pr A B Pr(½ 100) Pr(50, ½) Pr A B Pr A, B Pr B Pr B Pr(200 ½) Marginal Probability Pr B Pr B A1 Pr A1 Pr B An Pr An Pr( 1 ) Pr(100) Are there multiple ways to conceptually define such problems? Yes Relative Frequency Approach (statistics) Set Theory Approach (formal math) Venn Diagrams (pictures based on set theory)All ways to derive equations that form desired probabilities . The Relative Frequency Approach is the slowest and requires the most work!Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202010 of 25ECE 3800

A Priori and A Posteriori Probability (Sec. 2.2 Bayes Theorem)The probabilities defined for the expected outcomes, Pr Ai , are referred to as a prioriprobabilities (before the event). They describe the probability before the actualexperiment or experimental results are known.After an event has occurred, the outcome B is known. The probability of the eventbelonging to one of the expected outcomes can be defined asPr Ai B or from beforePr Ai B Pr B Ai Pr Ai Pr Ai B Pr B Pr Ai B Pr B Ai Pr Ai , for Pr B 0Pr B Using the concept of total probabilityPr B Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An We also have the following formsPr Ai B Pr B Ai Pr Ai Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An orPr A j B Pr B A j Pr A j n Pr B A Pr A i 11 Pr B A j Pr A j Pr B 1This probability is referred to as the a posteriori probability (after the event).It is also referred to as Bayes Theorem.Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202011 of 25ECE 3800

ExampleMore ResistorsBin 1Bin 2Bin 3Bin 4Bin 5Bin 6Subtotal10 ohm5000200800120010003700100 ohm300400600200800023001000 0020008600What is the probability of selecting a 10 ohm resistor from a random bin?Given Bin marginal probability Pr Bin # 16Pr 10 Bin1 5001000Pr 10 Bin 2 01000Pr 10 Bin3 2001000Pr 10 Bin 4 8001600Pr 10 Bin5 12002000Pr 10 Bin6 10002000Pr B Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An Pr B Pr B 500 10 1 200 1 800 1 1200 1 1000 1 1000 6 1000 6 1000 6 1600 6 2000 6 2000 65 1 0 1 2 1 5 1 6 1 5 1 23 1 0.383310 6 10 6 10 6 10 6 10 6 10 6 10 6Assuming a 10 ohm resistor is selected, what is the probability it came from bin 3?Pr Ai B Pr B Ai Pr Ai Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An Pr Bin3 10 Pr 10 Bin3 Pr Bin3 Pr 10 Bin1 Pr Bin1 Pr 10 Bin6 Pr Bin6 2 1Pr Bin3 10 10 6 0.086960.3833Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202012 of 25ECE 3800

Digital TransmissionsA digital communication system sends a sequence of 0 and 1, each of which are receivedat the other end of a link. Assume that the probability that 0 is received correctly is 0.90and that a 1 is received correctly is 0.90. Alternately, the probability that a 0 or 1 is notreceived correctly is 0.10 (the cross-over probability, ). Within the sequence, theprobability that a 0 is sent is 60% and that a one is sent is 40%. [S is Send and R isReceive]Pr S1 0.40Pr S 0 0.60Pr R1 S1 0.90 1 Pr R0 S 0 0.90 1 Pr R1 S 0 0.10 Pr R0 S1 0.10 Figure 1.7-1a) What is the probability that a zero is received?Total Probability: Pr B Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An Pr R0 Pr R0 S 0 Pr S 0 Pr R0 S1 Pr S1 Pr R0 0.90 0.60 0.10 0.40Pr R0 0.54 0.04 0.58b) What is the probability that a one is received?Pr R1 Pr R1 S 0 Pr S 0 Pr R1 S1 Pr S1 Pr R1 0.10 0.60 0.90 0.40Pr R1 0.06 0.36 0.42Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202013 of 25ECE 3800

Digital Communications (continued)Pr S1 0.40Pr S 0 0.60Pr R0 S 0 0.90Pr R1 S1 0.90Pr R1 S 0 0.10Pr R0 S1 0.10c) What is the probability that a received zero was transmitted as a 0?Bayes TheoremPr Ai B Pr S 0 R0 Pr B Ai Pr Ai Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An Pr R0 S 0 Pr S 0 Pr R0 S 0 Pr S 0 Pr R0 Pr R0 S 0 Pr S 0 Pr R0 S1 Pr S1 Pr S 0 R0 Pr R0 S 0 Pr S 0 0.90 0.60 0.54 0.9310.580.580.58d) What is the probability that a received one was transmitted as a 1?Bayes TheoremPr Ai B Pr S1 R1 Pr B Ai Pr Ai Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An Pr R1 S1 Pr S1 Pr R1 S1 Pr S1 Pr R1 Pr R1 S 0 Pr S 0 Pr R1 S1 Pr S1 Pr S1 R1 Pr R1 S1 Pr S1 0.90 0.40 0.36 0.8570.420.420.42Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202014 of 25ECE 3800

Digital Communications (continued)Pr S1 0.40Pr S 0 0.60Pr R0 S 0 0.90Pr R1 S1 0.90Pr R1 S 0 0.10Pr R0 S1 0.10e) What is the probability that a received zero was transmitted as a 1?Bayes TheoremPr Ai B Pr S1 R0 Pr B Ai Pr Ai Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An Pr R0 S1 Pr S1 Pr R0 S1 Pr S1 Pr R0 Pr R0 S 0 Pr S 0 Pr R0 S1 Pr S1 Pr S1 R0 Pr R0 S1 Pr S1 0.10 0.40 0.04 0.0690.54 0.040.580.58Note: Pr S1 R0 1 Pr S 0 R0 1 0.931 0.069f) What is the probability that a received one was transmitted as a 0?Bayes TheoremPr Ai B Pr S 0 R1 Pr B Ai Pr Ai Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An Pr R1 S 0 Pr S 0 Pr R1 S 0 Pr S 0 Pr R1 Pr R1 S 0 Pr S 0 Pr R1 S1 Pr S1 Pr S 0 R1 Pr R1 S 0 Pr S 0 0.10 0.60 0.06 0.1430.420.420.42Note: Pr S 0 R1 1 Pr S1 R1 1 0.857 0.143Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202015 of 25ECE 3800

Digital Communications (continued)Pr S1 0.40Pr S 0 0.60Pr R0 S 0 0.90Pr R1 S1 0.90Pr R1 S 0 0.10Pr R0 S1 0.10e) What is the probability that a symbol is received in error?Pr Error Pr R0 S1 Pr S1 Pr R1 S 0 Pr S 0 Pr Error 0.10 0.40 0.10 0.60 0.04 0.06 0.10Alternately,Pr Error Pr S 0 R1 Pr R1 Pr S1 R0 Pr R0 Pr Error 0.143 0.42 0.069 0.58 0.060 0.040 0.100Which way is easier?Notice that you were told originally that there was a 0.10 chance of receiving a symbol inerror!Summary:A-priori ProbabilitiesPr S 0 0.60Pr S1 0.40Pr R0 S 0 0.90Pr R1 S1 0.90Pr R1 S 0 0.10Pr R0 S1 0.10Pr R0 0.58Computed Total ProbabilityPr R1 0.42Bayes Theorem (A-posteriori Probabilities)Pr S1 R1 0.857Pr S 0 R0 0.931Pr S1 R0 0.069Pr S 0 R1 0.143Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202016 of 25ECE 3800

Example 1.7‐2: Amyloid test: is it a good test for Alzheimer’s?(Stark & Woods Example)An amyloid test for Alzheimer’s disease had reported results/information for people 65and older. Alzheimer’s patients with disease 90% had amyloid proteinAlzheimer’s free patients 36% had amyloid proteinGeneral population facts for Alzheimer’s Total Alzheimer’s probability 10% Total non-Alzheimer’s probability 1-10% 90%The setup – a-priori probabilities (given)Pr am Alz 0.90 and Pr am nonAlz 0.36Pr Alz 0.10 and Pr nonAlz 0.90What we want to know – if someone had the amyloid protein, what is the probability theyhave Alzheimer’s?Pr Alz am ?Using Bayes TheoremPr Alz am Pr am Alz Pr Alz Pr am But we need to know Pr am determine the total probability of the amyloid proteinPr am Pr am Alz Pr Alz Pr am nonAlz Pr nonAlz Pr am 0.90 0.10 0.36 0.90 0.414ThereforePr Alz am 0.90 0.10 0.21740.414The diagnosis is better than 10%, but for completeness what about the nonAlzheimer’s population 0.36 0.90Pr nonAlz am 0.78260.414 too high a probability for a good testNotes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202017 of 25ECE 3800

Textbook Urn Models on p. 34-36Textbook comment, p. 36: “Mathematicians have studied urn models for three centuries.Although they appear simplistic, urns and marbles can model numerous real experiments.Their study has led to many advances in our knowledge of probability.”Assume two urns with different distribution of marbles Urn 1: 5 red, 5 blueUrn 2: 2 Red, 4 BlueSelect an urn with probability 𝑃𝑟 𝑈and 𝑃𝑟 𝑈1𝑃𝑟 𝑈What is the probability of selecting a Red marble?Using total probability𝑃𝑟 𝑅𝑃𝑟 𝑅 𝑈𝑃𝑟 𝑅 𝑃𝑟 𝑈5 2 10 3𝑃𝑟 𝑅 𝑈2 1 6 31319 𝑃𝑟 𝑈49What is the probability of selecting two consecutive Red marbles from the same urnwhen (1) the urn is selected at random and (2) the first marble is red?𝑃𝑟 𝑅 𝑅𝑃𝑟 𝑅 𝑅𝑃𝑟 𝑅𝑃𝑟 𝑅 𝑅 𝑈 𝑃𝑟 𝑈𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑃𝑟 𝑈To compute the elements we need to determine U1 and U2 based probabilities. First,𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑈𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑈 𝑃𝑟 𝑅 𝑈𝑃𝑟 𝑈and continuing𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑈 𝑃𝑟 𝑅 𝑈𝑃𝑟 𝑈𝑃𝑟 𝑅 𝑅 𝑈 𝑃𝑟 𝑅 𝑈but𝑃𝑟 𝑅 𝑈𝑃𝑟 𝑅 𝑅 𝑈51049Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202018 of 25ECE 3800

Therefore4 5 9 10𝑃𝑟 𝑅 𝑅 𝑈29Similarly,𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑈𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑈 𝑃𝑟 𝑅 𝑈𝑃𝑟 𝑈and𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅 𝑈 𝑃𝑟 𝑅 𝑈𝑃𝑟 𝑈𝑃𝑟 𝑅 𝑅 𝑈 𝑃𝑟 𝑅 𝑈With26𝑃𝑟 𝑅 𝑈15𝑃𝑟 𝑅 𝑅 𝑈1 2 5 6𝑃𝑟 𝑅 𝑅 𝑈115Finally𝑃𝑟 𝑅 𝑅𝑃𝑟 𝑅 𝑅𝑃𝑟 𝑅 𝑅𝑃𝑟 𝑅𝑃𝑟 𝑅 𝑅𝑃𝑟 𝑅𝑃𝑟 𝑅 𝑅𝑃𝑟 𝑅 𝑅 𝑈2 2 9 349 𝑃𝑟 𝑈𝑃𝑟 𝑅 𝑅 𝑈𝑃𝑟 𝑅1 1 15 3𝑃𝑟 𝑅 𝑅𝑃𝑟 𝑅23135494274914523604 5 313549 𝑃𝑟 𝑈23135490.3833Note that you also computed the probability of two consecutive red marbles from arandom urn23𝑃𝑟 𝑅 𝑅0.1704135Be careful, the definition of this term changes in the next variation! In the next problem,the urn isn’t necessarily the same when drawing the second marble!Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202019 of 25ECE 3800

Digital Transmissions Textbook version on p. 36-37𝑃𝑟 𝑋1𝑝𝑃𝑟 𝑌 𝑋1𝑃𝑟 𝑌 𝑋𝜀𝑃𝑟 𝑋𝜀𝑃𝑟 𝑌 𝑋𝑃𝑟 𝑌 𝑋𝑝1𝜈𝜈Determine total probability 𝑃𝑟 𝑌 and 𝑃𝑟 𝑌𝑃𝑟 𝑌𝑃𝑟 𝑌 𝑋𝑃𝑟 𝑌𝑃𝑟 𝑌𝑃𝑟 𝑌 𝑋𝑃𝑟 𝑌 𝑃𝑟 𝑋1𝜀 1 𝑃𝑟 𝑋𝜀 1𝑝𝑃𝑟 𝑌 𝑋𝑝𝜈 𝑝𝑃𝑟 𝑌 𝑋1 𝑃𝑟 𝑋 𝑃𝑟 𝑋𝜈 𝑝Bayes Theorem (A-posteriori Probabilities)𝑃𝑟 𝑋 𝑌𝑃𝑟 𝑌 𝑋 𝑃𝑟 𝑋𝑃𝑟 𝑌𝑃𝑟 𝑋 𝑌𝑃𝑟 𝑌 𝑋 𝑃𝑟 𝑋𝑃𝑟 𝑌𝑃𝑟 𝑋 𝑌𝑃𝑟 𝑌 𝑋 𝑃𝑟 𝑋𝑃𝑟 𝑌𝑃𝑟 𝑋 𝑌𝑃𝑟 𝑌 𝑋 𝑃𝑟 𝑋𝑃𝑟 𝑌1 𝜀 1 𝑝1 𝜀 1 𝑝𝜈 𝑝1𝜈 𝑝𝜀 1 𝑝𝜀 1𝜀 1 𝑝𝜀 11𝑝𝜈 𝑝𝑝1 𝜈 𝑝𝜈 𝑝1 𝜈 𝑝Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202020 of 25ECE 3800

Maximum a-posteriori estimationIf a Y 1 was received, decide that a X 1 was sent if𝑃𝑟 𝑋 𝑌𝑃𝑟 𝑋 𝑌𝑃𝑟 𝑌 𝑋 𝑃𝑟 𝑋𝑃𝑟 𝑌𝑃𝑟 𝑌 𝑋 𝑃𝑟 𝑋𝑃𝑟 𝑌𝑃𝑟 𝑌 𝑋𝑃𝑟 𝑌 𝑋or 𝑃𝑟 𝑋 𝑃𝑟 𝑋which lead to𝜀 11𝑝𝜀 1 𝑝𝜀 1 𝑝1 𝜈 𝑝𝜈 𝑝1 𝜈 𝑝1𝜈 𝑝𝜀 1𝑝𝜀𝑝or11𝜈Using the bit error probability from before, let 𝜀𝜈𝑝1𝑝𝑝0.11 0.1190.10.111So you should believe what you see is what you get as long as the expectedprobability of a 1 is𝑝0.1The probability that a one was sent is greater than 0.1 Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202021 of 25ECE 3800

If a Y 0 was received, decide that a X 0 was sent if(Note that my derivation differs from a homework solution set answer for the sameconcept)The maximum a-posteriori estimation condition that a 0 was set can be stated as𝑃𝑟 𝑋 𝑌𝑃𝑟 𝑋 𝑌𝑃𝑟 𝑌 𝑋 𝑃𝑟 𝑋𝑃𝑟 𝑌𝑃𝑟 𝑌 𝑋 𝑃𝑟 𝑋𝑃𝑟 𝑌𝑃𝑟 𝑌 𝑋𝑃𝑟 𝑌 𝑋or 𝑃𝑟 𝑋 𝑃𝑟 𝑋and1 𝜀 1 𝑝1 𝜀 1 𝑝𝜈 𝑝1𝜈 𝑝𝜀 1 𝑝𝜈 𝑝or1𝑝𝜀1𝑝Using the bit error probability from before, let 𝜀𝜈𝜈911𝜀0.1𝑝𝜈1𝑝So you should believe what you see is what you get as long as the expectedprobability of a 1 is0.9𝑝1or𝑝0.1Therefore, combining the two inequalities we would hope that 0.9𝑝0.1Or if the bit error probabilities were not given1𝜀𝜈𝑝1𝜀𝑝1𝜈Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202022 of 25ECE 3800

2.6 Diamond NetworkMore Law of total Probability work (LTP)The vertical link is a new consideration in our paths! By conditioning on the vertical link,we can derive two separate problems and then find a general solution.If we consider Link3 to be broken and connected, we can establish the total probabilityTotal Probability: Pr B Pr B A1 Pr A1 Pr B A2 Pr A2 Pr B An Pr An 𝑃𝑟 𝑆 𝐷𝑃𝑟 𝑆 𝐷 𝐿1 𝑃𝑟 𝐿1𝑃𝑟 𝑆 𝐷 𝐿0 𝑃𝑟 𝐿0When Link 3 1, the problem reduces toAnd𝑃𝑟 𝑆 𝐷 𝐿1𝑃𝑟 𝐿 𝐿 𝐿 𝐿For equal probability, independent link closer (p)𝑃𝑟 𝐿 𝐿 𝐿 𝐿𝑃𝑟 𝐿 𝐿 𝑃𝑟 𝐿 𝐿with𝑃𝑟 𝐿 𝐿𝑃𝑟 𝐿𝑃𝑟 𝐿𝑃𝑟 𝐿 𝐿𝑝𝑝𝑝𝑃𝑟 𝐿 𝐿𝑃𝑟 𝐿𝑃𝑟 𝐿𝑃𝑟 𝐿 𝐿𝑝𝑝𝑝 2 𝑝𝑝then𝑃𝑟 𝐿 𝐿 𝐿 𝐿𝑃𝑟 𝐿 𝐿2 𝑝 𝐿 𝐿4 𝑝𝑝4 𝑝𝑝Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202023 of 25ECE 3800

We next need the link open/failed which can be diagramed as0𝑃𝑟 𝑆 𝐷 𝐿0𝑃𝑟 𝑆 𝐷 𝐿𝑃𝑟 𝐿 𝐿𝑃𝑟 𝐿 𝐿 𝐿 𝐿𝑃𝑟 𝐿 𝐿𝑃𝑟 𝑆 𝐷 𝐿0𝑝𝑃𝑟 𝐿 𝐿 𝐿 𝐿𝑝𝑝The diamond network solution then becomes𝑃𝑟 𝑆 𝐷𝑃𝑟 𝑆 𝐷 𝐿𝑃𝑟 𝑆 𝐷1 𝑃𝑟 𝐿14 𝑝4 𝑝𝑝4 𝑝4 𝑝𝑝𝑃𝑟 𝑆 𝐷2 𝑝2 𝑝𝑃𝑟 𝑆 𝐷𝑃𝑟 𝑆 𝐷 𝐿 𝑝2 𝑝2 𝑝𝑝5 𝑝𝑝0 𝑃𝑟 𝐿 12 𝑝0𝑝𝑝2 𝑝To use some real numbers For p 0.9𝑃𝑟 𝑆 𝐷2 0.812 0.729𝑃𝑟 𝑆 𝐷5 0.65612 0.590490.97848For p 0.5𝑃𝑟 𝑆 𝐷2 0.252 0.125𝑃𝑟 𝑆 𝐷5 0.06252 0.031250.5Summary Conditional probability captures the notion of partial information.Bayes Theorem is a widely discussed and used methodology in probability.Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202024 of 25ECE 3800

The diamond network solution Matlab plot𝑃𝑟 𝑆 𝐷2 𝑝%%% Diamond Network Probability% equally probable switch closure, pclear;close all;2 𝑝5 𝑝2 𝑝p (0:0.01:1)';Pr 2*p. 2 2*p. 3-5*p. 4 2*p. itch Prob. (p)')title('Diamond Network Connection')gridDiamond Network 00.10.20.30.40.50.6Switch Prob. (p)0.70.80.91Notes and figures are based on or taken from materials in the course textbook: Charles Boncelet,Probability, Statistics, and Random Signals, Oxford University Press, February 2016.B.J. Bazuin, Fall 202025 of 25ECE 3800

Subtotal 80/150 0.533 50/150 0.333 20/150 0.133 150/150 1.0 These are called the joint probabilities when all unique attributes must be considered. (Concept of total probability things that sum to 1.0) Conditional Probabilities: When one attributes probability is determined based

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