Week 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable

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1 Week 3 Coditioal probabilities, Bayes formula, WEEK 3 page 1 Expected value of a radom variable We recall our discussio of 5 card poker hads. Example 13 : a) What is the probability of evet A that a 5 card poker had cotais oe or more aces. This is the probability of the complemet of the evet that the had cotais o aces. P( o aces ) = 48 5 / Thus P( A) = P( 1 or more ace) = 1 P( o aces ) = / 52 5 where the umerator is the umber of hads havig 1 or more aces (i.e. ot havig zero aces ). We could also have computed this by otig that the evet 1 or more aces is the uio of the 4 evets : exactly 1 ace (hece 4 o-aces chose from 48 o-aces), exactly two aces (hece 3 oaces), exactly 3 aces or exactly 4 aces. By our basic coutig priciples we the must have = b) What is the probability of the evet B that we are dealt a full house cosistig of 3 aces ad 2 kigs? The probability is the umber of ways to choose 3 aces from the 4 aces i the deck times the ways to choose 2 kigs from 4 over the total umber of 5 card poker hads or / c) What is the probability of ay full house (3 of oe kid two of aother)? This is the same as i b) above except there are 13 ways to choose the kid that we have 3 of ad 12 ways to choose the kid we have 2 of (sice it ca't be the kid we already picked). So probability / 52 5 d) What is the coditioal probability P(B A ) of the full house i b) if we are told by the dealer before he gives us our 5 cards that our had has at least 1 ace i it already. This is a coditioal probability problem coditioed o the evet A that our had has at least oe ace i it. I this case we saw i part a) that the umber of such hads with oe or more ace is the total umber of hads mius the umber havig o aces or hads. These hads costitutes the reduced sample space A for the coditioal evet problem (i.e. the evet A that our had has at least oe ace i it i this example). We ca assume that our full house is the radomly selected from these hads with each such had beig equally likely. This yields the umber of full houses aces over kigs foud i the umerator of b) above over the size of the reduced sample space or P B A = = P B A P A 5 48 To see the last equality, ote that by dividig both the umerator ad deomiator above by the total umber of poker hads 52 5 hads this is the same as the aswer i b) divided by the aswer i a) i.e. it is equal to P( B ) / P( A) ad sice the full house cotaiig 3 aces that is the evet B is cotaied i the evet A that at least oe ace occurs we also have B=B A. This is how we will defie coditioal probability i geeral. I.e. we have motivated the followig defiitio.

2 WEEK 3 page 2 Defiitio of coditioal probability : for ay two evets A ad B with P(A) o-zero we defie P B A = P B A P A. If we let p(b)= P B A oe checks that for fixed A the set fuctio measure p(b) satisfies the 3 axioms eeded for it to be a probability measure. Namely it satisfies 1) 0 p A 1 2) p = p A = P A =1 (i.e. sice A= A the sample space has measure 1 ) P A 3) coutable additivity for p( ) follows directly from the same property of P( ). Thus a coditioal probability measure really is a probability measure. The above defiitio ca be re-writte as the multiplicatio rule for coditioal probabilities : P A B = P A P B A. For three evets (ad similarly for more tha three) this takes the form P A B C =P A P B A P C A B, which ca be verified by goig back to the defiitio of coditioal probabilities. If we thik of evets which are ordered i time, this says that the probability of a sequece of evets occurrig may be writte as a product where at each step of the product we coditio the ext evet o the previous evets that are assumed to have already occurred. For the three evets above first A occurs ad the B occurs (give A already has) ad the C occurs (give A ad B have already occurred). Example 1 : To calculate the probability of selectig 3 aces i a row from a radomly shuffled 52 card deck, we have doe such problems directly by a coutig argumet. Namely the umber of ways to do this is the umber of ways to choose 3 aces from the 4 i the deck ad to get the probability we divide by the umber of ways to select 3 cards from 52. I.e. P( select three aces i a row from a radomly shuffled 52 card deck ) 3 / 52 = 4 3 = We could also obtai the same result usig the multiplicatio rule for coditioal probabilities amely lettig A i be the evet that the i th card selected is a ace, we have 4 P A 1 A 2 A 3 =P A 1 P A 2 A 1 P A 3 A 1 A 2 = That is, for the first ace there are 4 ways (aces) to choose a ace out of 52 equally likely cards to select, but havig chose a ace there are ow 3 ways to choose a ace from the remaiig 3 aces left i the 51 equally likely cards remaiig, ad fially 2 ways to choose the third ace out of the remaiig 2 aces i the 50 equally likely cards remaiig give that a ace was obtaied i each of the previous two selectios. Example 2 : problem 3.70 b) from the text : durig the moth of May i a certai tow the probability of the evet R k 1 that day k+1 is raiy give R k that day k was is.80. That is assumig days are either raiy or suy (ot raiy) P R k 1 R k =.80. This implies P S k 1 R k =.20 is the probability of a suy day give the previous day was raiig. We are also told that P R k 1 S k =.60 is the probability that the ext day will be raiy give the previous was suy. What the is the probability that for some 5 cosecutive days i May i this tow a raiy day is followed by two more raiy days, the a suy day ad the a raiy day. We are ot told the probability that a particular day is raiy so we must iterpret this as the probability that give the first day is raiy

3 that the secod ad third are too, the fourth suy ad the fifth raiy. WEEK 3 page 3 By the multiplicatio rule for coditioal probabilities we have P R 5 S 4 R 3 R 2 R 1 = P R 5 S 4 R 3 R 2 R 1 P S 4 R 3 R 2 R 1 P R 3 R 2 R 1 P R 2 R 1 ow we use the property (so-called Markov property) that the ext evet oly depeds o the previous oe ad othig earlier (oe says that the Markov chai has a memory of oe. to get = P R 5 S 4 P S 4 R 3 P R 3 R 2 P R 2 R 1 = =.0768 for our desired coditioal probability. Remark : A Markov chai geeralizes this example to the case where there may be more tha the two states raiy or suy. For example i the fiite state space case for a 5 state chai, the chai is described by a 5 by 5 trasitio matrix of coditioal probabilities. For example the row 1 colum 3 etry of the matrix gives the coditioal probability of goig to state 3 give that the previous state was state 1. If we wat to kow the probability of goig from state 2 to state 5 i 4 time steps we look at the (2,5) etry of the fourth power of the trasitio matrix. If oe wats to remember the last two states (memory of two) there is a trick oe uses. This situatio ca still be described by a Markov chai where ow we elarge the state space from 5 states to all 25 pairs of states (which we could label as states 1 through 25). Now the trasitio matrix is a 25 by 25 matrix of 625 trasitio probabilities. Similarly if we wat to remember the previous three states our elarged state space would the cosist of all 125 triples of states oe through 5 (so 5 cubed or 125 states) etc. To lear more cosider takig Math 632 Itroductio to Stochastic (radom) Processes. Idepedet evets : Ituitively what we mea whe we say two evets A ad B are idepedet such as two cosecutive flips of a fair coi, is that beig told that evet B has occurred (the first flip yielded a head say) should ot ifluece the probability of A occurrig (the secod flip is a tail say) or i symbols P A B =P A. Equivaletly by the defiitio of coditioal probability, this is true if ad oly if P A B =P A P B ( <-- pairwise idepece ) i.e. the probability of the itersectio of the two evets factors as the product of the probabilities of the idividual evets. More geerally, we say a collectio of evets are idepedet if the probability of the itersectio of the evets i ay sub-collectio of two or more of the evets factors as the product of the probabilities of the idividual evets: P A i1 A i2... A ik =P A i1 P A i2... P A ik. It is possible for three evets to fail to be idepedet ( ot idepedet = depedet ) eve whe ay two of the evets are pairwise idepedet. Example 3: Three evets which are depedet (ot idepedet) but which are pairwise idepedet : For a simple example of this cosider flippig a fair coi twice. Let A= the first flip yields a head={hh, HT}, B= the secod flip yields a head={th, HH}, C= exactly oe head occurs i the two flips={th, HT} Note that the itersectio of the three evets is the empty set which has probability 0, we have P A B C =P =0 P A P B P C =1/8 sice the idividual evets each have probability P A =P B = P C =1/2 so their product is 1/8. Sice 0 is ot equal to 1/8, A, B, ad C are ot idepedet evets. but we claim the probability of the itersectio of ay two of these evets is P A B =P B C =P A C =1 /4, so these are pairwise idepedet evets sice clearly P A B =P A P B etc. Example 4 : Cosider 5 rolls of a fair (six-sided) die. The probability of rollig a 3 for ay particular

4 WEEK 3 page 4 roll is 1/6 while the probability of ot rollig a 3 is 5/6 by the probability of the complemet. Fid the probability of rollig exactly two 3's i 5 rolls. Usig idepedece, the probability of ay particular sequece of 5 rolls, which we view as 5 idepedet evets, two of which ivolve rollig a 3, ad the other three ivolve rollig aythig else, is the product of the idividual probabilities1/6 times 1/6 times 5/6 times 5/6 times 5/6. But there are 5 choose 2 ways that we could have selected the particular two rolls i which the 3 occurred. Thus by the sum rule for probabilities of disoit evets : P( two threes i 5 rolls) = = 5 2 1/6 2 5/6 3. This is a example of a biomial radom variable where the probability of success (a 3 is rolled) for each trial (roll) is p=1/6 ad the probability of failure is (1-p) = 5/6. Similar reasoig gives that the probability of exactly k successes i idepedet trials each havig success probability p is the biomial probability b( k;, p) = P( exactly k successes i idepedet trials each havig success probability p ) k pk = 1 p k Example 5 : Cosider a system where parts 1 ad 2 operate idepedetly of oe aother each with failure probability.1 but are ruig i parallel so that the system diagram looks like 1 / \ / \ \ / \ 2 / Lettig A ={ compoet 1 operates successfully}, B= {compoet 2 operates successfully}, the P( system succeeds ) = P( either 1 or 2 operates) = P A B =P A P B P A P B = =.99 so the combied system is operatioal 99% of the time. A slightly more complicated example ivolves a similar system like 1 3 / \ / \ \ / \ 2 / where ow with evet C = {compoet 3 operates successfully}, with C idepedet of A ad B P( system succeeds ) = P A C B =P A C P B P A B C = =.981 usig idepedece of the three evets ad assumig compoet 3 also fails with probability.1. The success probability is slightly lower tha before sice ow both compoets 1 ad 3 must work.

5 properly for the top series to work. WEEK 3 page 5 Example 6 : Problem 3.75 : A tree diagram for coditioal probabilities is a useful device.. Figure 3.16 of the text used for exercise 3.75 is the followig B A B A / \ / \.4 / \ A.4 / \ A / which ca be filled i as /.70 \ \ \.6\.8 \ B A \ B A \ \ \ A \ A.2.2 The iterpretatio is that P B =.4, P A B =.30, P A B =.20 from which we ifer P B =.6, P A B =.7, P A B =.8 (usig the law of the probability for the complemet ) a) P A =P B P A B P B P A B = (.4)(.3) + (.6)(.8) =.60 b) P A B P B P A B P B A = = =.4.3 = 1 P A P A.60 5 =.20 by part a) usig the multiplicatio rule for coditioal probabilities. Note that the origial diagram gave us P A B ad what was wated i part b) was to reverse the order of coditioig, that is to fid P B A. This is the situatio where Bayes' theorem applies. Oe has a collectio of mutually exclusive (disoit) evets which exhaust the sample space. I this case the sample space is a disoit uio of B ad B. c) Similarly P B A = P B P A B P A = =.70 Part a) is referred to as the rule of total probability (or rule of elimiatio). It is used to get the deomiator i part b). Parts b) ad c) are kow as Bayes' theorem. To get the rule of total probability we ote that the disoit partitio of the sample space such as B B= also partitios ay set A= A B A B as a disoit uio. Sice the uio is disoit, the third axiom of probability gives P A =P A B P A B. We the re-write each probability o the right via the multiplicatio rule for coditioal probabilities (essetially the defiitio of coditioal probability). This gives the rule of total probability which gives the deomiator i Bayes' theorem. More geerally if E 1 E 2... E k = is a disoit uio which exhausts the sample space the Bayes' theorem says P E l D = P E P D E l l k P E P D E. The probabilities P E l l=1,.., k are =1 called the priors (which reflect our best kowledge before the experimet D). The we collect some ew data ad update these to get the posterior probabilities P E l D. Certai types of probability distributios have the property that the prior ad posterior both have a similar form except that certai real umber parameters characteristic of the distributio chage i ways which are easy to compute. The if we get some ew data i we ca regard the old posterior as the ew prior ad use the ew data

6 to update to get the ew posterior ad so o. WEEK 3 page 6 Example 7 : A Ace Electroics dealer sells 3 brads of televisios : 50% are the first brad which are Italia made, 30% are the secod brad which are Frech made, ad 20% are the third brad which are Swiss made. Each TV comes with a 1 year warraty. We kow that 25% of the Italia TVs will require repair work uder warraty, 20% of the Frech will require repairs uder warraty, ad 10% of the Swiss TVs will eed repair uder warraty. a) What is the probability that a radomly selected customer has bought a Italia TV that will eed repairs uder warraty? b) What is the probability that a radomly selected customer has a TV that will eed repairs uder warraty? c) If the customer returs to the store with a TV that eeds repairs uder warraty, what is the probability that it is a Italia TV? Frech? Swiss? Lettig A be the evet that a radomly selected TV is Italia, B that it is Frech, C that it is Swiss ad R that it will eed repairs uder warraty. We are give that P A =.50, P B =.30, P C =.20, P R A =.25, P R B =.20, P R C =.10 Part a) asks for P A R =P A P R A = =.125=1/8. by the product rule for coditioal probabilities. Part b) wats P(R) which uses the rule of total probability P R = P R A P R B P R C =P A P R A P B P R B P C P R C = =.205. Part c) asks for P A P R A P A R = P R.205 =.29 ad P C R =.20 =.125 =.61 usig Bayes' theorem ad agai.205 for P B R = =.10 which could also be obtaied as 1 P A R P B R = =.10. Example 8 : Suppose a certai diagostic test is 98% accurate both o those who do ad those that do't have a disease. If.3%=.003 of the populatio has cacer, fid the probability that a radomly selected tested perso has cacer (C) give that the test is positive (+). The give iformatio says P C =.003, P C =.98 = P C from which we deduce that P C =.997, P C =.02=P C The by Bayes' theorem, P C P C P ( C ) = P C P C P C P C = This is slightly larger 23 tha 15%. Thus the posterior probability of havig the disease (give a positive test result) is over 150 chaces out of 1000 up from the origial prior probability of.003 ( 3 chaces i 1000 ) prior to testig. A positive test result oly gives a 15% chace of havig the disease due to the fact that the chace of havig it i the populatio as a whole is so small. Expectatio of a radom variable (also called its expected value or mea value) Cosider the followig game: we flip a fair coi 5 times. If a head occurs for the first time o the th flip the game pays us wiigs amout W =W =a =2 dollars for 1 5 where is a particular outcome of the experimet that is a sequece of 5 heads or tails such that the first head occurs o the th flip ad usig idepedece of flips this happes with probability p = P :W =a = 1 2 (= P W =a for short ). If o heads occur i 5 flips we'll take =6 so that the game pays us the grad prize of W =2 6 =$ 64 ad this occurs with probability 1/32. Now the 64 dollar questio is : How much are you willig to pay to play this game so that o average you

7 will break eve? WEEK 3 page 7 If we play the game a large umber of times the by the relative frequecy iterpretatio of probabilities, the umber of times s we wo amout a dollars (successes s i wiig the th amout ) is approximately p (equivaletly the probability p s is approximately the relative fractio of times we wo amout a ). Thus if we play the game idepedetly may times wiig amout W k o the k th time that we played, our log term average or sample mea wiigs W= W k k=1 is approximately a p or cacelig the 's ad deotig our average wiigs by E[ W ] (the expected value of the radom variable W ) we fid for the defiitio of the expectatio of ay discrete radom variable W takig possible values a (from a coutable set of values) with probabilities p : E [W ]= a p = a P W =a. I our particular example whe the sample size gets big but the sum over is for 1 6 this gives E [W ]=2 1/2 4 1/ 4 8 1/8 16 1/ / /32 = =$7. Thus we should be willig to pay $7 each time we play the game, assumig we play more tha 32 times, log eough to wi the grad prize which we expect to happe aroud 1/32 of the time. The situatio is a little differet if istead of stoppig at the 5 th flip we flip the coi 15 times with grad prize 2 16 dollars. It is easy eough to calculate the expectatio i this case usig the above formula. But realistically ot all of us would wat to wait aroud o average 2 15 times which could take several years before we see the rare evet of wiig the grad prize which occurs with probability 1/ Ecoomists speak of a utility fuctio which describes how much playig such a game is worth to us persoally ad which may vary from perso to perso depedig o our tastes i gamblig ad how much we are willig to risk. I oe of the homeworks problem 3.90 ivolvig expected values, a compay pays some per uit cost of C dollars ad sells the item at a per uit sales price S=S 1. If a fixed umber k of items are stocked for the day the the cost of the k uits is a fixed amout k C. The demad (how may uits customers desire to purchase that day) is assumed to be a radom variable where p gives the probability that the demad equals uits that day ad the sales price resultig from a give demad is the a radom variable X give by X =S if k 1 else X =S k if k. (I.e. The actual sales caot exceed the umber i stock.) The the expected profit is the expected sales price mius the fixed cost or E [ P ]=E [ X ] C k= p S p S k C k. =0 k That is for demad less tha the umber k i stock our sales price for the uits sold is k 1 S ad this occurs with probability p while if the demad is greater tha or equal to the umber k i stock we sell k items at a price S k ad this occurs with probability p. Note we could have re-writte k p (sice the sum of the probabilities equals 1) ad the the above is the fixed cost k C =k C equivalet to the expected profit where the profit is the radom variable price radom variable mius the fixed cost for k items i stock. X C k which is the sales

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