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


 Oliver Sharp
 1 years ago
 Views:
Transcription
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 oaces chose from 48 oaces), 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) ozero 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 rewritte 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 (socalled 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 subcollectio 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 (sixsided) 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 (1p) = 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 rewrite 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 rewritte 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
Discrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13
EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may
More informationI. Chisquared Distributions
1 M 358K Supplemet to Chapter 23: CHISQUARED DISTRIBUTIONS, TDISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad tdistributios, we first eed to look at aother family of distributios, the chisquared distributios.
More information3 Basic Definitions of Probability Theory
3 Basic Defiitios of Probability Theory 3defprob.tex: Feb 10, 2003 Classical probability Frequecy probability axiomatic probability Historical developemet: Classical Frequecy Axiomatic The Axiomatic defiitio
More informationHypergeometric Distributions
7.4 Hypergeometric Distributios Whe choosig the startig lieup for a game, a coach obviously has to choose a differet player for each positio. Similarly, whe a uio elects delegates for a covetio or you
More informationChapter 6: Variance, the law of large numbers and the MonteCarlo method
Chapter 6: Variace, the law of large umbers ad the MoteCarlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value
More informationMARTINGALES AND A BASIC APPLICATION
MARTINGALES AND A BASIC APPLICATION TURNER SMITH Abstract. This paper will develop the measuretheoretic approach to probability i order to preset the defiitio of martigales. From there we will apply this
More informationIn nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008
I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces
More informationHere are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.
This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at http://tutorial.math.lamar.edu/terms.asp. The olie versio
More information1. C. The formula for the confidence interval for a population mean is: x t, which was
s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : pvalue
More informationThe Stable Marriage Problem
The Stable Marriage Problem William Hut Lae Departmet of Computer Sciece ad Electrical Egieerig, West Virgiia Uiversity, Morgatow, WV William.Hut@mail.wvu.edu 1 Itroductio Imagie you are a matchmaker,
More informationEngineering 323 Beautiful Homework Set 3 1 of 7 Kuszmar Problem 2.51
Egieerig 33 eautiful Homewor et 3 of 7 Kuszmar roblem.5.5 large departmet store sells sport shirts i three sizes small, medium, ad large, three patters plaid, prit, ad stripe, ad two sleeve legths log
More informationDUBLIN INSTITUTE OF TECHNOLOGY KEVIN STREET, DUBLIN 8. Probability Based Learning: Introduction to Probability REVISION QUESTIONS *** SOLUTIONS ***
DUBLIN INSTITUTE OF TECHNOLOGY KEVIN STREET, DUBLIN 8 Probability Based Learig: Itroductio to Probability REVISION QUESTIONS *** *** MACHINE LEARNING AT DIT Dr. Joh Kelleher Dr. Bria Mac Namee *** ***
More informationKey Ideas Section 81: Overview hypothesis testing Hypothesis Hypothesis Test Section 82: Basics of Hypothesis Testing Null Hypothesis
Chapter 8 Key Ideas Hypothesis (Null ad Alterative), Hypothesis Test, Test Statistic, Pvalue Type I Error, Type II Error, Sigificace Level, Power Sectio 81: Overview Cofidece Itervals (Chapter 7) are
More informationLesson 17 Pearson s Correlation Coefficient
Outlie Measures of Relatioships Pearso s Correlatio Coefficiet (r) types of data scatter plots measure of directio measure of stregth Computatio covariatio of X ad Y uique variatio i X ad Y measurig
More informationCHAPTER 7: Central Limit Theorem: CLT for Averages (Means)
CHAPTER 7: Cetral Limit Theorem: CLT for Averages (Meas) X = the umber obtaied whe rollig oe six sided die oce. If we roll a six sided die oce, the mea of the probability distributio is X P(X = x) Simulatio:
More informationProperties of MLE: consistency, asymptotic normality. Fisher information.
Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout
More informationExample 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).
BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook  Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly
More informationHypothesis testing. Null and alternative hypotheses
Hypothesis testig Aother importat use of samplig distributios is to test hypotheses about populatio parameters, e.g. mea, proportio, regressio coefficiets, etc. For example, it is possible to stipulate
More informationConfidence Intervals. CI for a population mean (σ is known and n > 30 or the variable is normally distributed in the.
Cofidece Itervals A cofidece iterval is a iterval whose purpose is to estimate a parameter (a umber that could, i theory, be calculated from the populatio, if measuremets were available for the whole populatio).
More informationDepartment of Computer Science, University of Otago
Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS200609 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly
More informationSoving Recurrence Relations
Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree
More informationx : X bar Mean (i.e. Average) of a sample
A quick referece for symbols ad formulas covered i COGS14: MEAN OF SAMPLE: x = x i x : X bar Mea (i.e. Average) of a sample x i : X sub i This stads for each idividual value you have i your sample. For
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More informationInfinite Sequences and Series
CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...
More informationCS103X: Discrete Structures Homework 4 Solutions
CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible sixfigure salaries i whole dollar amouts are there that cotai at least
More informationMath C067 Sampling Distributions
Math C067 Samplig Distributios Sample Mea ad Sample Proportio Richard Beigel Some time betwee April 16, 2007 ad April 16, 2007 Examples of Samplig A pollster may try to estimate the proportio of voters
More information5 Boolean Decision Trees (February 11)
5 Boolea Decisio Trees (February 11) 5.1 Graph Coectivity Suppose we are give a udirected graph G, represeted as a boolea adjacecy matrix = (a ij ), where a ij = 1 if ad oly if vertices i ad j are coected
More informationRepeating Decimals are decimal numbers that have number(s) after the decimal point that repeat in a pattern.
5.5 Fractios ad Decimals Steps for Chagig a Fractio to a Decimal. Simplify the fractio, if possible. 2. Divide the umerator by the deomiator. d d Repeatig Decimals Repeatig Decimals are decimal umbers
More informationTHE ARITHMETIC OF INTEGERS.  multiplication, exponentiation, division, addition, and subtraction
THE ARITHMETIC OF INTEGERS  multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,
More informationThe following example will help us understand The Sampling Distribution of the Mean. C1 C2 C3 C4 C5 50 miles 84 miles 38 miles 120 miles 48 miles
The followig eample will help us uderstad The Samplig Distributio of the Mea Review: The populatio is the etire collectio of all idividuals or objects of iterest The sample is the portio of the populatio
More informationPresent Value Factor To bring one dollar in the future back to present, one uses the Present Value Factor (PVF): Concept 9: Present Value
Cocept 9: Preset Value Is the value of a dollar received today the same as received a year from today? A dollar today is worth more tha a dollar tomorrow because of iflatio, opportuity cost, ad risk Brigig
More informationAsymptotic Growth of Functions
CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll
More informationA probabilistic proof of a binomial identity
A probabilistic proof of a biomial idetity Joatho Peterso Abstract We give a elemetary probabilistic proof of a biomial idetity. The proof is obtaied by computig the probability of a certai evet i two
More informationCenter, Spread, and Shape in Inference: Claims, Caveats, and Insights
Ceter, Spread, ad Shape i Iferece: Claims, Caveats, ad Isights Dr. Nacy Pfeig (Uiversity of Pittsburgh) AMATYC November 2008 Prelimiary Activities 1. I would like to produce a iterval estimate for the
More information.04. This means $1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth
Questio 1: What is a ordiary auity? Let s look at a ordiary auity that is certai ad simple. By this, we mea a auity over a fixed term whose paymet period matches the iterest coversio period. Additioally,
More informationTHE ABRACADABRA PROBLEM
THE ABRACADABRA PROBLEM FRANCESCO CARAVENNA Abstract. We preset a detailed solutio of Exercise E0.6 i [Wil9]: i a radom sequece of letters, draw idepedetly ad uiformly from the Eglish alphabet, the expected
More informationSolving Logarithms and Exponential Equations
Solvig Logarithms ad Epoetial Equatios Logarithmic Equatios There are two major ideas required whe solvig Logarithmic Equatios. The first is the Defiitio of a Logarithm. You may recall from a earlier topic:
More informationx(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More informationTrigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is
0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values
More informationAQA STATISTICS 1 REVISION NOTES
AQA STATISTICS 1 REVISION NOTES AVERAGES AND MEASURES OF SPREAD www.mathsbox.org.uk Mode : the most commo or most popular data value the oly average that ca be used for qualitative data ot suitable if
More informationA Mathematical Perspective on Gambling
A Mathematical Perspective o Gamblig Molly Maxwell Abstract. This paper presets some basic topics i probability ad statistics, icludig sample spaces, probabilistic evets, expectatios, the biomial ad ormal
More informationOverview of some probability distributions.
Lecture Overview of some probability distributios. I this lecture we will review several commo distributios that will be used ofte throughtout the class. Each distributio is usually described by its probability
More information8.1 Arithmetic Sequences
MCR3U Uit 8: Sequeces & Series Page 1 of 1 8.1 Arithmetic Sequeces Defiitio: A sequece is a comma separated list of ordered terms that follow a patter. Examples: 1, 2, 3, 4, 5 : a sequece of the first
More information1 Computing the Standard Deviation of Sample Means
Computig the Stadard Deviatio of Sample Meas Quality cotrol charts are based o sample meas ot o idividual values withi a sample. A sample is a group of items, which are cosidered all together for our aalysis.
More informationLecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)
18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the BruMikowski iequality for boxes. Today we ll go over the
More information5.4 Amortization. Question 1: How do you find the present value of an annuity? Question 2: How is a loan amortized?
5.4 Amortizatio Questio 1: How do you fid the preset value of a auity? Questio 2: How is a loa amortized? Questio 3: How do you make a amortizatio table? Oe of the most commo fiacial istrumets a perso
More informationGCSE STATISTICS. 4) How to calculate the range: The difference between the biggest number and the smallest number.
GCSE STATISTICS You should kow: 1) How to draw a frequecy diagram: e.g. NUMBER TALLY FREQUENCY 1 3 5 ) How to draw a bar chart, a pictogram, ad a pie chart. 3) How to use averages: a) Mea  add up all
More informationChapter 5 O A Cojecture Of Erdíos Proceedigs NCUR VIII è1994è, Vol II, pp 794í798 Jeærey F Gold Departmet of Mathematics, Departmet of Physics Uiversity of Utah Do H Tucker Departmet of Mathematics Uiversity
More informationChapter 5 Discrete Probability Distributions
Slides Prepared by JOHN S. LOUCKS St. Edward s Uiversity Slide Chapter 5 Discrete Probability Distributios Radom Variables Discrete Probability Distributios Epected Value ad Variace Poisso Distributio
More informationDetermining the sample size
Determiig the sample size Oe of the most commo questios ay statisticia gets asked is How large a sample size do I eed? Researchers are ofte surprised to fid out that the aswer depeds o a umber of factors
More informationExample Consider the following set of data, showing the number of times a sample of 5 students check their per day:
Sectio 82: Measures of cetral tedecy Whe thikig about questios such as: how may calories do I eat per day? or how much time do I sped talkig per day?, we quickly realize that the aswer will vary from day
More information{{1}, {2, 4}, {3}} {{1, 3, 4}, {2}} {{1}, {2}, {3, 4}} 5.4 Stirling Numbers
. Stirlig Numbers Whe coutig various types of fuctios from., we quicly discovered that eumeratig the umber of oto fuctios was a difficult problem. For a domai of five elemets ad a rage of four elemets,
More information5: Introduction to Estimation
5: Itroductio to Estimatio Cotets Acroyms ad symbols... 1 Statistical iferece... Estimatig µ with cofidece... 3 Samplig distributio of the mea... 3 Cofidece Iterval for μ whe σ is kow before had... 4 Sample
More informationsum of all values n x = the number of values = i=1 x = n n. When finding the mean of a frequency distribution the mean is given by
Statistics Module Revisio Sheet The S exam is hour 30 miutes log ad is i two sectios Sectio A 3 marks 5 questios worth o more tha 8 marks each Sectio B 3 marks questios worth about 8 marks each You are
More informationDiscrete Random Variables and Probability Distributions. Random Variables. Chapter 3 3.1
UCLA STAT A Applied Probability & Statistics for Egieers Istructor: Ivo Diov, Asst. Prof. I Statistics ad Neurology Teachig Assistat: Neda Farziia, UCLA Statistics Uiversity of Califoria, Los Ageles, Sprig
More informationSequences and Series
CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their
More information3. Covariance and Correlation
Virtual Laboratories > 3. Expected Value > 1 2 3 4 5 6 3. Covariace ad Correlatio Recall that by takig the expected value of various trasformatios of a radom variable, we ca measure may iterestig characteristics
More informationThe second difference is the sequence of differences of the first difference sequence, 2
Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for
More informationSimple Annuities Present Value.
Simple Auities Preset Value. OBJECTIVES (i) To uderstad the uderlyig priciple of a preset value auity. (ii) To use a CASIO CFX9850GB PLUS to efficietly compute values associated with preset value auities.
More information15.075 Exam 3. Instructor: Cynthia Rudin TA: Dimitrios Bisias. November 22, 2011
15.075 Exam 3 Istructor: Cythia Rudi TA: Dimitrios Bisias November 22, 2011 Gradig is based o demostratio of coceptual uderstadig, so you eed to show all of your work. Problem 1 A compay makes highdefiitio
More informationCase Study. Normal and t Distributions. Density Plot. Normal Distributions
Case Study Normal ad t Distributios Bret Halo ad Bret Larget Departmet of Statistics Uiversity of Wiscosi Madiso October 11 13, 2011 Case Study Body temperature varies withi idividuals over time (it ca
More informationLesson 15 ANOVA (analysis of variance)
Outlie Variability betwee group variability withi group variability total variability Fratio Computatio sums of squares (betwee/withi/total degrees of freedom (betwee/withi/total mea square (betwee/withi
More informationBetting on Football Pools
Bettig o Football Pools by Edward A. Beder I a pool, oe tries to guess the wiers i a set of games. For example, oe may have te matches this weeked ad oe bets o who the wiers will be. We ve put wiers i
More informationCS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations
CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad
More informationA Resource for Freestanding Mathematics Qualifications Working with %
Ca you aswer these questios? A savigs accout gives % iterest per aum.. If 000 is ivested i this accout, how much will be i the accout at the ed of years? A ew car costs 16 000 ad its value falls by 1%
More informationThe Euler Totient, the Möbius and the Divisor Functions
The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA 01075 1 Ackowledgemets This work was supported by the Mout Holyoke College fellowship
More information7 b) 0. Guided Notes for lesson P.2 Properties of Exponents. If a, b, x, y and a, b, 0, and m, n Z then the following properties hold: 1 n b
Guided Notes for lesso P. Properties of Expoets If a, b, x, y ad a, b, 0, ad m, Z the the followig properties hold:. Negative Expoet Rule: b ad b b b Aswers must ever cotai egative expoets. Examples: 5
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits
More informationTIEE Teaching Issues and Experiments in Ecology  Volume 1, January 2004
TIEE Teachig Issues ad Experimets i Ecology  Volume 1, Jauary 2004 EXPERIMENTS Evirometal Correlates of Leaf Stomata Desity Bruce W. Grat ad Itzick Vatick Biology, Wideer Uiversity, Chester PA, 19013
More informationG r a d e. 2 M a t h e M a t i c s. statistics and Probability
G r a d e 2 M a t h e M a t i c s statistics ad Probability Grade 2: Statistics (Data Aalysis) (2.SP.1, 2.SP.2) edurig uderstadigs: data ca be collected ad orgaized i a variety of ways. data ca be used
More informationUniversity of California, Los Angeles Department of Statistics. Distributions related to the normal distribution
Uiversity of Califoria, Los Ageles Departmet of Statistics Statistics 100B Istructor: Nicolas Christou Three importat distributios: Distributios related to the ormal distributio Chisquare (χ ) distributio.
More informationSECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES
SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,
More informationChapter 7  Sampling Distributions. 1 Introduction. What is statistics? It consist of three major areas:
Chapter 7  Samplig Distributios 1 Itroductio What is statistics? It cosist of three major areas: Data Collectio: samplig plas ad experimetal desigs Descriptive Statistics: umerical ad graphical summaries
More informationThe analysis of the Cournot oligopoly model considering the subjective motive in the strategy selection
The aalysis of the Courot oligopoly model cosiderig the subjective motive i the strategy selectio Shigehito Furuyama Teruhisa Nakai Departmet of Systems Maagemet Egieerig Faculty of Egieerig Kasai Uiversity
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0.,
More information9.8: THE POWER OF A TEST
9.8: The Power of a Test CD91 9.8: THE POWER OF A TEST I the iitial discussio of statistical hypothesis testig, the two types of risks that are take whe decisios are made about populatio parameters based
More informationGrade 7. Strand: Number Specific Learning Outcomes It is expected that students will:
Strad: Number Specific Learig Outcomes It is expected that studets will: 7.N.1. Determie ad explai why a umber is divisible by 2, 3, 4, 5, 6, 8, 9, or 10, ad why a umber caot be divided by 0. [C, R] [C]
More informationChapter 5: Inner Product Spaces
Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples
More informationDefinition. A variable X that takes on values X 1, X 2, X 3,...X k with respective frequencies f 1, f 2, f 3,...f k has mean
1 Social Studies 201 October 13, 2004 Note: The examples i these otes may be differet tha used i class. However, the examples are similar ad the methods used are idetical to what was preseted i class.
More informationChapter 7 Methods of Finding Estimators
Chapter 7 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 011 Chapter 7 Methods of Fidig Estimators Sectio 7.1 Itroductio Defiitio 7.1.1 A poit estimator is ay fuctio W( X) W( X1, X,, X ) of
More informationA PROBABILISTIC VIEW ON THE ECONOMICS OF GAMBLING
A PROBABILISTIC VIEW ON THE ECONOMICS OF GAMBLING MATTHEW ACTIPES Abstract. This paper begis by defiig a probability space ad establishig probability fuctios i this space over discrete radom variables.
More informationMaximum Likelihood Estimators.
Lecture 2 Maximum Likelihood Estimators. Matlab example. As a motivatio, let us look at oe Matlab example. Let us geerate a radom sample of size 00 from beta distributio Beta(5, 2). We will lear the defiitio
More informationBINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients
652 (1226) Chapter 12 Sequeces ad Series 12.5 BINOMIAL EXPANSIONS I this sectio Some Examples Otaiig the Coefficiets The Biomial Theorem I Chapter 5 you leared how to square a iomial. I this sectio you
More informationPROCEEDINGS OF THE YEREVAN STATE UNIVERSITY AN ALTERNATIVE MODEL FOR BONUSMALUS SYSTEM
PROCEEDINGS OF THE YEREVAN STATE UNIVERSITY Physical ad Mathematical Scieces 2015, 1, p. 15 19 M a t h e m a t i c s AN ALTERNATIVE MODEL FOR BONUSMALUS SYSTEM A. G. GULYAN Chair of Actuarial Mathematics
More informationIncremental calculation of weighted mean and variance
Icremetal calculatio of weighted mea ad variace Toy Fich faf@cam.ac.uk dot@dotat.at Uiversity of Cambridge Computig Service February 009 Abstract I these otes I eplai how to derive formulae for umerically
More informationUsing Excel to Construct Confidence Intervals
OPIM 303 Statistics Ja Stallaert Usig Excel to Costruct Cofidece Itervals This hadout explais how to costruct cofidece itervals i Excel for the followig cases: 1. Cofidece Itervals for the mea of a populatio
More informationTrading the randomness  Designing an optimal trading strategy under a drifted random walk price model
Tradig the radomess  Desigig a optimal tradig strategy uder a drifted radom walk price model Yuao Wu Math 20 Project Paper Professor Zachary Hamaker Abstract: I this paper the author iteds to explore
More informationNormal Distribution.
Normal Distributio www.icrf.l Normal distributio I probability theory, the ormal or Gaussia distributio, is a cotiuous probability distributio that is ofte used as a first approimatio to describe realvalued
More informationChapter 14 Nonparametric Statistics
Chapter 14 Noparametric Statistics A.K.A. distributiofree statistics! Does ot deped o the populatio fittig ay particular type of distributio (e.g, ormal). Sice these methods make fewer assumptios, they
More informationElementary Theory of Russian Roulette
Elemetary Theory of Russia Roulette iterestig patters of fractios Satoshi Hashiba Daisuke Miematsu Ryohei Miyadera Itroductio. Today we are goig to study mathematical theory of Russia roulette. If some
More informationDivide and Conquer. Maximum/minimum. Integer Multiplication. CS125 Lecture 4 Fall 2015
CS125 Lecture 4 Fall 2015 Divide ad Coquer We have see oe geeral paradigm for fidig algorithms: the greedy approach. We ow cosider aother geeral paradigm, kow as divide ad coquer. We have already see a
More informationUC Berkeley Department of Electrical Engineering and Computer Science. EE 126: Probablity and Random Processes. Solutions 9 Spring 2006
Exam format UC Bereley Departmet of Electrical Egieerig ad Computer Sciece EE 6: Probablity ad Radom Processes Solutios 9 Sprig 006 The secod midterm will be held o Wedesday May 7; CHECK the fial exam
More informationUnit 20 Hypotheses Testing
Uit 2 Hypotheses Testig Objectives: To uderstad how to formulate a ull hypothesis ad a alterative hypothesis about a populatio proportio, ad how to choose a sigificace level To uderstad how to collect
More informationCHAPTER 11 Financial mathematics
CHAPTER 11 Fiacial mathematics I this chapter you will: Calculate iterest usig the simple iterest formula ( ) Use the simple iterest formula to calculate the pricipal (P) Use the simple iterest formula
More informationSum and Product Rules. Combinatorics. Some Subtler Examples
Combiatorics Sum ad Product Rules Problem: How to cout without coutig. How do you figure out how may thigs there are with a certai property without actually eumeratig all of them. Sometimes this requires
More informationZTEST / ZSTATISTIC: used to test hypotheses about. µ when the population standard deviation is unknown
ZTEST / ZSTATISTIC: used to test hypotheses about µ whe the populatio stadard deviatio is kow ad populatio distributio is ormal or sample size is large TTEST / TSTATISTIC: used to test hypotheses about
More informationListing terms of a finite sequence List all of the terms of each finite sequence. a) a n n 2 for 1 n 5 1 b) a n for 1 n 4 n 2
74 (4 ) Chapter 4 Sequeces ad Series 4. SEQUENCES I this sectio Defiitio Fidig a Formula for the th Term The word sequece is a familiar word. We may speak of a sequece of evets or say that somethig is
More information0.7 0.6 0.2 0 0 96 96.5 97 97.5 98 98.5 99 99.5 100 100.5 96.5 97 97.5 98 98.5 99 99.5 100 100.5
Sectio 13 KolmogorovSmirov test. Suppose that we have a i.i.d. sample X 1,..., X with some ukow distributio P ad we would like to test the hypothesis that P is equal to a particular distributio P 0, i.e.
More information4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
SECTION 2.6 THE RATIO TEST 79 2.6. THE RATIO TEST We ow kow how to hadle series which we ca itegrate (the Itegral Test), ad series which are similar to geometric or pseries (the Compariso Test), but of
More informationWHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER?
WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I tought Topological Groups at the Göttige Georg August Uiversity. This
More information