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Date Created: 09/28/15
STAT 430510 Probability Hui Nie Lecture 2 May 27th 2009 Review 0 The basic principle of counting provides us a powerful tool for counting the number of ways that things can happen 0 An ordered sequence of k distinct objects taken from a set of n objects is called a permutation of size k 0 The total number of permutations of size k from 17 objects is given by 17 Pkquot nik Combinations 0 An unordered subset of size k from a set of n distinct objects is called a combination 0 The number of combinations of size k from n distinct objects is denoted by lt Z gt Example 0 From a group of 5 women and 7 men how many different committees consisting of 2 women and 3 men can be formed 5 7 7 Solutlonlt2gtlt3gt7350 o What if 2 of the men are feuding and refused to serve on the committee together Solutionlt2gtltltzgtelt2gtltfgtgtsoo Example 0 Consider a set of n antennas of which m are defective and n 7 m are functional and assume that all of the detectives and all of the lunctionals are considered indistinguishable How many linear orderings are there in which no two detectives are consecutive 0 Solution lt nim1 m Useful Combinatorial Identity A useful combinatorial identity lt gtltgtlt 1gtw The Binomial Theorem The Binomial Theorem H yquot ilt 7gtX yquot r0 lt 7 gt is called binomial coefficients Example 0 Find the coefficient of X2y in X y3 3 7 OSOIUIIOHlt2gt73 Example 0 How many different letter arrangements can be formed from the letters PEPPER 6 3 1 oSqutIonlt3gtlt2gtlt1gt Example o A set of n distinct items is to be divided into r distinct groups of respective sizes n1 r12 n where 2L1 n 17 How many different divisions are possible 0 Solution n nimniminzHnimin27iny17 nl W 2 3 m 7 n1ln2In Multinomial Coefficients n o If 2 n 17 define by 4 n17n277nr n 7 n quot17 72w 7nr imlnglmml n o are known as multlnomlal coeffcrents 1717 27 7nr n represents the number of possible 1717 27 7m divisions of n distinct objects into r distinct groups of respective sizes n1 n Example 0 Ten children are to be divided into an A team and a B team of 5 each The A team will play in one league and the B team in another How many different divisions are possible 0 Solution lt gt 252 Example 0 In order to play a game of basket ball ten children at a plyground divided themselves into two teams of 5 each How many different games are possible lt 10 gt 5 5 0 Solution 5153 126 The Multinomial Theorem The Multinomial Theorem 7 n n n X1X2 Xr NWTm X11X22quot er quot1w J7r quot139quotnyn Example 0 Find the coefficient of X3y222 in X y Z7 7 o Solutlon lt 3272 gt Example 0 Find the coefficient of Xeyz in 3X2 y Z5 5 3 0 Solution 37171 gt33 STAT 430510 Probability Hui Nie Lecture 13 June 181h 2009 Introduction 0 Many random variables are naturally related to each other 0 Forecast and actual weather 9 Height and weight a Lifetimes of a system and a component 0 These random variables are usually studied together instead of individually to The joint behavior of several random variables is described by their joint distribution Joint Distribution o For two random variables X and Y the joint cumulative probability distribution function PX y is defined for each pair of numbers Xy by FXy PXg xand Yg y o The distributions of X and Y can be obtained from the joint cdf PX y FXX PX g X PXg X Ylt oo FXoo FAY PYS Y PXlt 007 YE Y HWY 0 The distributions FX and Fy are called marginal distributions of X and Y Joint Distribution Continued 0 All joint probability statements about X and Y can be answered in terms of their joint distribution function 0 Example PX1 lt X S X2y1 lt y S ya FX27J2FX1J1FX2J17FX17J2 Joint pmf of Discrete Random Variable 0 When X and Y are two discrete random variables the joint pmf pXy is defined for each pair of numbers xy by pXy PX X7 Y y 0 Let A be any set consisting of pairs of xy values Then PX7Y Al Z PX7Y xyeA Marginal pmf of Discrete Random Variable o The marginal pmf s of X and Y denoted by pXx and py y respectively are given by PXX mey y My mey Example 0 Suppose that 15 percent of the families in a certain community have no children 20 percent have 1 child 35 percent have 2 children and 30 percent have 3 Suppose further that in each family each child is equally likely independently to be a boy or a girl It a family is chosen at random from this community then B the number of boys and G the number of girls in this family will have the joint probability mass function as below i i l 0 l i l i i 2 i l l Column sum P 0 Joint pdf of Continuous Random Variable o X and Y are two continuous rv39s Then fX y is the joint pdf for X and Y if for any 2dimensional set A PXYeA Xy6AfXydXdy Joint cdf of Continuous Random Variable o If A is a rectangle then d b PMXucYw nnww C a o The joint cdf is b a HamWXaYm4KampWWW Marginal pdf of Continuous Random Variable o The marginal pdf s of X and Y denoted by fXx and fy y respectively are given by fXX fXydy for 7 00 lt X lt oo fyy fXydX for 7 00 lt y lt oo o The marginal cdf s of X and Y denoted by FX and Fy are given by FXX PX lt x PX x Ylt oo Fxoo FYY PYSY PXlt 007 YSY POOJ Example 0 The joint density function of X and Y is given by X y Ze xe zy 0lt xlt oo0ltylt oo 7 0 otherwise Compute a PXgt 1 Ylt 1 b PX lt Y c The marginal density of X Example Continued 0 PX gt 1Ylt 1 f01f1 2e xe 2ydxdy e 1 7 e 3 o PX lt Y f0 joy Ze xe zydxdy 13 0 fXX f0 Ze xe zydy e x Joint Distribution 0 The joint distribution of n rv39s X1 Xn can be defined in the same manner as n 2 o The joint cumulative probability distribution function FX1 Xn is defined as FX17 7Xn PX1 X17 7Xn Xn Joint Distribution 0 If X1X2 Xn are discrete rv39s the joint pmi is pX17quot 7Xn PX1X17 7Xn Xn o n rv39s X1X2 Xn are said to be jointly continuous if there exists a function fX1 Xn such that for all measurable set A in Hquot PX1 X e A 3X1XHEAIX1 XndX1dx STAT 430510 Probability Hui Nie Lecture 10 June 151h 2009 Introduction o The set of possible values for discrete random variable is either finite or countany infinite 0 However there also exist random variables whose set of possible values is uncountable Definition of Continuous Random Variable o X is a continuous random variable if there exists a nonnegative function f defined for all real X e 700 00 having the property that for any set B of real numbers PX e B fXdX B o The function fis called the probability density function of random variable X o The cumulative distribution function is given by Properties of Continuous Random Variable o 1 PX E 700700 ff fXdX o Pa X g b j fXdX o PX a fXdX O o PX lt a PX g a fXdX Example 0 The amount of time in hours that a computer functions before breaking down is a continuous random variable with probability density function given by f Aeix1007 X Z 0 0 o xlto What is the probability that a a computer will function between 50 and 150 hours before breaking down b it will function for fewer than 100 hours Relationship between cdf and pdf 0 Fa PX e 700 a fix fXdX Fa ma Example o If X is continuous with distribution function FX and density function fX find the density function of Y 2X 0 Determine Fy by Fya PY g a P2X g a PX g a2 FXa2 o Differentiating Fy gives fya nga2 Proposition 521 o If X is a continuous random variable with probability density function fX then for any realvalued function g EgX gmfmdx Expected Value and Variance of Continuous FLV For continuous random variable X with pdf fX o EX ff XfXdX o EXz ff X2fxdx o VarX EX2 7 EX2 o 8000 m Properties of Expected Value and Variance o EaX b aEX b where a and b are constants 0 VaraX b a2 VarX where a and b are constants Example 0 Find EX and VarX when the density function X is 00 2X O X 1 7 0 otherwise 0 EX fo x2de 23 0 EXz fo x22de 12 0 VarX EXz 7 EX2 118 Example 0 The density function X is given by 007 10gxg1 7 0 otherwise Find EeX o EeXf019X1dXei1 Example 0 A stick of length 1 is split at a point U that is uniformly distributed over 01 Determine the expected length of the piece that contains the point p 0 g p g 1 Example 0 Suppose that it you are 3 minutes early for an appointment then you incur the cost cs and it you are 3 minutes late then you incur the cost ks Suppose also that the travel time from where you presently are to the location of your appointment is a continuous random variable having probability density function f Determine the time at which you should depart it you want to minimize your expected cost STAT 430510 Probability Hui Nie Lecture 16 June 24th 2009 Review 0 Sum of Independent Normal Random Variables 0 Sum of Independent Poisson Random Variables 0 Sum of Independent Binomial Random Variables o Conditional Distributions Discrete Case Conditional Distributions Continuous Case 0 Let X and Y be jointly continuous rv39s Then for any X value for which fXx gt 0 the conditional pdf of Y given X X is We y fypdyix Xm fooltyltoo o For any set A PY e AiX x YiXOiXWy MW A A fXX o If X and Y are independent then finOix x amp Yy W Example 0 The joint density of X and Y is given by 12 7 X27xiy0ltxlt10ltylt1 fX7Y 0 otherwise Compute the conditional density of X given that Y y where 0 lt y lt 1 thYO iY One Discrete FLV and one Continuous FLV 0 Suppose that X is a continuous random variable having probability density function fand N is a discrete random variable Then the conditional distribution of X given that N n is PN X n X inNO i WWO Example 0 Consider n m trials having a common probability of success Suppose however that this success probability is not fixed in advance but is chosen from a uniform 01 population What is the conditional distribution of the success probability given that the n m trials result in n successes Example Solution 0 Let X denote the probability that a given trials is a success 0 Let N be the number of successes PwmXmampm fX NX n lt n2mgtx 1ixm HNm cx Xquot17 X 7 o The conditional density is a beta distribution with parameters n 1m 1 Joint Probability Distribution of Functions of FlV s 0 X1 and X2 are jointly continuous random variables with joint probability density function thXZ 0 Want to find the joint distribution of Y1 g1X1X2 and Y2 92X17X2 Two Conditions 0 Condition 1 The equations y1 g1X1X2 and y2 g2X1 X2 can be uniquely solved for X1 and X2 in terms of y1 and y2 with solutions given by X1 h1y1y2 and X2 h2y1y2 0 Condition 2 The functions g1 and 532 have continuous partial derivatives at all points X1 X2 and are such that the 2 x 2 determinant 3i 3i X X JX17X2 39 393 7amp0 TM TX2 at all points X17X2 Change of FtV s 0 Under two conditions above random variables Y1 and Y2 are jointly continuous with joint density function given by Y1Y2Y17Y2 r1X2X17X2iJX17X2i 1 where X1 h1y1y2 X2 h2y17y2 Example 0 Let X1 and X2 be jointly continuous random variables with probability density function thXZ Let Y1 X1 X2 Y2 X1 7 X2 find the joint density function of Y1 and Y2 in terms of thXZ Example Solution 0 Let g1x1X2 X1 X2 and g2x1X2 X17 X2 Then 1 JX17X2 7 1 71 0 Equations g1x1X2 X1 X2 and g2x1X2 X1 7 X2 have X1 y1 y22 X2 y1 7 y22 as their solutions on1gtY2y17y2 l61gtX2y1y27y1Ey2 Generalization to n dimensions 0 Two Conditions y1lg1X177Xngyn gnx1 xn haveaunique solution X1 h1J17 7Jn 39an hnJ17quot397Jn JX17quot39 7Xn 0 0 Conclusion fy1ynJ17 J n fmXHX17 7Xn JX17 7Xn whereX hy17 7yni17 7n 71 Example 0 Let X1 X2 and X3 be independent standard normal random variables It Y1 X1 X2 X3 Y2 X1 7 X2 and Y3 X1 7 X3 compute the joint density function of Y1 Y2 Y3 1 1 1 1 71 0 1 0 71 0 X1 Y1 2Y31X2 Y1 23Y21 Y3 X3 Y1Y 2Y3 OJ 3 expV1 2Ja2 Y172 2Ya2 Y1V 2V322 327r32 expy123 2y223 2y 3 2y2y332 327r32 max3 STAT 430510 Probability Hui Nie Lecture 5 June 2nd 2009 Review 0 Sample Spaces and Events 0 Axioms and Properties of Probability 0 Sample Spaces Having Equally Likely Outcomes Conditional Probability o The Probability of an event measures how often it will occur 0 A conditional probability predicts how often an event will occur under specified conditions 0 Notation PE i F represents the conditional probability that event E occurs given that event F has occurred Definition 0 Definition If PF gt 0 then PE i F 7735 0 The quotconditionquot F contains partial knowledge Example o A coin is flipped twice Assuming that all four points in the sample space S H H H T T H T T are equally likely what is the conditional probability that both flips land on heads given that a the first flip lands on heads b at least one flip lands on heads Example o A stock market analyst feels that o The probability that a mutual fund will get increased contributions from investors is 06 o The probability becomes 09 if the stock market goes up 0 There is a probability of 05 that the stock market rises o A The stock market rises B The company receives increased contribution a Calculate PAB and PAU B Multiplication Rule 0 Conditional probability PEF PE i F 7 W o Multiplication Rule For any events E1 En PE1En PEPEglE1PEnlE1En1 0 Especially PEF PEPF l E PFPEl F Example 0 Sarah is undecided as to whether to take a Frensh course or a chemistry course She estimates that her probability of receiving an A grade would be 05 in a French course and 23 in a chemistry course If Sarah decides to base her decision on the flip of a fair coin what is the probability that she gets an A in chemistry Example 0 An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each compute the probability that each pile has exactly 1 ace The Law of Total Probability 9 Suppose that the events A A are a partition of the sample space S that is A1 A are mutually exclusive and U A S Then for any event B PB PBlA1PA1 PBlAnPAn Bayes s Formula 0 If PB gt 0 then w PA B 7 PB APA PBA PA o If PB gt O and A A are a partition of the sample space S then PBAPA PAB 39 271 PBAPA Example 0 An insurance company believes that people can be divided into two classes those who are accident prone and those who are not The company39s statistics show that an accidentprone person will have an accident at some time within a fixed 1year period with probability 04 whereas this probability decreases to 02 for a person who is not accident prone If we assume that 30 percent of the population is accident prone what is the probability that a new policyholder will have an accident within a year of purchasing a policy Example 0 Suppose that a new policyholder has an accident within a year of purchasing a policy What is the probability that he or she is accident prone Example o A lab test yields 2 possible results positive or negative 99 of a particular disease will produce a positive result but 2 of people without the disease will also produce a positive result Suppose that 01 of the population actually has the disease What is the probability that a person chosen at random will have the disease given that the person39s blood yields a positive result Example 0 Ddisease positive test result Want PD o PD0001 PDC 0999 PDO99 and PD 002 0 Applying Bayes Rule PDPD PDPD PD PDC 099 x 0001 099 x 0001 002 x 0999 PDH 47
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