Econ 240ALast TimeProblem 6.61PowerPoint PresentationSlide 5Probability of a heart attack in the next ten yearsRandom VariablesOutlineOutline (Cont.)ConceptFlipping a Coin OnceFlipping a coin onceFlipping a coin twice: 4 elementary outcomesFlipping a Coin TwiceContinuing with the variance in kSlide 16Frequency Distribution for the Number of HeadsSlide 18Slide 19Three Flips of a Fair CoinSlide 21Three Flips of a CoinSlide 23Slide 24The Probability of Getting k HeadsExpected Value of a discrete random variableExpected Value of the sum of random variablesExpected Number of Heads After Two FlipsCont.Variance of a discrete random variableSlide 31The variance of the sum of discrete random variablesThe variance of the sum if x and y are independentVariance of the number of heads after two flipsSlide 35Slide 36The Field PollField PollSlide 39Slide 40Slide 41Slide 42Slide 43Slide 441Econ 240APower Four112Last Time•Probability3Problem 6.61•A survey of middle aged men reveals that 28% of them are balding at the crown of their head. Moreover, it is known that such men have an 18% probability of suffering a heart attack in the next ten years. Men who are not balding in this way have an 11% probability of a heart attack. Find the probability that a middle aged man will suffer a heart attack in the next ten years.4Middle Aged menBaldP (Bald and MA) = 0.28Not Bald5Middle Aged menBaldP (Bald and MA) = 0.28Not BaldP(HA/Bald and MA) = 0.18P(HA/Not Bald and MA)= 0.116Probability of a heart attack in the next ten years•P(HA) = P(HA and Bald and MA) + P(HA and Not Bald and MA)•P(HA) = P(HA/Bald and MA)*P(BALD and MA) + P(HA/Not BALD and MA)* P(Not Bald and MA)•P(HA) = 0.18*0.28 + 0.11*0.72 = 0.054 + .0792 = 0.12967Random Variables•There is a natural transition or easy segue from our discussion of probability and Bernoulli trials last time to random variables•Define k to be the random variable # of heads in 1 flip, 2 flips or n flips of a coin•We can find the probability that k=0, or k=n by brute force using probability trees. We can find the histogram for k, its central tendency and its dispersion8Outline•Random Variables & Bernoulli Trials•example: one flip of a coin–expected value of the number of heads–variance in the number of heads•example: two flips of a coin•a fair coin: frequency distribution of the number of heads–one flip–two flips9Outline (Cont.)•Three flips of a fair coin, the number of combinations of the number of heads•The binomial distribution•frequency distributions for the binomial•The expected value of a discrete random variable•the variance of a discrete random variable10Concept•Bernoulli Trial–two outcomes, e.g. success or failure–successive independent trials–probability of success is the same in each trial•Example: flipping a coin multiple times11Flipping a Coin OnceHeads, k=1Tails, k=0Prob. = pProb. = 1-pThe random variable k is the number of headsit is variable because k can equal one or zeroit is random because the value of k depends on probabilities of occurrence, p and 1-p12Flipping a coin once•Expected value of the number of heads is the value of k weighted by the probability that value of k occurs–E(k) = 1*p + 0*(1-p) = p•variance of k is the value of k minus its expected value, squared, weighted by the probability that value of k occurs–VAR(k) = (1-p)2 *p +(0-p)2 *(1-p) = VAR(k) = (1-p)*p[(1-p)+p] =(1-p)*p13Flipping a coin twice: 4 elementary outcomesheadstailsheadstailsheadstailsh, hh, tt, ht, th, h; k=2h, t; k=1t, h; k=1t, t; k=0Prob =pProb =pProb =1-pProb =1-pProb=pProb=1-p14Flipping a Coin Twice•Expected number of heads–E(k)=2*p2 +1*p*(1-p) +1*(1-p)*p + 0*(1-p)2 E(k) = 2*p2 + p - p2 + p - p2 =2p–so we might expect the expected value of k in n independent flips is n*p•Variance in k–VAR(k) = (2-2p)2 *p2 + 2*(1-2p)2 *p(1-p) + (0-2p)2 (1-p)215Continuing with the variance in k–VAR(k) = (2-2p)2 *p2 + 2*(1-2p)2 *p(1-p) + (0-2p)2 (1-p)2–VAR(k) = 4(1-p)2 *p2 +2*(1 - 4p +4p2)*p*(1-p) + 4p2 *(1-p)2–adding the first and last terms, 8p2 *(1-p)2 + 2*(1 - 4p +4p2)*p*(1-p)–and expanding this last term, 2p(1-p) -8p2 *(1-p) + 8p3 *(1-p)–VAR(k) = 8p2 *(1-p)2 + 2p(1-p) -8p2 *(1-p)(1-p)–so VAR(k) = 2p(1-p) , or twice VAR(k) for 1 flip16•So we might expect the variance in n flips to be np(1-p)17Frequency Distribution for the Number of Heads•A fair coin18O heads1 head1/2probability# of headsOne Flip of the Coin19012 # of headsprobability1/41/2Two Flips of a Fair Coin20Three Flips of a Fair Coin•It is not so hard to see what the value of the number of heads, k, might be for three flips of a coin: zero, one ,two, three•But one head can occur two ways, as can two heads•Hence we need to consider the number of ways k can occur, I.e. the combinations of branching probabilities where order does not countHTHTpHT1-pp1 - ppHTHTHTHTp1-pp1-pThree flips of a coin; 8 elementary outcomes3 heads2 heads2 heads1 head2 heads1 head1 head0 heads22Three Flips of a Coin•There is only one way of getting three heads or of getting zero heads•But there are three ways of getting two heads or getting one head•One way of calculating the number of combinations is Cn(k) = n!/k!*(n-k)!•Another way of calculating the number of combinations is Pascal’s triangle232401 21/82/83/8Probability3 # of headsThree Flips of a Coin25The Probability of Getting k Heads•The probability of getting k heads (along a given branch) in n trials is: pk *(1-p)n-k•The number of branches with k heads in n trials is given by Cn(k)•So the probability of k heads in n trials is Prob(k) = Cn(k) pk *(1-p)n-k•This is the discrete binomial distribution where k can only take on discrete values of 0, 1, …kExpected Value of a discrete random variable•E(x) =• the expected value of a discrete random variable is the weighted average of the observations where the weight is the frequency of that observationniixpix0)]([*)(27Expected Value of the sum of random variables•E(x + y) = E(x) + E(y)Expected Number of Heads After Two Flips•Flip One: kiI heads•Flip Two: kjII heads•Because of independence p(kiI and kjII) = p(kiI)*p(kjII) •Expected number of heads after two flips: E(kiI + kjII) = (kiI + kjII) p(kiI)*p(kjII) •E(kiI + kjII) = kiI p(kiI)* p(kjII) +1010 ji10i10jCont.•E(kiI + kjII) = kiI p(kiI)* p(kjII) + kjII *p(kjII) p(kiI)• E(kiI +
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