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ECMT5001 Principles of EconometricsLecture 2Semester 1,2010In This Lecture:nRandom variablesnDiscrete random variablesnContinuous random variablesnProbabilities for normal random variablesRandom VariablesRandom VariablesnA random variable is a variable whose value is unknown until it is observed nThe outcome of an experiment is a random variable before the experiment takes placenWe use random variables to assign numerical values to the outcomes of an experimentRandom VariablesnA random variable has a set of possible values it can takene.g.,number of heads occurring when tossing two coins is a random variablenSet of possible values:0,1,2nDistinguish two types:qDiscrete&Continuous random variablenA discrete random variable can take only a limited or countable number of values(e.g.no.of children,TVs owned)nA continuous random variable can take any value in an interval.It has an uncountable number of possible values(GDP in 2008,income,Stock price index)Random VariablesNotationnWhen we refer to a random variable whose outcome is unknown we generally denote it with a capital letter(upper case)e.g.XnWhen we refer to one specific outcome of an experiment of the random variable we denote it the lower case version of the variable name e.g.xnWhen we have n outcomes of an experiment we use a subscript to distinguish them,e.g.x1,x2,x3,xn.Probability Density FunctionnThe probability density function(pdf)summarises the probabilities attached to possible outcomes of a random variablenA pdf can be presented as a table,a graph,or a formula,f(x)nFor discrete random variables the pdf is also called the probability mass function(pmf)nWe shall use the term probability density function for both continuous and discrete random variablesCumulative Distribution FunctionnAnother way to represent probabilities is the cumulative distribution function(cdf)nThe cdf for a random variable X gives the probability that X is less than or equal to a specific value xF(x)=P(X x)Discrete Random VariablesDiscrete Random VariablesnFor a discrete random variable the probability that a random variable X takes the value x isf(x)=P(X=x)nf(x)is a probability so 0 f(x)1 for all values of xnIf X can only take n values x1,xn,the probabilities must sum to 1:f(x1)+f(x2)+f(xn)=1,i.e.nIn this lecture we shall consider a few commonly used discrete distributions:qBernoulliqBinomialqPoisson Discrete Random VariablesBernoulli Random VariablenRandom variable X is a Bernoulli variable if it only takes values 0 or 1nUsed for experiments with only 2 possible outcomes(labelled success(1)or fail(0)nExamples:qHeads,TailsqRent,OwnqEmployed,unemployedqPass,Fail,qPurchase,Dont PurchaseqMale,Femalepdf for Bernoulli Variablen Probability density function in equation form:f(x;p)=px(1 p)1-x x=0,1 0 p 1 f(0;p)=P(x=0)=1-p f(1;p)=P(x=1)=pnHere p is a parameter giving the probability of successnE.g.If p=0.6 we have:f(0)=P(X=0)=0.4f(1)=P(X=1)=0.6Graphical Representation of Bernoulli pdfBernoulli pdf in a TableEmployment Statusxf(x)Unemployed00.4Employed10.6Binomial Random VariablesnA Binomial random variable is obtained if we repeat the Bernoulli experiment a number(n)of times nLet X=number of successes in n independent trialsn The pdf for the new binomial random variable X isq f(x;p,n)=px(1 p)n-x n!/x!(n-x)!q x=0,1,2,.,nq 0 p 1Binomial DistributionnNote that the binomial distribution has two parameters:qp the probability of a success in each trial,andqn the number of trialsnAlso,the symbol“!”means“factorial”nn!=n*(n-1)*(n-2)*2*1,and 0!=1(by definition)Calculation for n=10,X=6Binomial ExamplenLet X be the number of households with 1 phone in a random sample of n householdsnSuppose n=2nIf the probability that a single randomly drawn house has 1 phone is p thennP(X=0)=f(0;p;2)=(1-p)2P(X=1)=f(1;p;2)=2p(1-p)P(X=2)=f(2;p;2)=p2Binomial Example for n=2&p=0.4Example:Successful TradesnA long-term trader compares the annual return on stocks in his portfolio with that of the All Ordinaries Index representing the market average.He believes his chosen stocks return more than the All Ords 70%of the time.nLet X be a random variable counting the number of stocks in his 5-stock portfolio that beat the All Ords in 2008.nIf his belief is correct find the probability that 4 or more of his stocks beat the All Ords index this yearExample:Successful TradesPoisson Random VariablesnThe Poisson distribution is used to model the probability of“rare events”that occur randomly in time,distance or spacene.g.qNumber of accidents on NSW roads in a weekqNumber of arrivals a the train station ticketing office every 5 minutesqNumber of wombats encountered on a trip from Sydney to CanberraqNumber of days in a given year in which the All Ords index has a move of over 1%Poisson CharacteristicsnA Poisson experiment consists of counting the number,x,of times a particular event occurs during a given unit of time,or a given area,or volume,(or weight,or distance,or any other unit of measurement)nThe probability that an event occurs in a given unit of time,area,or volume is the same for all unitsnThe number of events that occur in one unit of time area or volume is independent of the number that occur in other unitsPoisson pdfnThe pdf for a Poisson distribution is given bywhere x=Number of rare events per unit of time,distance or space=Mean value of xe=2.71828Poisson ExamplenLet X be the number of calls received by a customer help-desk in a ten-minute periodnIf X is Poisson distributed with the mean number of calls in a ten-minute period being equal to 5,find the following:a)Probability of no calls b)Probability of only 1 callc)Probability of at least 2 callsPoisson Examplea)b)c)Continuous Random VariablesContinuous Random Variables nContinuous random variables can take any value in an interval on the number linenThey have(at least theoretically)an uncountable(infinite)number of valuesnThe probability of any specific value is zero(1/)nInstead we talk about outcomes being in a certain rangeContinuous Random VariablesnConsider for example the population of salaries for all Australian residentsnWhat is the probability that a randomly selected individual has a salary less than$100,000?nTo answer this question we need to know the proportion of all salaries that are below$100,000nThis proportion is the area under the relative frequency histogram that lies to the left of$100,000(see slide)Hypothetical Salaries DistributionProportion of salaries below$100,000Continuous Random VariablesnFor example,if the shaded area under the hypothetical relative frequency distribution is 80%of the total area under the curve,then the probability that a randomly selected individuals salary is less than$100,000 is 0.8nHowever,without the full population we will not know the exact shape of the distributionnIn this case,we postulate a model,i.e.select a smooth curve as a model for the population relative frequency distributionContinuous Random VariablesnTo find the probability that a particular observation lies in a particular interval we use the model and find the area under the curve that falls over that intervalnOf course we need to be sure the population relative frequency and our model are very similarnWe shall see in later lectures why we believe that the models we use are good approximations to realityContinuous Random VariablesnOur models are continuous functions,defined for the range of values our underlying random variable of interest can takenThe model is the probability density function(pdf)of the random variable:f(x)nTo find the probability of observing an outcome within a range of values we integrate the pdf across that rangeContinuous random variablesnProbabilities are found by integrating the pdf over the range:nIntegration gives the area under the curvenThe total area under the pdf must be 1:Cumulative Distribution FunctionnAnother way to represent probabilities is the cumulative distribution function(cdf)nThe cdf for a random variable X,F(x),gives the probability that X is less than or equal to a specific value xnWe shall consider two continuous distributions lecture:qUniform distributionqNormal distributionnOthers well come across laterq Students t distribution,q Chi-square distribution,q F distributionContinuous Random VariablesUniform DistributionnSuppose you were to random select a number x from the interval a x bnx is called a uniform random variable and its distribution is called the uniform distributionnThe shape of the distribution is a rectangle(see following slide)nThe height of the rectangle is 1/(b-a)(which guarantees the area under the rectangle is one)Uniform Distribution1/(b-a)f(x)xThe Uniform DistributionabUniform DistributionnProbability density function f(x)=1/(b-a)a x b=0 elsewherenWhat is the probability that a randomly selected number is less than c(where a c b)?(see next slide)Uniform DistributionP(X c)Area of rectangle=baseheight=(c-a)1/(b-a)=(c-a)/(b-a)Uniform ExamplenConsider randomly choosing a number between 2 and 6.What is the probability that the chosen number is between 3 and 5?nf(x)=1/(6-2)=0.25 2 x 6 =0 elsewhereUniform ExampleP(3.5 X 5)=0.375Normal DistributionnThe most widely-used and well-known distribution in statistics/econometrics is the Normal distributionnThe Normal distribution is commonly referred to as the bell-curve due to its familiar shapenProposed by Gauss(1777-1855),it remarkably provides an adequate distribution for the modelling of many variables as we shall see later in the courseNormal Density Functionn PropertiesqBell-shapedqSpreads out more as gets largerqSymmetric around the mean,P(X )=P(X )=0.5qTotal area under the curve=1Probabilities for Normal Random VariablesNormal DistributionnThe pdf for a normal random variable with mean and variance 2 is given bywhere -x 0nIf X follows a normal distribution we write it as X N(,2)Normal DistributionnUnfortunately,integration of the pdf for the normal distribution is intractablenThat is,there is no simple mathematical solution as there is with the uniform distributionnHowever,probabilities for the standard normal distribution have been tabulated for various values and we can use these to obtain probabilities for any normal distributionStandard Normal DistributionnThe standard normal random variable is usually denoted Z and an observed outcome is denoted znThe standard normal distribution is a normal distribution with mean=0 and variance=1i.e.Z N(0,1)nProbabilities for the standard normal cumulative distribution are tabulated(see following slide)Standard Normal Table*-page 1z0123456789-30.00130.00130.00130.00120.00120.00110.00110.00110.0010.001-2.90.00190.00180.00180.00170.00160.00160.00150.00150.00140.0014-2.80.00260.00250.00240.00230.00230.00220.00210.00210.0020.0019-2.70.00350.00340.00330.00320.00310.0030.00290.00280.00270.0026-2.60.00470.00450.00440.00430.00410.0040.00390.00380.00370.0036-2.50.00620.0060.00590.00570.00550.00540.00520.00510.00490.0048-2.40.00820.0080.00780.00750.00730.00710.00690.00680.00660.0064-2.30.01070.01040.01020.00990.00960.00940.00910.00890.00870.0084-2.20.01390.01360.01320.01290.01250.01220.01190.01160.01130.011-2.10.01790.01740.0170.01660.01620.01580.01540.0150.01460.0143-20.02280.02220.02170.02120.02070.02020.01970.01920.01880.0183-1.90.02870.02810.02740.02680.02620.02560.0250.02440.02390.0233-1.80.03590.03510.03440.03360.03290.03220.03140.03070.03010.0294-1.70.04460.04360.04270.04180.04090.04010.03920.03840.03750.0367-1.60.05480.05370.05260.05160.05050.04950.04850.04750.04650.0455-1.50.06680.06550.06430.0630.06180.06060.05940.05820.05710.0559-1.40.08080.07930.07780.07640.07490.07350.07210.07080.06940.0681-1.30.09680.09510.09340.09180.09010.08850.08690.08530.08380.0823-1.20.11510.11310.11120.10930.10750.10560.10380.1020.10030.0985-1.10.13570.13350.13140.12920.12710.12510.1230.1210.1190.117-10.15870.15620.15390.15150.14920.14690.14460.14230.14010.1379-0.90.18410.18140.17880.17620.17360.17110.16850.1660.16350.1611-0.80.21190.2090.20610.20330.20050.19770.19490.19220.18940.1867-0.70.2420.23890.23580.23270.22960.22660.22360.22060.21770.2148-0.60.27430.27090.26760.26430.26110.25780.25460.25140.24830.2451-0.50.30850.3050.30150.29810.29460.29120.28770.28430.2810.2776-0.40.34460.34090.33720.33360.330.32640.32280.31920.31560.3121-0.30.38210.37830.37450.37070.36690.36320.35940.35570.3520.3483-0.20.42070.41680.41290.4090.40520.40130.39740.39360.38970.3859-0.10.46020.45620.45220.44830.44430.44040.43640.43250.42860.4247-00.50.4960.4920.4880.4840.48010.47610.47210.46810.4641*Appendix G.1,Wooldridge(2009)Standard Normal Table*-page 2z012345678900.50.5040.5080.5120.5160.51990.52390.52790.53190.53590.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.57530.20.57930.58320.58710.5910.59480.59870.60260.60640.61030.61410.30.61790.62170.62550.62930.63310.63680.64060.64430.6480.65170.40.65540.65910.66280.66640.670.67360.67720.68080.68440.68790.50.69150.6950.69850.70190.70540.70880.71230.71570.7190.72240.60.72570.72910.73240.73570.73890.74220.74540.74860.75170.75490.70.7580.76110.76420.76730.77040.77340.77640.77940.78230.78520.80.78810.7910.79390.79670.79950.80230.80510.80780.81060.81330.90.81590.81860.82120.82380.82640.82890.83150.8340.83650.838910.84130.84380.84610.84850.85080.85310.85540.85770.85990.86211.10.86430.86650.86860.87080.87290.87490.8770.8790.8810.8831.20.88490.88690.88880.89070.89250.89440.89620.8980.89970.90151.30.90320.90490.90660.90820.90990.91150.91310.91470.91620.91771.40.91920.92070.92220.92360.92510.92650.92790.92920.93060.93191.50.93320.93450.93570.9370.93820.93940.94060.94180.94290.94411.60.94520.94630.94740.94840.94950.95050.95150.95250.95350.95451.70.95540.95640.95730.95820.95910.95990.96080.96160.96250.96331.80.96410.96490.96560.96640.96710.96780.96860.96930.96990.97061.90.97130.97190.97260.97320.97380.97440.9750.97560.97610.976720.97720.97780.97830.97880.97930.97980.98030.98080.98120.98172.10.98210.98260.9830.98340.98380.98420.98460.9850.98540.98572.20.98610.98640.98680.98710.98750.98780.98810.98840.98870.9892.30.98930.98960.98980.99010.99040.99060.99090.99110.99130.99162.40.99180.9920.99220.99250.99270.99290.99310.99320.99340.99362.50.99380.9940.99410.99430.99450.99460.99480.99490.99510.99522.60.99530.99550.99560.99570.99590.9960.99610.99620.99630.99642.70.99650.99660.99670.99680.99690.9970.99710.99720.99730.99742.80.99740.99750.99760.99770.99770.99780.99790.99790.9980.99812.90.99810.99820.99820.99830.99840.99840.99850.99850.99860.998630.99870.99870.99870.99880.99880.99890.99890.99890.9990.999*Appendix G.1,Wooldridge(2009)Using Standard Normal TablesnThe standard normal tables we use give the cumulative distribution functionnIf you look up a number say z*,the table providesP(-Z 1.32)c.P(0.08 Z 1.67)d.P(-1.35 Z 0.8)Exercisesa.Note P(Z -1.8)=P(Z 1.32)=Exercisesc.P(0.08 Z 1.67)=P(Z 1.67)-P(Z 0.08)P(Z 1.67)=0.9525(from tables)P(Z 0.08)=0.5319(from tables)P(0.08 Z 1.67)=0.9525-0.5319=0.4206Exercisesd.P(-1.35 Z 0.8)=Properties of the NormalnA linear transformation of a normal variable results in another normal variable.nThat is,if X N(,2)and Y=a+bX,then Y is also a normal variablenSpecificallyY N(a+b,b22)new meannew varianceStandardising a normal variablenConsider the transformation Y=a+bXnA special case occurs when a=-/,b=1/nThe variable Y will have a mean equal to:a+b=-/+1/=-/+/=0nand variance equal to b22=(1/)2 2=1nAnd we get a Standard Normal variable,i.e.Y=Z N(0,1)Standardising a normal variablenSo with a=-/,b=1/we havenSo to convert a normal random variable,X,to a standard normal variable Z we simply subtract from X its mean and divide by its standard deviationnWe can standardise any normal variablenProbabilities for standard normal variables are tabulated(as seen above)nAll normal variables must be transformed to a standard normal variable to use tablesnWe can then analyse any normal distribution with the one table(rather than requiring an infinite number of tables!)Standardising a normal variableConverting from X to ZnWhen finding probabilities for a non-standard normal r.v.,say XN(,2)we must first convert it to a standard normal variable,Z,using the formulaConverting from X to ZnWe can then look up the probability P(Z z*)in the Standard Normal tablesnExamples:nIf X N(60,100)i.e.=60,2=100find the following:e.P(X 65)f.P(45 X 70)Converting from X to Z-Examplese.X N(60,100)(from tables)X N(60,100)Z N(0,1)Converting from X to Z-Examplesf.X N(60,100)Reverse CalculationnIf we are given a probability we can also work backwards to find the value z*(or x*)nTry the following:g.If P(Z z*)=0.9678,find z*h.If XN(10,25)and P(X x*)=0.0694,find x*Reverse Calculation-Examplesg.If P(Z z*)=0.9678,find z*Reverse Calculation-Examplesh.If XN(10,25)and P(X x*)=0.0694,find x*From tables:P(Z -1.48)=0.0694,So z*=-1.48Probability=0.0694X N(10,25)Z N(0,1)z*=-1.48x*=2.6
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