Table of contentsOutlineProbabilityRandom variablesPMFs and PDFsCDFs, survival functions and quantilesLecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesLecture 2Brian CaffoDepartment of BiostatisticsJohns Hopkins Bloomberg School of Public HealthJohns Hopkins UniversityOctober 21, 2007Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesTable of contents1 Table of contents2 Outline3 Probability4 Random variables5 PMFs and PDFs6 CDFs, survival functions and quantilesLecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesOutline•Define probability calculus•Basic axioms of probability•Define random variables•Define density and mass functions•Define cumulative distribution functions and survivorfunctions•Define quantiles, percentiles, mediansLecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesProbability measuresA probability measure, P, is a real valued function from thecollection of possible events so that the following hold1. For an event E ⊂ Ω, 0 ≤ P(E ) ≤ 12. P(Ω) = 13. If E1and E2are mutually exclusive eventsP(E1∪ E2) = P(E1) + P(E2).Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesAdditivityPart 3 of the definition implies finite additivityP(∪ni =1Ai) =nXi =1P(Ai)where the {Ai} are mutually exclusive.This is usually extended to countable additivityP(∪∞i =1Ai) =∞Xi =1P(Ai)Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesNote•P is defined on F a collection of subsets of Ω•Example Ω = {1, 2, 3} thenF = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} .•When Ω is a continuous set, the definition gets muchtrickier. In this case we assume that F is sufficiently richso that any set that we’re interested in will be in it.Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesConsequencesYou should be able to prove all of the following:•P(∅) = 0•P(E ) = 1 − P(Ec)•P(A ∪ B) = P(A) + P(B) − P(A ∩ B)•if A ⊂ B then P(A) ≤ P(B)•P (A ∪ B) = 1 − P(Ac∩ Bc)•P(A ∩ Bc) = P(A) − P(A ∩ B)•P(∪ni =1Ei) ≤Pni =1P(Ei)•P(∪ni =1Ei) ≥ maxiP(Ei)Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesExampleProof that P(E) = 1 − P(Ec)1 = P(Ω)= P(E ∪ Ec)= P(E ) + P(Ec)Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesExampleProof that P(∪ni =1Ei) ≤Pni =1P(Ei)P(E1∪ E2) = P(E1) + P(E2) − P(E1∩ E2)≤ P(E1) + P(E2)Assume the statement is true for n − 1 and consider nP(∪ni =1Ei) ≤ P(En) + P(∪n−1i =1Ei)≤ P(En) +n−1Xi =1P(Ei)=nXi =1P(Ei)Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesExampleThe National Sleep Foundation (www.sleepfoundation.org)reports that around 3% of the American population has sleepapnea. They also report that around 10% of the NorthAmerican and European population has restless leg syndrome.Similarly, they report that 58% of adults in the US experienceinsomnia. Does this imply that 71% of people will have at leastone sleep problems of these sorts?Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesExample continuedAnswer: No, the events are not mutually exclusive. Toelaborate let:A1= {Person has sleep apnea}A2= {Person has RLS}A3= {Person has insomnia}Then (work out the details for yourself)P(A1∪ A2∪ A3) = P(A1) + P(A2) + P(A3)− P(A1∩ A2) − P(A1∩ A3) − P(A2∩ A3)+ P(A1∩ A2∩ A3)= .71 + Other stuffwhere the “Other stuff” has to be less than 0Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesExample : LA Times from Ricepage 26Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesRandom variables•A random variable is a numerical outcome of anexperiment.•The random variables that we study will come in twovarieties, discrete or continuous.•Discrete random variable are random variables that takeon only a countable number of possibilities.•P(X = k)•Continuous random variable can take any value on the realline or some subset of the real line.•P(X ∈ A)Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesExamples of random variables•The (0 − 1) outcome of the flip of a coin•The outcome from the roll of a die•The BMI of a subject four years after a baselinemeasurement•The hypertension status of a subject randomly drawn froma populationLecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesPMFA probability mass function evaluated at a value corresponds tothe probability that a random variable takes that value. To bea valid pmf a function, p, must satisfy1 p(x) ≥ 0 for all x2Pxp(x) = 1The sum is taken over all of the possible values for x.Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesExampleLet X be the result of a coin flip where X = 0 represents tailsand X = 1 represents heads.p(x) = (1/2)x(1/2)1−xfor x = 0, 1Suppose that we do not know whether or not the coin is fair;Let θ be the probability of a headp(x) = θx(1 − θ)1−xfor x = 0, 1Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesPDFA probability density function (pdf), is a function associatedwith a continuous random variableAreas under pdfs correspond to probabilities for thatrandom variableTo be a valid pdf, a function f must satisfy1 f (x) ≥ 0 for all x2R∞−∞f (x)dx = 1Lecture 2Brian CaffoTable ofcontentsOutlineProbabilityRandomvariablesPMFs andPDFsCDFs, survivalfunctions andquantilesExampleAssume that the time in years from diagnosis until death ofpersons with a specific kind of cancer follows a density likef (x) =(e−x /55for x > 00 otherwiseMore compactly