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16/7/2007 Unit 2 - Stat 571 - Ramón V. León 1Unit 2: Review of ProbabilityStatistics 571: Statistical MethodsRamón V. León6/7/2007 Unit 2 - Stat 571 - Ramón V. León 2Approaches to Probability• Approaches to probability– Classical approach–Frequentist– Personal or subjective approach– Axiomatic approach• Basic ideas of axiomatic approach– Sample space– Events– Union– Intersection– Complement– Disjoint or mutually exclusive events–Inclusion26/7/2007 Unit 2 - Stat 571 - Ramón V. León 3Axioms of Probability• Axioms: –P(A)≥0– P(S) = 1 where S is the sample space–P(A ∪ B) = P(A) + P(B) if A and B are mutually exclusive events• Theorems about probability can be proved using these axioms• These theorems can be used in probability calculations– E.g. assuming all elements of the sample space are equally likely– Counting arguments used. (Take a look at Birthday Problem on Page 13.)6/7/2007 Unit 2 - Stat 571 - Ramón V. León 4Conditional Probability and Independence• Conditional probability– P(A | B) = P (A ∩ B) / P(B)• Events A and B are mutually independent if P (A | B) = P(A)– Implies P (A ∩ B) = P(A)P(B)36/7/2007 Unit 2 - Stat 571 - Ramón V. León 5Tossing Two Dice(6,6)(6,5)(6,4)(6,3)(6,2)(6,1)6(5,6)(5,5)(5,4)(5,3)(5,2)(5,1)5(4,6)(4,5)(4,4)(4,3)(4,2)(4,1)4(3,6)(3,5)(3,4)(3,3)(3,2)(3,1)3(2,6)(2,5)(2,4)(2,3)(2,2)(2,1)2(1,6)(1,5)(1,4)(1,3)(1,2)(1,1)1654321First Die OutcomeSecond Die OutcomeSample space has 6 x 6 = 36 outcomes6/7/2007 Unit 2 - Stat 571 - Ramón V. León 6Conditional Probability ExampleP(A)=8/36P(B)=18/3646/7/2007 Unit 2 - Stat 571 - Ramón V. León 7AIDS Example100009900100941094055TestNegative59049595Test positiveNot AIDSAIDSP(A) = 100/10000 =.01 P(+|A) = 95/100 =.95 P(-|~A) = 9405/9900 =.95P(A|+) = 95/590 =.16The usual way of solving this problem uses Bayes TheoremGivenConclude6/7/2007 Unit 2 - Stat 571 - Ramón V. León 8What Does a Positive HIV Test Means?56/7/2007 Unit 2 - Stat 571 - Ramón V. León 9Bayes Theorem Consequences() (|)PAPAB⎯⎯→(|) (|)PAB PBA≠() (| )PAPAData⎯⎯→6/7/2007 Unit 2 - Stat 571 - Ramón V. León 10Independence Example66/7/2007 Unit 2 - Stat 571 - Ramón V. León 11Random Variables•A random variable (r.v.) associates a unique numerical value with each outcome in the sample space• Example:• Discrete random variables: number of possible values is finite or countably infinite: x1, x2, x3, x4, x5, x6, …• Probability mass (density) function (p.m.f. or p.d.f.)– f(x) = P(X= x )• Cumulative distribution function (c.d.f.)– F(x) = P (X ≤ x) = 10X⎧=⎨⎩if coin toss results in a headif coin toss results in a tail()kxfk≤∑6/7/2007 Unit 2 - Stat 571 - Ramón V. León 12Discrete Random Variable Example76/7/2007 Unit 2 - Stat 571 - Ramón V. León 13Graphs of Probability Mass (Density) Function and Probability Distribution Function6/7/2007 Unit 2 - Stat 571 - Ramón V. León 14Continuous Random Variables()0fx≥•An r.v. is continuous if it can assume any value from one or more intervals of real numbers•Probability density function f(x) :() 1fxdx∞−∞=∫()()baPaXb fxdx≤≤=∫for anyab≤86/7/2007 Unit 2 - Stat 571 - Ramón V. León 15Cumulative Distribution FunctionThe cumulative distribution function (c.d.f.), denoted by F(x) , for a continuous random variable is given by: () ( ) ()xFxPXx fydy−∞=≤=∫It follows that ()()dF xfxdx=xf(x)F(x)6/7/2007 Unit 2 - Stat 571 - Ramón V. León 16Exponential Distribution Example96/7/2007 Unit 2 - Stat 571 - Ramón V. León 17Mean and Variance of Random Variables: Discrete Case2( ) ( ), ( ) ( ( )) ( )EX xfx VarX x EX fx==−∑∑6/7/2007 Unit 2 - Stat 571 - Ramón V. León 18Mean and Variance of Sum of Two Dice Tosses106/7/2007 Unit 2 - Stat 571 - Ramón V. León 19Expected Value or Mean of Random VariablesThe expected value or mean of a discrete r. v. X denoted by E(X), μX, or simply μ, is defined as:11 2 2() () () ()...xEX xfx xfx xfxμ== = + +∑The expected value of a continuous r. v. is defined as:() ()EX xfxdxμ==∫0Mean of Exponetial Distribution 1()xEX xe dxλλλ∞−==∫6/7/2007 Unit 2 - Stat 571 - Ramón V. León 20Variance and Standard DeviationThe variance of an r.v. X, denoted by Var(X),2Xσ, or simply2σis defined as22() ( )Var X E Xσμ== −We can show that()22() ( ) ()VarX EX EX=−The standard deviation (SD) is the square root of the varianceChallenge exercise: Show that for the exponential distribution the standard deviation is 1/λ116/7/2007 Unit 2 - Stat 571 - Ramón V. León 21Variance of the Mean of independent, Identically Distributed Random Variables()()12121222()111niiniiniiXVar X VarnVar XnVar Xnnnnσσ===⎛⎞⎜⎟=⎜⎟⎝⎠⎛⎞=⎜⎟⎝⎠⎛⎞=⎜⎟⎝⎠⎛⎞=⎜⎟⎝⎠=∑∑∑by independencesince the r.v.’s are identically distributed 12We often refer to , ,..., as a randon samplewith replacement orfrom a very large populationnXX X6/7/2007 Unit 2 - Stat 571 - Ramón V. León 22Quantiles and PercentilesFor 01p≤≤the pthquantile (or the 100pthpercentile), denoted bypθ,of a continuous r.v. X is defined by the following equation:()()ppPXFpθθ≤==.5θis called the medianpF(x)θp126/7/2007 Unit 2 - Stat 571 - Ramón V. León 23Exponential Distribution Percentiles6/7/2007 Unit 2 - Stat 571 - Ramón V. León 24Jointly Distributed Random Variables(, ) joint probability mass functionfxy=320.16200=136/7/2007 Unit 2 - Stat 571 - Ramón V. León 25Marginal DistributionDiscrete: ( ) ( ) ( , )Continuous: ( ) ( ) ( , )yXgx PX x fxygx f x fxydy∞−∞=====∑∫6/7/2007 Unit 2 - Stat 571 - Ramón V. León 26Conditional Distribution(, )(|) ( | )()fxyfyx PY yX xgx== ==Conditional probability mass function (p.m.f.):(4,1)0.005(1| 4)(4) 0.315PX YPY XPX===== ==146/7/2007 Unit 2 - Stat 571 - Ramón V. León 27Independent Random Variables and are independent r.v.'s if ( , ) ( ) ( )(, )Note that ( | ) ( )()XYfxygxhyfxyfyx hygx===6/7/2007 Unit 2 - Stat 571 - Ramón V. León 28Covariance and CorrelationIf X and Y are independent then E(XY)=E(X)E(Y) so the covarianceis zero. The other direction is not true; zero covariance does not imply independence.(,) ( )( ) ( ) ()()XY X YCov X Y E X Y E XY E X E Yσμμ== − −= −Note that:() (,)EXY xyf x y dxdy∞∞−∞ −∞=∫∫(,)(,)var( ) var( )XYXYXYCov X Ycorr X YXYσρσσ== =Measures strength of linear association 11XYρ−≤≤156/7/2007 Unit 2 - Stat 571 - Ramón V. León 29Covariance

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