Continuous Random Variables Expected Values …

1 Continuous Random VariablesExpected Values and MomentsStatistics 110 Summer 2006 Copyrightc 2006 by Mark E. IrwinContinuous Random VariablesWhen defining a distribution for a Continuous RV, the PMF approach won tquite work since summations only work for a finite or a countably infinitenumber of items. Instead they are based on the followingDefinition:LetXbe a Continuous RV. TheProbability Density Function(PDF) is a functionf(x)on the range ofXthat satisfies the (x) f(x) 0 fis piecewise Continuous f(x)dx= 1 Continuous Random Variables1 For anya < b, the probability thatP[a < X < b]is the area under thedensity curve (x)abP[a < X < b] = baf(x)dxContinuous Random Variables2 Note thatf(a)isNOTthe probability of observingX=aasP[X=a] = aaf(x)dx= 0 Thus the probability that a Continuous RV takes on any particular value is0.

2 (While this might seem counterintuitive, things do work properly.) Aconsequence of this is thatP[a < X < b] =P[a X < b] =P[a < X b] =P[a X b]for Continuous RVs. Note that this won t hold for discrete Random Variables3 Note that for small , iffis Continuous atxP[x 2 X x+ 2]= x+ 2x 2f(u)du f(x) Xf(x)x 2x+ 2xSo the probability of seeing an outcome in a small interval aroundxisproportional tof(x). So the PDF is giving information of how likely anobservation Random Variables4As with the PMF and the CDF for discrete RVs, there is a relationshipbetween the PDF,f(x), and the CDF,F(x), for Continuous RVsF(x) =P[X x] = x f(u)duf(x) =F (x)assuming thatfis Continuous on this relationship, the probability for any reasonable event describinga RV can determined with the CDF as the probability of any interval satisfiesP[a < X b] =F(b) F(a)Note that this is slightly different than the formula given on page 47.

3 Theabove holds for any RV (discrete, Continuous , mixed). The form given onpage 47P[a X b] =F(b) F(a)only holds for Continuous Random Variables5 Example: Uniform RV on [0,1] (DenotedX U(0,1))What most people think of when we say pick a numberbetween 0 and 1. Any real number in the interval ispossible and equally likely, implying that any interval oflengthhmust have the same probability (which needsto beh). The PDF forXthen must bef(x) ={1 0 x 10x <0orx >1 of U(0,1)xf(x) Continuous Random Variables6 The CDF for aU(0,1)isF(x) = 0x <0x0 x 11x >1 of U(0,1)xF(x) Continuous Random Variables7 One way to think of the CDF is that you give a value of the RV and it givesa probability associated with it ( [X x]).}

4 It can also be useful to gothe other way. Give a probability and figure out which value of the RV isassociated with assume thatFis Continuous and strictly increasing in some intervalI( 0to the left ofIandF= 1to the right ofI) (noteImightbe unbounded). Under these assumptions the inverse functionF 1is welldefined (x=F 1(y)ifF(x) =y).Definition:ThepthQuantileof the distributionFis defined to be thevaluexpsuch thatF(xp) =porP[X xp] =pUnder the above assumptionsxp=F 1(p). Continuous Random Variables8 4 (x)xppSpecial cases of interest of theMedian(p=12) and the lower and upperQuartiles(p=14and=34) Continuous Random Variables9 Note.

5 Defining quantiles for discrete distributions is a bit tougher since theCDF doesn t take all Values between 0 and 1 (due to the jumps) for number of heads in 3 flipsx (number of heads)P[X <= x]The definition above can be extended to solving the simultaneous equationsP[X xp] pandP[X < xp] pContinuous Random Variables10 This can be though of as the place where the CDF jumps from belowptoabovep for number of heads in 3 flipsx (number of heads)P[X <= x] function for number of heads in 3 flipspxpContinuous Random Variables11 Expected Values and MomentsDefinition:TheExpected Valueof a Continuous RVX(with PDFf(x))isE[X] = xf(x)dxassuming that |x|f(x)dx <.

6 The Expected value of a distribution is often referred to as the mean of with the discrete case, the absolute integrability is a technical point,which if ignored, can lead to an example of a Continuous RV with infinite mean, see the Cauchydistribution (Example G, page 114) Expected Values and Moments12As with the discrete case,E[X]can be thought as a measure of center ofthe Random example, whenX U(0,1)E[X] = 10xdx= of U(0,1)xf(x) Expected Values and Moments13 Not surprisingly, expectations of functions of Continuous RVs satisfy theexpected relationshipE[g(X)] = g(x)f(x)dxFor example, ifX U(0,1),E[X2] = 10x2dx=13 This is often easier than figuring out the PDF ofY=g(X)and applyingthe definition as there is often some work to figure out the PDF ofY.

7 (Which we will do later, it does have its uses)As with discrete RVs,g(E[X])6=E[g(X)]in most cases. However, with alinear transformationY=a+bXE[a+bX] =a+bE[X] Expected Values and Moments14 Spread of a RV 2 1012XP[X=x] 2 1012XP[X=x] 0 1p(x)131313x-2 -1 0 1 2p(x)1929392919 Expected Values and Moments15 2 (x) 2 (x)f(x) ={ 1 x 10 Otherwisef(x) = +x4 2 x x40 x 20 OtherwiseAll these distributions haveE[X] = 0but the right hand side in each casehas a bigger spread. A common measure of spread is the Standard DeviationExpected Values and Moments16 Definition:Let =E[X], then theVarianceof the Random variableXisVar(X) =E[(X )2]provided the expectation DeviationofXisSD(X) = Var(X)For a discrete RV,Var(X) = i(xi )2p(xi)For a Continuous RVVar(X) = (x )2f(x)dxExpected Values and Moments17 The variance measures the Expected squared difference of an observationfrom the mean.}

8 While the interpretation of the standard deviation isn tquite easy, it can be thought of a measure of the typical spread of a can be shown that, assuming that the variance exists,Var(X) =E[X2] (E[X])2 This form is often useful for calculation : The variance is often denoted by 2and the standard deviationby . Expected Values and Moments18 For the examples 2 1012XP[X=x] 1 0 1p(x)131313 Var(X) = ( 1 0)213+ (0 0)213+ (1 0)213=23SD(X) = 23= Values and Moments19 2 1012XP[X=x] 2 1 0 1 2p(x)1929392919 Var(X) = ( 2 0)219+ ( 1 0)229+ (0 0)239+ (1 0)229+ (2 0)219=109SD(X) = 109= Values and Moments20 2 (x)f(x) ={ 1 x 10 OtherwiseVar(X) = 1 1(x 0)212dx=13SD(X) = 13= Values and Moments21 2 (x)f(x) = +x4 2 x x40 x 20 OtherwiseVar(X) = 2 20(x 0)2( x4)dx=43SD(X)}

9 = 43= Values and Moments22 What is the effect of a linear transformation (Y=a+bX) on the varianceand standard deviation?Var(a+bX) =b2 Var(X) SD(a+bX) =|b|SD(X)These two results are to be Expected . For example, if two possibleXvaluesdiffer byd=|x1 x2|, the correspondingYvalues differ by|b|d, suggestingthat we want the standard deviation to scale by a factor of|b|. Since thevariance measures squared spread, it needs to scale by a factor factoranot having an effect also makes sense. Addingato a randomvariable shifts the location of its distribution, but doesn t changes thedistance between corresponding pairs of Values and Moments23