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Properties of Expected values and Variance

Properties of Expected values and VarianceChristopher CrokeUniversity of PennsylvaniaMath 115 UPenn, Fall 2011 Christopher CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).You get from one integral to the other by careful uses consequence isE(aX+b) = (ax+b)f(x)dx=aE(X) +b.(It is not usually the case thatE(r(X)) =r(E(X)).)Similar facts old for discrete random CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).

Properties of Expected values and Variance Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011 Christopher Croke Calculus 115. Expected value Consider a random variable Y = r(X) for some function r, e.g. Y = X2 + 3 so in …

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1 Properties of Expected values and VarianceChristopher CrokeUniversity of PennsylvaniaMath 115 UPenn, Fall 2011 Christopher CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).You get from one integral to the other by careful uses consequence isE(aX+b) = (ax+b)f(x)dx=aE(X) +b.(It is not usually the case thatE(r(X)) =r(E(X)).)Similar facts old for discrete random CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).

2 You get from one integral to the other by careful uses consequence isE(aX+b) = (ax+b)f(x)dx=aE(X) +b.(It is not usually the case thatE(r(X)) =r(E(X)).)Similar facts old for discrete random CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).You get from one integral to the other by careful uses consequence isE(aX+b) = (ax+b)f(x)dx=aE(X) +b.(It is not usually the case thatE(r(X)) =r(E(X)).)Similar facts old for discrete random CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).

3 You get from one integral to the other by careful uses consequence isE(aX+b) = (ax+b)f(x)dx=aE(X) +b.(It is not usually the case thatE(r(X)) =r(E(X)).)Similar facts old for discrete random CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).You get from one integral to the other by careful uses consequence isE(aX+b) = (ax+b)f(x)dx=aE(X) +b.(It is not usually the case thatE(r(X)) =r(E(X)).)Similar facts old for discrete random CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).

4 You get from one integral to the other by careful uses consequence isE(aX+b) = (ax+b)f(x)dx=aE(X) +b.(It is not usually the case thatE(r(X)) =r(E(X)).)Similar facts old for discrete random CrokeCalculus 115 Expected valueConsider a random variableY=r(X) for some functionr, + 3 so in this caser(x) =x2+ turns out (and wehave already used) thatE(r(X)) = r(x)f(x) is not obvious since by definitionE(r(X)) = xfY(x)dxwherefY(x) is the probability density function ofY=r(X).You get from one integral to the other by careful uses consequence isE(aX+b) = (ax+b)f(x)dx=aE(X) +b.(It is not usually the case thatE(r(X)) =r(E(X)).)Similar facts old for discrete random CrokeCalculus 115 ProblemForXthe uniform distribution on [0,2] what isE(X2)?IfX1,X2,X3,..Xnare random variables andY=r(X1,X2,X3,..Xn) thenE(Y) = .. r(x1,x2,x3,..,xn)f(x1,x2,x3,..,xn) (x1,x2,x3,..,xn) is the joint probability density again our example of randomly choosing a pointin [0,1] [0,1].

5 We could letXbe the random variable of choosingthe first coordinate andYthe second. What isE(X+Y)?(note thatf(x,y) = 1.)Easy Properties of Expected values :IfPr(X a) = 1 thenE(X) (X b) = 1 thenE(X) CrokeCalculus 115 ProblemForXthe uniform distribution on [0,2] what isE(X2)?IfX1,X2,X3,..Xnare random variables andY=r(X1,X2,X3,..Xn) thenE(Y) = .. r(x1,x2,x3,..,xn)f(x1,x2,x3,..,xn) (x1,x2,x3,..,xn) is the joint probability density again our example of randomly choosing a pointin [0,1] [0,1].We could letXbe the random variable of choosingthe first coordinate andYthe second. What isE(X+Y)?(note thatf(x,y) = 1.)Easy Properties of Expected values :IfPr(X a) = 1 thenE(X) (X b) = 1 thenE(X) CrokeCalculus 115 ProblemForXthe uniform distribution on [0,2] what isE(X2)?IfX1,X2,X3,..Xnare random variables andY=r(X1,X2,X3,..Xn) thenE(Y) = .. r(x1,x2,x3,..,xn)f(x1,x2,x3,..,xn) (x1,x2,x3,..,xn) is the joint probability density again our example of randomly choosing a pointin [0,1] [0,1].

6 We could letXbe the random variable of choosingthe first coordinate andYthe second. What isE(X+Y)?(note thatf(x,y) = 1.)Easy Properties of Expected values :IfPr(X a) = 1 thenE(X) (X b) = 1 thenE(X) CrokeCalculus 115 ProblemForXthe uniform distribution on [0,2] what isE(X2)?IfX1,X2,X3,..Xnare random variables andY=r(X1,X2,X3,..Xn) thenE(Y) = .. r(x1,x2,x3,..,xn)f(x1,x2,x3,..,xn) (x1,x2,x3,..,xn) is the joint probability density again our example of randomly choosing a pointin [0,1] [0,1].We could letXbe the random variable of choosingthe first coordinate andYthe second. What isE(X+Y)?(note thatf(x,y) = 1.)Easy Properties of Expected values :IfPr(X a) = 1 thenE(X) (X b) = 1 thenE(X) CrokeCalculus 115 ProblemForXthe uniform distribution on [0,2] what isE(X2)?IfX1,X2,X3,..Xnare random variables andY=r(X1,X2,X3,..Xn) thenE(Y) = .. r(x1,x2,x3,..,xn)f(x1,x2,x3,..,xn) (x1,x2,x3,..,xn) is the joint probability density again our example of randomly choosing a pointin [0,1] [0,1].

7 We could letXbe the random variable of choosingthe first coordinate andYthe second. What isE(X+Y)?(note thatf(x,y) = 1.)Easy Properties of Expected values :IfPr(X a) = 1 thenE(X) (X b) = 1 thenE(X) CrokeCalculus 115 Properties ofE(X)A little more surprising (but not hard and we have already used):E(X1+X2+X3+..+Xn) =E(X1) +E(X2) +E(X3) +..+E(Xn).Another way to look at binomial random variables;LetXibe 1 if theithtrial is a success and 0 if a thatE(Xi) = 0 q+ 1 p= binomial variable (the number of successes) isX=X1+X2+X3+..+XnsoE(X) =E(X1) +E(X2) +E(X3) +..+E(Xn) = about products?Only works out well if the random variablesareindependent. IfX1,X2,X3,..Xnare independent randomvariables then:E(n i=1Xi) =n i=1E(Xi).Christopher CrokeCalculus 115 Properties ofE(X)A little more surprising (but not hard and we have already used):E(X1+X2+X3+..+Xn) =E(X1) +E(X2) +E(X3) +..+E(Xn).Another way to look at binomial random variables;LetXibe 1 if theithtrial is a success and 0 if a thatE(Xi) = 0 q+ 1 p= binomial variable (the number of successes) isX=X1+X2+X3+.

8 +XnsoE(X) =E(X1) +E(X2) +E(X3) +..+E(Xn) = about products?Only works out well if the random variablesareindependent. IfX1,X2,X3,..Xnare independent randomvariables then:E(n i=1Xi) =n i=1E(Xi).Christopher CrokeCalculus 115 Properties ofE(X)A little more surprising (but not hard and we have already used):E(X1+X2+X3+..+Xn) =E(X1) +E(X2) +E(X3) +..+E(Xn).Another way to look at binomial random variables;LetXibe 1 if theithtrial is a success and 0 if a thatE(Xi) = 0 q+ 1 p= binomial variable (the number of successes) isX=X1+X2+X3+..+XnsoE(X) =E(X1) +E(X2) +E(X3) +..+E(Xn) = about products?Only works out well if the random variablesareindependent. IfX1,X2,X3,..Xnare independent randomvariables then:E(n i=1Xi) =n i=1E(Xi).Christopher CrokeCalculus 115 Properties ofE(X)A little more surprising (but not hard and we have already used):E(X1+X2+X3+..+Xn) =E(X1) +E(X2) +E(X3) +..+E(Xn).Another way to look at binomial random variables;LetXibe 1 if theithtrial is a success and 0 if a thatE(Xi) = 0 q+ 1 p= binomial variable (the number of successes) isX=X1+X2+X3+.

9 +XnsoE(X) =E(X1) +E(X2) +E(X3) +..+E(Xn) = about products?Only works out well if the random variablesareindependent. IfX1,X2,X3,..Xnare independent randomvariables then:E(n i=1Xi) =n i=1E(Xi).Christopher CrokeCalculus 115 Properties ofE(X)A little more surprising (but not hard and we have already used):E(X1+X2+X3+..+Xn) =E(X1) +E(X2) +E(X3) +..+E(Xn).Another way to look at binomial random variables;LetXibe 1 if theithtrial is a success and 0 if a thatE(Xi) = 0 q+ 1 p= binomial variable (the number of successes) isX=X1+X2+X3+..+XnsoE(X) =E(X1) +E(X2) +E(X3) +..+E(Xn) = about products?Only works out well if the random variablesareindependent. IfX1,X2,X3,..Xnare independent randomvariables then:E(n i=1Xi) =n i=1E(Xi).Christopher CrokeCalculus 115 Properties ofE(X)A little more surprising (but not hard and we have already used):E(X1+X2+X3+..+Xn) =E(X1) +E(X2) +E(X3) +..+E(Xn).Another way to look at binomial random variables;LetXibe 1 if theithtrial is a success and 0 if a thatE(Xi) = 0 q+ 1 p= binomial variable (the number of successes) isX=X1+X2+X3+.

10 +XnsoE(X) =E(X1) +E(X2) +E(X3) +..+E(Xn) = about products?Only works out well if the random variablesareindependent. IfX1,X2,X3,..Xnare independent randomvariables then:E(n i=1Xi) =n i=1E(Xi).Christopher CrokeCalculus 115 Properties ofVar(X)Problem:Considerindependentrando m variablesX1,X2, andX3, whereE(X1) = 2,E(X2) = 1, andE(X3)=0. ComputeE(X21(X2+ 3X3)2).There is not enough information!Also assumeE(X21) = 3,E(X22) = 1, andE(X23) = aboutVar(X):Var(X) = 0 means the same as: there is acsuch thatPr(X=c) = 1. 2(X) =E(X2) E(X)2(alternative definition) 2(aX+b) =a2 2(X).Proof: 2(aX+b) =E[(aX+b (a +b))2]=E[(aX a )2]=a2E[(X )2]=a2 2(X).ForindependentX1,X2,X3,..,Xn 2(X1+X2+X3+..+Xn) = 2(X1)+ 2(X2)+ 2(X3)+..+ 2(Xn).Christopher CrokeCalculus 115 Properties ofVar(X)Problem:Considerindependentrando m variablesX1,X2, andX3, whereE(X1) = 2,E(X2) = 1, andE(X3)=0. ComputeE(X21(X2+ 3X3)2).There is not enough information!

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