01 Random Variables Slides

01 Random Variables Slides - Random Variables Introduction...

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Unformatted text preview: Random Variables Introduction to Econometrics Lecture Douglas G. Steigerwald UC Santa Barbara August 2010 D. Steigerwald (UCSB) Random Variables August 2010 1 / 11 My Research Group 1 marine reserves D. Steigerwald (UCSB) Random Variables August 2010 2 / 11 My Research Group 1 marine reserves 2 conservation banking D. Steigerwald (UCSB) Random Variables August 2010 2 / 11 My Research Group 1 marine reserves 2 conservation banking 3 policing: safety or revenue D. Steigerwald (UCSB) Random Variables August 2010 2 / 11 My Research Group 1 marine reserves 2 conservation banking 3 policing: safety or revenue 4 judicial independence and growth D. Steigerwald (UCSB) Random Variables August 2010 2 / 11 My Research Group 1 marine reserves 2 conservation banking 3 policing: safety or revenue 4 judicial independence and growth 5 legislative redistricting and polarization D. Steigerwald (UCSB) Random Variables August 2010 2 / 11 My Research Group 1 marine reserves 2 conservation banking 3 policing: safety or revenue 4 judicial independence and growth 5 legislative redistricting and polarization 6 speeding and gasoline prices D. Steigerwald (UCSB) Random Variables August 2010 2 / 11 My Research Group 1 marine reserves 2 conservation banking 3 policing: safety or revenue 4 judicial independence and growth 5 legislative redistricting and polarization 6 speeding and gasoline prices 7 charity ratings and donations D. Steigerwald (UCSB) Random Variables August 2010 2 / 11 My Research Group 1 marine reserves 2 conservation banking 3 policing: safety or revenue 4 judicial independence and growth 5 legislative redistricting and polarization 6 speeding and gasoline prices 7 charity ratings and donations 8 anti-competitive behavior in Hawaii D. Steigerwald (UCSB) Random Variables August 2010 2 / 11 Summation Notation De…nition: Summation operator ∑ ∑2=1 xj pj = x1 p1 + x2 p2 j Product 2 2 ∑j =1 xj pj = x1 p1 D. Steigerwald (UCSB) = (x1 p1 + x2 p2 )2 x1 p1 + x1 p1 x2 p2 + x2 p2 x1 p1 + x2 p2 x2 p2 = x1 p1 ∑2=1 xj pj + x2 p2 ∑2=1 xj pj j j = ∑2 =1 xk pk ∑2=1 xj pj j k = ∑2 =1 ∑2=1 xk pk xj pj j k Random Variables August 2010 3 / 11 Location: Mean De…nition: Expectation operator E random variable X takes J distinct values E (X ) = ∑J=1 xj P (X = xj ) := µX j Example: fair die E (X ) = ∑6=1 j j 1 6 = 3.5 mean is not one of the distinct values D. Steigerwald (UCSB) Random Variables August 2010 4 / 11 Dispersion Variance measures dispersion Var (X ) = E (X µX )2 = ∑J=1 (xj µX )2 P (X = xj ) := σ2 j X Var (X ) 0 X - investment return in dollars Var (X ) - units are dollars2 q standard deviation: σX = two investments: X and Z prefer X or Z ? D. Steigerwald (UCSB) σ2 - units are dollars X µX = µZ Random Variables σ2 < σ2 X Z August 2010 5 / 11 Covariance Covariance measures linear relation Cov (X , Z ) = E [(X µX ) (Z µZ )] = ∑J=1 ∑K=1 (xj µX ) (zk µZ ) P (X = xj , Z = zk ) j k Cov (X , Z ) could be negative X , Z - investment returns in dollars Cov (X , Z ) - units are dollar2 replace square root (Cov can be negative) with correlation Cor (X , Z ) = 1 Cor (X , Z ) D. Steigerwald (UCSB) Cov (X ,Z ) σX σZ 1 - unitless measure Random Variables August 2010 6 / 11 Conditional Mean E (XU jX = xj ) = xj E (U jX = xj ) U takes k distinct values E (U jX = xj ) = ∑K=1 uk P (U = uk jX = xj ) k if E (U jX = xj ) = 0 for all j , then E (U jX ) = 0 E (U jX ) = 0 ) E (U ) = 0 E (XU ) = 0 E (U ) = E [E (U jX )] = 0 D. Steigerwald (UCSB) Random Variables August 2010 7 / 11 Conditional Mean E (XU jX = xj ) = xj E (U jX = xj ) U takes k distinct values E (U jX = xj ) = ∑K=1 uk P (U = uk jX = xj ) k if E (U jX = xj ) = 0 for all j , then E (U jX ) = 0 E (U jX ) = 0 ) E (U ) = 0 E (XU ) = 0 E (U ) = E [E (U jX )] = 0 E (XU ) = E [X E (U jX )] = 0 D. Steigerwald (UCSB) Random Variables August 2010 7 / 11 Law of Iterated Expectations E (U ) = EX [E (U jX )] = 0 1 E (U ) = ∑K=1 uk P (U = uk ) k D. Steigerwald (UCSB) Random Variables August 2010 8 / 11 Law of Iterated Expectations E (U ) = EX [E (U jX )] = 0 1 2 E (U ) = ∑K=1 uk P (U = uk ) k = ∑K=1 uk ∑J=1 P (U = uk , X = xj ) j k D. Steigerwald (UCSB) Random Variables August 2010 8 / 11 Law of Iterated Expectations E (U ) = EX [E (U jX )] = 0 1 2 3 E (U ) = ∑K=1 uk P (U = uk ) k = ∑K=1 uk ∑J=1 P (U = uk , X = xj ) j k = ∑K=1 uk ∑J=1 P (U = uk jX = xj ) P (X = xj ) j k D. Steigerwald (UCSB) Random Variables August 2010 8 / 11 Law of Iterated Expectations E (U ) = EX [E (U jX )] = 0 1 2 3 4 E (U ) = ∑K=1 uk P (U = uk ) k = ∑K=1 uk ∑J=1 P (U = uk , X = xj ) j k = ∑K=1 uk ∑J=1 P (U = uk jX = xj ) P (X = xj ) j k = ∑J=1 P (X = xj ) ∑K=1 uk P (U = uk jX = xj ) j k D. Steigerwald (UCSB) Random Variables August 2010 8 / 11 Law of Iterated Expectations E (U ) = EX [E (U jX )] = 0 1 2 3 4 5 E (U ) = ∑K=1 uk P (U = uk ) k = ∑K=1 uk ∑J=1 P (U = uk , X = xj ) j k = ∑K=1 uk ∑J=1 P (U = uk jX = xj ) P (X = xj ) j k = ∑J=1 P (X = xj ) ∑K=1 uk P (U = uk jX = xj ) j k = ∑J=1 P (X = xj ) E (U jX = xj ) j D. Steigerwald (UCSB) Random Variables August 2010 8 / 11 Law of Iterated Expectations E (U ) = EX [E (U jX )] = 0 1 2 3 4 5 6 E (U ) = ∑K=1 uk P (U = uk ) k = ∑K=1 uk ∑J=1 P (U = uk , X = xj ) j k = ∑K=1 uk ∑J=1 P (U = uk jX = xj ) P (X = xj ) j k = ∑J=1 P (X = xj ) ∑K=1 uk P (U = uk jX = xj ) j k = ∑J=1 P (X = xj ) E (U jX = xj ) j = EX E (U jX = xj ) D. Steigerwald (UCSB) Random Variables August 2010 8 / 11 Conditional Variance U takes k distinct values Var (U jX = xj ) = ∑K=1 [uk k E (U jX = xj )]2 P (U = uk jX = xj ) if Var (U jX = xj ) = σ2 for all j , then Var (U jX ) = σ2 Var (U jX ) = σ2 ) Var (U ) = σ2 D. Steigerwald (UCSB) Random Variables August 2010 9 / 11 Relating Variance: Conditional to Unconditional 1 Var (U ) = ∑K=1 [uk k D. Steigerwald (UCSB) E (U )]2 P (U = uk ) Random Variables August 2010 10 / 11 Relating Variance: Conditional to Unconditional 1 2 Var (U ) = ∑K=1 [uk E (U )]2 P (U = uk ) k = ∑J=1 P (X = xj ) ∑K=1 [uk E (U )]2 P (U = uk jX = xj ) j k D. Steigerwald (UCSB) Random Variables August 2010 10 / 11 Relating Variance: Conditional to Unconditional 1 2 3 Var (U ) = ∑K=1 [uk E (U )]2 P (U = uk ) k = ∑J=1 P (X = xj ) ∑K=1 [uk E (U )]2 P (U = uk jX = xj ) j k [uk E (U )]2 = ([uk D. Steigerwald (UCSB) E (U jX = xj )] Random Variables [E (U ) E (U jX = xj )]) August 2010 2 10 / 11 Relating Variance: Conditional to Unconditional 1 2 3 4 Var (U ) = ∑K=1 [uk E (U )]2 P (U = uk ) k = ∑J=1 P (X = xj ) ∑K=1 [uk E (U )]2 P (U = uk jX = xj ) j k [uk E (U )]2 = ([uk E (U jX = xj )] plug each term into point 2 D. Steigerwald (UCSB) Random Variables [E (U ) E (U jX = xj )]) August 2010 2 10 / 11 Relating Variance: Conditional to Unconditional 1 2 3 4 5 Var (U ) = ∑K=1 [uk E (U )]2 P (U = uk ) k = ∑J=1 P (X = xj ) ∑K=1 [uk E (U )]2 P (U = uk jX = xj ) j k [uk E (U )]2 = ([uk E (U jX = xj )] plug each term into point 2 all but …rst term vanish (equal 0) D. Steigerwald (UCSB) Random Variables [E (U ) E (U jX = xj )]) August 2010 2 10 / 11 Relating Variance: Conditional to Unconditional 1 2 3 4 5 6 Var (U ) = ∑K=1 [uk E (U )]2 P (U = uk ) k = ∑J=1 P (X = xj ) ∑K=1 [uk E (U )]2 P (U = uk jX = xj ) j k [uk E (U )]2 = ([uk E (U jX = xj )] [E (U ) E (U jX = xj )])2 plug each term into point 2 all but …rst term vanish (equal 0) Var (U ) = 2 ∑J=1 P (X = xj ) ∑K=1 [uk E (U jX = xj )] P (U = uk jX = xj ) j k D. Steigerwald (UCSB) Random Variables August 2010 10 / 11 Relating Variance: Conditional to Unconditional 1 2 3 4 5 6 7 Var (U ) = ∑K=1 [uk E (U )]2 P (U = uk ) k = ∑J=1 P (X = xj ) ∑K=1 [uk E (U )]2 P (U = uk jX = xj ) j k [uk E (U )]2 = ([uk E (U jX = xj )] [E (U ) E (U jX = xj )])2 plug each term into point 2 all but …rst term vanish (equal 0) Var (U ) = 2 ∑J=1 P (X = xj ) ∑K=1 [uk E (U jX = xj )] P (U = uk jX = xj ) j k = EX Var (U jX ) D. Steigerwald (UCSB) Random Variables August 2010 10 / 11 Vanishing Terms [E (U ) ∑J=1 j E (U jX = xj )] P (X = xj ) ∑K=1 [E (U ) k D. Steigerwald (UCSB) 2 E (U jX = xj )] Random Variables 2 P (U = uk jX = xj ) August 2010 11 / 11 Vanishing Terms [E (U ) ∑J=1 j 1 E (U jX = xj )] P (X = xj ) ∑K=1 [E (U ) k = ∑J=1 P (X = xj ) [E (U ) j D. Steigerwald (UCSB) 2 E (U jX = xj )] 2 P (U = uk jX = xj ) E (U jX = xj )]2 ∑K=1 P (U = uk jX = xj ) k Random Variables August 2010 11 / 11 Vanishing Terms [E (U ) ∑J=1 j 1 2 E (U jX = xj )] P (X = xj ) ∑K=1 [E (U ) k 2 E (U jX = xj )] 2 P (U = uk jX = xj ) = 2 ∑J=1 P (X = xj ) [E (U ) E (U jX = xj )] ∑K=1 P (U = uk jX = xj ) j k = 2 2 2E (U ) E (U jX = xj ) ∑J=1 P (X = xj ) E (U ) + E (U jX = xj ) j D . Steigerwald (UCSB) Random Variables August 2010 11 / 11 Vanishing Terms [E (U ) ∑J=1 j 1 2 3 E (U jX = xj )] P (X = xj ) ∑K=1 [E (U ) k 2 E (U jX = xj )] 2 P (U = uk jX = xj ) = 2 ∑J=1 P (X = xj ) [E (U ) E (U jX = xj )] ∑K=1 P (U = uk jX = xj ) j k = 2 2 2E (U ) E (U jX = xj ) ∑J=1 P (X = xj ) E (U ) + E (U jX = xj ) j = E (U )2 + ∑J=1 P (X = xj ) E (U jX = xj )2 j D . Steigerwald (UCSB) Random Variables 2E (U ) E (U jX = xj ) August 2010 11 / 11 Vanishing Terms [E (U ) ∑J=1 j 1 2 3 4 E (U jX = xj )] P (X = xj ) ∑K=1 [E (U ) k 2 E (U jX = xj )] 2 P (U = uk jX = xj ) = 2 ∑J=1 P (X = xj ) [E (U ) E (U jX = xj )] ∑K=1 P (U = uk jX = xj ) j k = 2 2 2E (U ) E (U jX = xj ) ∑J=1 P (X = xj ) E (U ) + E (U jX = xj ) j = E (U )2 + ∑J=1 P (X = xj ) E (U jX = xj )2 j = E (U )2 + E (U )2 D. Steigerwald (UCSB) 2E (U )2 Random Variables 2E (U ) E (U jX = xj ) August 2010 11 / 11 ...
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This note was uploaded on 12/26/2011 for the course ECON 140a taught by Professor Staff during the Fall '08 term at UCSB.

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