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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 anticompetitive 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.
 Fall '08
 Staff
 Econometrics

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