Session12 Solution

Exercise Session 12, Solutions, December 1st ,2006 Mathematics for Economics and Finance Prof: Norman Schürho¤ TAs: Zhihua Chen (Cissy), Natalia Guseva Problem 1 OLS estimator in …nite sample. Recall b = (X 0 X) 1. Find the sampling error. (Hint: b

1

X 0Y

)

2. Show OLS estimator is unbiased. (Hint: E(b j X) = ) 3. Find the variance of b for given X: Solution 1. The sampling error: = (X 0 X) 1 X 0 y = (X 0 X) 1 X 0 (X + ") = (X 0 X) 1 X 0 X + (X 0 X) {z } |

b

1

X 0"

=In

=

(X 0 X)

1

X 0"

2. To show E(b j X) = is equivalent to show E(b set A = (X 0 X) 1 X 0 ;then E(b A is X m easurable

=

AE("

j X) = 0;

j X) = E(A" j X) j X) = 0 by Assumption E(" j X) = 0

3. V ar(b

j = = =

X) = V ar(b j X)....since is true value, V ar( ) = 0; Cov( ; b) = 0: V ar(A" j X) A is X measurable 0 AV ar(" j X)A ......Notes: E(" j X) = 0 AE(""0 j X)A0 0

= A 2 In A0 ......By assumption E("" j X) = 2 = (X 0 X) 1 X 0 [(X 0 X) 1 X 0 ]0 2 = (X 0 X) 1 X 0 X [(X 0 X)0 ] 1 | {z } In

=

2

0

(X X)

1

1

2

In homoskedasticity

Problem 2 Let ( ; F; P) be a probability space. There exists two random variables X and Y . If we can observe X,what we can Rsay about Y ? One way is to 2 use mean square error, M SE = E(Y g(X))2 = fY (!) g(X(!))g dP (!). Show that g (X) = E(Y j X) is the best solution. Solution = E[Y g(X)]2 = E[Y E(Y j X) + E(Y j X) g(X)]2 = E[Y E(Y j X)]2 + E[E(Y j X) g(X)]2 +2Ef[Y E(Y j X)][E(Y j X) g(X)]g {z } |

M SE

(3)

(3)

= 2EfE[(Y

E(Y j X))

(E(Y j X) | {z

g(X)) }

j X]g

X m easurable, take out one exp ectation

= 2Ef(E(Y j X) = 2Ef(E(Y j X)

g(X))E[(Y E(Y j X)) j X]g g(X))(E(Y j X) E(Y j X))g {z } | =0

=

0

) M SE

=

E[Y

E(Y j X)]2 + E[E(Y j X) | {z 0

)

E[Y

2

g(X)]

E[Y

E(Y j X)]2

g(X)]2 }

which implies g (X) = E(Y j X) minimizes MSE. Problem 3 Variable X is normally distributed with mean

2

and variance

.

1. Assume 2 = 80: The observed value of the sample mean X of a random sample of size 20 is 81:2. Find a 95% con…dence interval for . 2

2. Assume mately. Solution if X s N ( ;

= 9:Find n such that P r(X

2

) then X s N ( ;

1. CI95% ( ) = (X p

F

1

2

n

1<

< X + 1) = 0:9 approxi-

)

(0:975) pn ; X + F

1

p

1:96 p80 ; 20

(0:975) pn ) = (81:2

81:2 + 1:96 p80 ) = (77:28; 85:12) 20 2. Pr(X

1 <

< X + 1) = 0:9 =) CI90% ( ) = (X

1

F

1

(0:95) pn ;

X + F (0:95) pn ) = (X 1; X + 1) =) p F 1 (0:95) pn = 1 =) n = F 1 (0:95) = 1:645 3 = 4:935 =) n t 24 2

Problem 4 Show for any two random variables X and Y , V arX = E(V ar(X j Y )) + V ar(E(X j Y )): By de…nition, we have

Solution V arX

= E(X EX)2 = E[X E(X j Y ) + E(X j Y ) | {z } add and subtract 2

= Ef(X +2 (X = E (X |

= 2EfE[(X

2

E(X j Y )) + (E(X j Y ) EX) E(X j Y )) (E(X j Y ) EX)g 2

E(X j Y )) + E (E(X j Y ) {z } | {z (1)

+2E[(X |

(3)

EX]2

(2)

E(X j Y )) (E(X j Y ) {z (3)

E(X j Y ))

(E(X j Y ) | {z

2

EX) }

EX)]...by linearity property }

EX) }

j Y ]g by law of iterated exp.

EX cons.& E(XjY ) Y m easable

=

2Ef(E(X j Y )

=

2Ef(E(X j Y )

EX) E[(X |

E(X j Y )) j Y ]g Take out what is known {z }

from now on, fo cus this part

EX) [E(X j Y )

E(E(X j Y ) j Y )] | {z }

g...by linearity property

=E(XjY ) By taking out what is known

=

2Ef(E(X j Y )

=

0:

EX) [E(X j Y ) E(X j Y )]g | {z } =0

(1)

= E[E (X

2

E(X j Y )) j Y ]....by law of iterated expectation 2

= EfE[X 2 + (E(X j Y )) 2

2XE(X j Y ) j Y ]g

= E[E(X j Y ) + (E(X j Y ))2

2

2

2 (E(X j Y )) ]...by linearity & take out known

= E[E(X 2 j Y ) (E(X j Y )) ] = E[V ar(X j Y )] (2) = E[E 2 (X j Y ) + E 2 (X) 2E(X)E(X j Y )] = E(E 2 (X j Y )) + EE 2 (X) 2E(X)E(E(X j Y )) | {z } | {z } =E 2 X

2

= E(E (X j Y ))

=EX

2

(E(E(X j Y ))) = V ar(E(X j Y )):

) V arX = E(V ar(X j Y )) + V ar(E(X j Y )):

3

Problem 5 Suppose the distribution of Y conditional on X = x is N (x; x2 ) and the marginal distribution of X is uniform (0; 1):Find EY; V arY; and Cov(X; Y ): Solution Notes the pdf for X is fX (x) =

1 1 0

= 1;because X is uniform in (0; 1):

law of iterated exp.

EY

=

E(E(Y j X)) | {z } =X

= EX =

Z1

x fX (x)dx =

0

V arY

Z1

x 1dx =

x2 1 1 j = 2 0 2

0

= E(V ar(Y j X)) + V ar(E(Y j X)) use Q3 result | | {z } {z } X2

X

2

= EX + V arX 2 = EX 2 + EX 2 (EX) 2

2EX 2 (EX) Z1 = 2 x2 fX (x)dx

=

1 2

2

0

=

Cov(X; Y )

2

x3 1 j 3 0

1 5 = 4 12

= E(XY )

EXEY

x

1 1 ...law of iterated expectation 2 2

= E(E(XY j X)) = EXE(Y j X) | {z }

1 ......take out what is known 4

=X

= EX =

Z1

2

1 4

x2 fX (x)dx

1 1 = 4 3

0

4

1 1 = : 4 12