advertisement

Stat 511 Lecture 5 "Handout" The following are equivalent regarding c ∈ ℜk 1) ∃ a ∈ℜn such that a′Xβ = c′β ∀β 2) c ∈ C ( X′ ) ( c′ is a linear combination of the rows of X ) 3) Xβ1 = Xβ 2 ⇒ c′β1 = c′β 2 Condition 3) is a condition about the lack of ambiguity of the value of c′β . Condition 1) is c′β = a′Xβ = a′EY = Ea′Y so that a′Y can be used to estimate c′β in an unbiased fashion. Proof: First suppose that 2) holds. c ∈ C ( X′ ) ⇒ ∃ a such that c′ = aX . So for this a , c′β = a′Xβ ∀β and 1) holds. Next suppose that 1) holds, i.e. that ∃ a ∈ ℜn such that a′Xβ = c′β ∀β . Suppose Xβ1 = Xβ 2 . For this a , c′β1 = a′Xβ1 = a′Xβ 2 = c′β 2 and 3) holds. Finally, suppose that 3) holds. It is "obviously" equivalent to write X ( β1 − β 2 ) = 0 ⇒ c′ ( β1 − β 2 ) = 0 ∀ β1 , β 2 That is, it is equivalent to 3) to write Xd = 0 ⇒ c′d = 0 ∀ d Then, the claim that 3) ⇒ 2) is the claim that {c | [ Xd = 0 ⇒ c′d = 0 ∀ d] holds} ⊂ C ( X′) Suppose that c is such that [ Xd = 0 ⇒ c′d = 0 ∀ d ] holds. Write c = PX′c + ( I − PX′ ) c and let d* = ( I − PX′ ) c . We can then argue that d* = 0 as follows. Clearly, d* ∈ C ( X′ ) so it ⊥ must be that c′d* = 0 from the condition [ ]. But c′d* = c′ ( I − Px′ ) c = c′c − c′Px′c so that c′c = c′Px′c . But c′c = c′ ( PX′ + ( I − PX′ ) ) c = c′PX′c + c′ ( I − PX′ ) c = c′PX′c + c′ ( I − PX′ )′ ( I − PX′ ) c Then since c′ ( I − PX′ )′ ( I − PX′ ) c = d*′d* , we have d*′d* = 0 so that d* = 0 . Thus c = PX′c so that c ∈ C ( X′ ) . ,