Theorem Let br be a sequence of real-valued random variables. Ifbr - b, then br-b Proposition: Given g:Rk→Rl(k,1∈M) and any sequence of random k× 1 vector bT such that bT b, where b is k x 1, if g is continous at b, then g br)-+g(b) amp If XIT - PCI and X2T P.C2, then(XIT+X2r)P(c1 +C2). This follows im- tely, since g(XIT, X2T)=(XIT+ X2r)is a continous function of(XIT, X2T) Ex amp Consider an alternative estimator of the mean given by YT=1/(T-lI2taIY This can be written as cITYT, where cIT =T/(T-1) and YT =(1/T)2Yt Under general condition, the sample mean is a consistent estimator of the popu- lation mean, implying that Yr-+ u. It is also easy to verify that cIr -+1. Since CiTYT is a continous function of CIr and Yr, it follows that cirYT P,1-u=u Y is alos a consistent estimator of u Definition (i). The sequence br) is at most of order T in probability, denoted bT=Op(Ta) if for every e>0 there exist a finite△>0,andN∈ N such that for all T≥N, Pr{s:T-br(s)>△}< (ii). The sequence (br is of order smaller than TA in probability, denoted br=op(T ), if T-Ab Lemma(Product rule) Let Ar be l x k and let br be k x 1. If AT =op(1)and br =Op(), then ArbT=op(1) Proof: Each element of Arbr is the sums of the product of Op(T )op(T)=Op(To+o) Op(1) and therefore is op(1)Theorem: Let {bT } be a sequence of real-valued random variables. If bT a.s. −→ b, then bT p −→ b. Proposition: Given g : Rk → Rl (k, l ∈ N ) and any sequence of random k × 1 vector bT such that bT p −→ b, where b is k × 1, if g is continous at b, then g(bT) p −→ g(b). Example: If X1T p −→ c1 and X2T p −→ c2, then (X1T + X2T ) p −→ (c1 + c2). This follows im￾mediately, since g(X1T , X2T ) ≡ (X1T +X2T ) is a continous function of (X1T , X2T ). Example: Consider an alternative estimator of the mean given by Y¯ ∗ T = [1/(T −1)] PT t=1 Yt . This can be written as c1T Y¯ T , where c1T ≡ [T/(T − 1)] and Y¯ T ≡ (1/T) PT t=1 Yt . Under general condition, the sample mean is a consistent estimator of the popu￾lation mean, implying that Y¯ T p −→ µ. It is also easy to verify that c1T → 1. Since c1T Y¯ T is a continous function of c1T and Y¯ T , it follows that c1T Y¯ T p −→ 1 · µ = µ. Thus Y¯ ∗ T is alos a consistent estimator of µ. Definition: (i). The sequence {bT } is at most of order T λ in probability, denoted bT = Op(T λ ), if for every ε > 0 there exist a finite 4ε > 0, and Nε ∈ N such that for all T ≥ Nε, Pr{s : |T −λ bT (s)| > 4ε} < ε. (ii). The sequence {bT } is of order smaller than T λ in probability, denoted bT = op(T λ ), if T −λ bT p −→ 0. Lemma (Product rule): Let AT be l × k and let bT be k × 1. If AT = op(1) and bT = Op(1), then AT bT = op(1). Proof: Each element of AT bT is the sums of the product of Op(T 0 )op(T 0 ) = op(T 0+0) = op(1) and therefore is op(1). 5