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Lemma(Product rule Recall that if AT =op(1)and br=Op(1), then Arbr=0p(1). Hence, if AT+0 and br -+Z, then Lemma(Asymptotic equivalence Let ar) and br be two sequence of random vectors. If ar- br -+0 and Z The results is helpful in situation in which we wish to find the asymptotic distribution of ar but cannot do so directly. Often, however, it is easy to find a br that has a known asymptotic distribution and that satisfies at-br-0 This Lemma then ensures that ar has the same limiting distribution as br and we say that ar is"asymptotically equivalent"to bT Lemma given g:Rk→R(k,l∈) and any sequence of random k× I vector br such that bm_L *Z, where z is k x 1, if g is continous(not dependent on T)at z, then b g(z) Suppose that Xr- N(O, 1)Then the square of Xr asymptotically behaves as the square of a N(O, 1)variables: XF-x( Let xr be a sequence of random(n x 1) vector with xT -+c, and let yrI ector with y constructed from the sum xr+yr converges in distribution to c+y and the sequence constructed from the product xryr converges in distribution to cy Ex Let iXr) be a sequence of random(m x n)matrixwith XT C, and let Then the limitting distribution of Xryr is the same as that of Cy; that is N(Cu, CQ2C/) (C1Lemma (Product rule): Recall that if AT = op(1) and bT = Op(1), then AT bT = op(1). Hence, if AT p −→ 0 and bT d −→ Z, then AT bT p −→ 0. Lemma (Asymptotic equivalence): Let {aT } and {bT } be two sequence of random vectors. If aT − bT p −→ 0 and bT d −→ Z, then aT d −→ Z. The results is helpful in situation in which we wish to find the asymptotic distribution of aT but cannot do so directly. Often, however, it is easy to find a bT that has a known asymptotic distribution and that satisfies aT − bT p −→ 0. This Lemma then ensures that aT has the same limiting distribution as bT and we say that aT is ”asymptotically equivalent” to bT . Lemma: Given g : Rk → Rl (k, l ∈ N ) and any sequence of random k × 1 vector bT such that bT L−→ z, where z is k ×1, if g is continous (not dependent on T) at z, then g(bT) L−→ g(z). Example: Suppose that XT L−→ N(0, 1) Then the square of XT asymptotically behaves as the square of a N(0, 1) variables: X2 T L−→ χ 2 (1). Lemma: Let {xT } be a sequence of random (n × 1) vector with xT p −→ c , and let {yT } be a sequence of random (n × 1) vector with yT L−→ y. Then the sequence constructed from the sum {xT + yT } converges in distribution to c + y and the sequence constructed from the product {x 0 T yT } converges in distribution to c 0y. Example: Let {XT } be a sequence of random (m × n) matrixwith XT p −→ C , and let {yT } be a sequence of random (n × 1) vector with yT L−→ y ∼ N(µ, Ω). Then the limitting distribution of XT yT is the same as that of Cy; that is XT yT L−→ N(Cµ, CΩC0 ). Lemma (Cramer δ ): 8
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