茆诗松高等数理统计课后习题第三次作业

(3)ξn∗ = Op(an), ηn∗ = op(bn), ξn = ξn∗/an, η = ηn∗/bn, by (2) ξnηn = op(1). Thus we have Op(an) · op(bn) = op(anbn).
(4)By (1), ξn∗a−n 1 + ηn∗b−n 1 = Op(1). ∀ε > 0, ∃M s.t. P
+ ξn∗
ηn∗
an + ηn∗ | ≥ M ≤ P | ξn∗ + ηn∗ | ≥ M < ε
an ∨ bn
an bn
. Then we know, ξn∗ + ηn∗ = Op(an ∨ bn).
(5)ηn∗ b−n 1 →P 0. Thus we have, ∀ε > 0, P (|ηn∗ b−n 1|r ≥ ε) = P (|ηn∗ b−n 1| ≥ ε1/r) → 0, as n goes to infinity, which means (ηn∗b−n 1)r →P 0.
(3)The proof is similar to (2).
(4) A sequence of random variables {Xn}, with respective distribution function {Fn} is said
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to be bounded in probability if for every ε > 0 there exist Mε and Nε such that Fn(Mε) − Fn(−Mε) > 1 − ε ∀n > Nε.
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Necessity. By Xn’s convergence, E(it Xn) → E(it X) for ∀t. Let t = ta. Then we have, Ei(ta) Xn → Ei(ta )X. That is to say, Eit(a Xn) → Eit(a X).
3. Xn →L X, Yn →P c§y²µ (1)Xn + Yn →L X + c; (2)˜ ž§c = 0, XnYn−1 →L Xc−1; (3)˜ ž§XnYn →L Xc; (4)Xn = Op(1). (1)By conclusion of the last problem, we only need to prove that for ∀a, a (Xn + Yn) →L a (X + c). Let Xn∗ = a Xn, Yn∗ = a Yn, X∗ = a X, c∗ = a c. We prove Xn∗ + Yn∗ →L X∗ + c∗. For ∀z, P (Xn∗ + Yn∗ ≤ z) = P (Xn∗ + Yn∗ ≤ z, |Yn∗ − c∗| > ε) + P (Xn∗ + Yn∗ ≤ z, |Yn∗ − c∗| ≤ ε) ≤ P (|Yn∗ − c∗| > ε) + P (Xn∗ ≤ z − c∗ + ε). Thus we have, lim supn P (Xn∗ + Yn∗ ≤ z) ≤ F (z − c∗ + ε). Also, P (Xn∗ + Yn∗ ≤ z) ≥ P (Xn∗ + Yn∗ ≤ z, |Yn∗ − c∗| ≤ ε) ≥ P (Xn∗ ≤ z − c∗ − ε, |Yn∗ − c∗| ≤ ε) ≥ P (Xn∗ ≤ z − c∗ − ε) − P (|Yn∗ − c ∗ | > ε). Thus lim infn P (Xn∗ + Yn∗ ≤ z) ≥ F (z − c∗ − ε). Let ε → 0, we have limn P (Xn∗ + Yn∗ ≤ z) = F (z − c∗). So far, we have proved that for ∀a, a (Xn + Yn) →L a (X + c).
(1)ξn = Op(1) and ηn = op(1), then we have, ξn = Op(1) ⇔ ∀Mn → ∞ P (|ξn| > Mn) → 0 as n → ∞. Also, P (|ξn + ηn| ≥ Mn) ≤ P (|ξn| ≥ Mn/2) + P (|ηn| ≥ Mn/2). Since ηn = op(1), P (|ηn| ≥ Mn/2) → 0 as n → ∞, we have, lim supn→∞ P (|ξn + ηn| ≥ Mn) = 0, i.e. limn→∞ P (|ξn + ηn| ≥ Mn) = 0. By the necessary and sufficient condition mentioned above, we have, ξn + ηn = Op(1).
(2)Suppose M satisfies ∀ε∗, ∃M s.t. P (|ξn| > M ) < ε∗.P (|ξnηn| ≥ ε) = P (|ξnηn| ≥ ε, |ξn| ≥ M ) + P (|ξnηn| ≥ ε, |ξn| < M ) ≤ P (|ξn| ≥ M ) + P (|ηn| ≥ ε/M ). lim supn→∞ P (|ξnηn| ≥ ε) ≤ ε∗. P (|ξnηn| ≥ ε) → 0 as n goes to infinity.
2. Xn, X ´‘Å•þ§y²Xn →L X ¿‡^‡´ ∀a, a Xn →L a X.
Sufficiency. If ∀a, a Xn →L a X, then for ∀t, ∀a, we have E(ita Xn) → E(ita X). Let t = 1. We get for ∀a, E(ia Xn) → E(ia X).
(2)Suppose that c > 0 otherwise −Yn → −c. ∀z, P (XnYn−1 ≤ z) = P (XnYn−1 ≤ z, |Yn − c| > ε) + P (XnYn−1 ≤ z, |Yn − c| ≤ ε) ≤ P (|Yn − c| > ε) + P (Xn ≤ z(c + ε)). lim supn P (XnYn−1 ≤ z) ≤ F (z(c + ε)). Similarly, P (XnYn−1 ≤ z) ≥ P (XnYn−1 ≤ z, |Yn − c| ≤ ε) ≥ P (Xn ≤ z(c − ε)) − P (|Yn − c| > ε). Thus, lim infn P (XnYn−1 ≤ z) ≤ F (z(c − ε)). ε is arbitrary, limn P (XnYn−1 ≤ z) = F (zc). The right hand side of the equation is the c.d.f of Xc−1. Thus, c = 0, XnYn−1 →L X−1.
p ÚOÆ1ngŠ’‰Y½J« 1. an > 0, bn > 0§ y²±e(ص (1)Op(1) + op(1) = Op(1); (2)Op(1) · op(1) = op(1); (3)Op(an) · op(bn) = op(anbn); (4)Op(an) · op(bn) = Op(an ∨ bn); (5)[op(an)r] = op(arn).
≥ |F (Mε) − F (−Mε)| − (|Fn(Mε) − F (Mε)| + |F (−Mε) − Fn(−Mε)|) > 1 − ε.
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The notation Xn = Op(1) will be used. We only need to prove that each element of the vector is Op(1).Since limnFn(x) = F (x), ∀ε > 0, ∃Nε, n > Nε, |Fn(X) − F (x)| < ε/4. Also, ∀ε, we can find Mε s.t.F (Mε) − F (−Mε) > 1 − ε/2. Thus we have, Fn(Mε) − Fn(−Mε) = Fn(Mε) − F (Mε) + F (−Mε) − Fn(−Mε) + F (Mε) − F (−Mε)