Probablity 2 - Exercises - Mathematics, Exercises of Mathematics

Let Xn be a sequence of random variables such that Xn  0 for all n. Suppose that P(Xn > t)  ( 1 t )k where k > 1. Derive an upper bound on E(Xn).

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2011/2012

Uploaded on 03/08/2012

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Homework 2
36-705
Due: Thursday Sept 15 by 3:00
1. Let Xnbe a sequence of random variables such that Xn0 for all n. Suppose that
P(Xn> t)(1
t)kwhere k > 1. Derive an upper bound on E(Xn).
2. Let X1, . . . , XnUnif(0,1). Let Y= max1inXi.
(i) Bound E(Y) using the method we derived in lecture notes 2.
(ii) Find an exact expression for E(Y). Compare the result to part (i).
3. An improvement on Hoeffding’s inequality is Bernstein’s inequality. Let X1, . . . , Xnbe
iid, with mean µ, Var(Xi) = σ2and |Xi| c. Then Bernstein’s inequality says that
P|Xnµ|> 2 exp n2
2σ2+ 2c/3.
(When σis sufficiently small, this bound is tighter than Hoeffding’s inequality.) Let
X1, . . . , XnUniform(0,1) and An= [0,1/n]. Let pn=P(XiAn) and let
bpn=1
n
n
X
i=1
IAn(Xi).
(i) Use Hoeffding’s inequality and Bernstein’s inequality to bound
P(|bpnpn|> ).
(ii) Show that the bound from Bernstein’s inequality is tighter.
(iii) Show that Hoeffding’s inequality implies bpnpn=Oq1
nbut that Bernstein’s
inequality implies bpnpn=OP(1/n).
4. Show that Xn=oP(an) and Yn=OP(bn) implies that XnYn=oP(anbn).
1

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Homework 2 36- Due: Thursday Sept 15 by 3:

  1. Let Xn be a sequence of random variables such that Xn ≥ 0 for all n. Suppose that P(Xn > t) ≤ (^1 t )k^ where k > 1. Derive an upper bound on E(Xn).
  2. Let X 1 ,... , Xn ∼ Unif(0, 1). Let Y = max 1 ≤i≤n Xi. (i) Bound E(Y ) using the method we derived in lecture notes 2. (ii) Find an exact expression for E(Y ). Compare the result to part (i).
  3. An improvement on Hoeffding’s inequality is Bernstein’s inequality. Let X 1 ,... , Xn be iid, with mean μ, Var(Xi) = σ^2 and |Xi| ≤ c. Then Bernstein’s inequality says that P (|Xn − μ| > )^ ≤ 2 exp

− n 2 2 σ^2 + 2c/ 3

(When σ is sufficiently small, this bound is tighter than Hoeffding’s inequality.) Let X 1 ,... , Xn ∼ Uniform(0, 1) and An = [0, 1 /n]. Let pn = P(Xi ∈ An) and let p̂n =^1 n ∑^ n i=

IAn (Xi).

(i) Use Hoeffding’s inequality and Bernstein’s inequality to bound P(|̂pn − pn| > ). (ii) Show that the bound from Bernstein’s inequality is tighter. (iii) Show that Hoeffding’s inequality implies ̂pn − pn = O

(√^1

n

but that Bernstein’s inequality implies p̂n − pn = OP (1/n).

  1. Show that Xn = oP (an) and Yn = OP (bn) implies that XnYn = oP (anbn).