10. Asymptotics

Question 1

Convergence in probability

Answer 1

lim Pr{|Xn-X| < epsilon }=1

Question 2

Convergence in mean of order k

Answer 2

lim E{|Xn-X|^k}=0

if k=2 then we have the quadratic mean

Question 3

Convergence in distribution

Answer 3

lim Fn(x) = F(x)

Question 4

Relationship between convergences

Answer 4

• convergence in k-th mean implies convergence in probability which implies convergence in distribution
• convergence in distribution to a real number implies convergence in probability

Question 5

Proof that conv. in k-th mean implies conv. in prob

Answer 5

E{|Xn-X|^k}= E[|Y|^k] = int |y|f_Y
= int (|y|=> j) + int (|y| < j)
=> j^k int (|y|=> j) f_Y = j^k Pr{|Y| => j}
then 0<= Pr{|Xn-X| => e} <=1/e^k E{|Xn-X|^e}

Question 6

Central Limit th.

Answer 6

(X* - m)/sqr(v/n) ~ N(0,1)

if X* is the sample mean

Question 7

Slutsky th.

Answer 7

Let Xn and Yn be 2 sequences, with Xn convergent in distr. to X and Yn convergent in prob. to y then:

• Xn +/- Yn = X+/-y
• XnYn = Xy
• Xn/Yn = X/y

Question 8

Mann-Wald th.

Answer 8

if Xn is st:
sqr(n) (Xn - @) ~ N(0, v(@))
with g a differentiable funct, non vanishing in the first derivative, then:
sqr(n)(g(Xn) - g(@)) ~ N(0, v(@)*[g’(@)]^2)

it allows to construct an asymp. normal estimator from another

Question 9

Consistency

Answer 9

An estimator for g(@) is consistent if it’s convergent in probability (or in quadratic mean) to g(@)

Question 10

Asymptotically normal estimator

Answer 10

if for some V, we have that sqr(n) (Tn - @) ~ N(0, V)