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Quant版 - 问一个convergence of random variables的问题
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话题: random话题: converge
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1 (共1页)
f*********1
发帖数: 117
1
1. convergence in distribution;
2. convergence in probability;
1 doesn't imply 2.
但俺想不出converge in distribution 但是不converge in probability的例子了
谁能给列举几个?
多谢!
f****e
发帖数: 590
2
a sequence of iid random variable

【在 f*********1 的大作中提到】
: 1. convergence in distribution;
: 2. convergence in probability;
: 1 doesn't imply 2.
: 但俺想不出converge in distribution 但是不converge in probability的例子了
: 谁能给列举几个?
: 多谢!

g****y
发帖数: 71
3
convergence in probability usually refers to the law of large numbers. e.g.
average of iid Xi should converge in p to its expectation.
convergence in distribution usually refers to the central limit theorem.
e.g. square root of n times (average of iid Xi minus its expectation) will
behave like a normal distribution. it converges in distribution to a normal
random variable. but indeed it's not a normal random variable even in the
limit.
f*********1
发帖数: 117
4
谢谢findle!
你的意思是the average of a sequence of iid RVs?

a sequence of iid random variable

【在 f****e 的大作中提到】
: a sequence of iid random variable
f****e
发帖数: 590
5
不是,就是iid
convergence in distribution就是说他们的cumulative function收敛,但是这些rv有
没有关系就不管了
convergence almost sure, in probability包括in Lp,都要求这些rv's之间要有联系

【在 f*********1 的大作中提到】
: 谢谢findle!
: 你的意思是the average of a sequence of iid RVs?
:
: a sequence of iid random variable

f*********1
发帖数: 117
6
thank u, gatsby!

.
normal

【在 g****y 的大作中提到】
: convergence in probability usually refers to the law of large numbers. e.g.
: average of iid Xi should converge in p to its expectation.
: convergence in distribution usually refers to the central limit theorem.
: e.g. square root of n times (average of iid Xi minus its expectation) will
: behave like a normal distribution. it converges in distribution to a normal
: random variable. but indeed it's not a normal random variable even in the
: limit.

n******r
发帖数: 1247
7
this is not right
a sequence of iid random variables converges with probability 1
an example for 1 doesn't imply 2 can be
toss a fair coin
X=1 if head, 0 if tail
Y=1 if tail, 0 if head
X Y have the same cumulative distribution, but for e<1,P(|X(w)-Y(w)|>e)=1
therefore no convergence in probability

【在 f****e 的大作中提到】
: a sequence of iid random variable
A***d
发帖数: 25
8

The following is an example:
Suppose Xn follows exponential distribution with df n/(n+1)exp{-nx/(n+1)}, X
follows exponential
distribution with df exp(-x). Further assume that Xn and X are independent.
Obviously, Xn converges in law to
X. However, by using the convolution formula, we can compute that P(|Xn-X|>=
e)->exp(-e)>0 for any e>0.

【在 f*********1 的大作中提到】
: 1. convergence in distribution;
: 2. convergence in probability;
: 1 doesn't imply 2.
: 但俺想不出converge in distribution 但是不converge in probability的例子了
: 谁能给列举几个?
: 多谢!

k*******d
发帖数: 1340
9
X_n are N(0,1)
X_0 is -X_n
clearly X_n -> X_0 in distribution since they have the same distribution;
However
P(|X_n-X_0|>\eps) = P(|2X_0| > \eps), where X_0 is N(0,1), clearly this prob
does not go to zero.
还有,converge in distribution不要求随机变量在同一个probability space上,
converge in probablity 要求在同一个prob space上,除非X_0是常数
Note that if X_n -->(in distr) X_0, where P(X_0 = c) = 1, c is a constant
then X_n --> X_0 in prob
1 (共1页)
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