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Mathematics版 - 大牛们帮忙看一下这道题(有关Order Statistics)
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话题: xi话题: iid话题: yi话题: y1话题: order
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1 (共1页)
f**********g
发帖数: 107
1
Let X be a random variable with mean >=0
Y is a random variable with mean >=0, independent of X.
Z=X+Y
Let X(n) and Z(n) be the first order statistics of X and Z respectively,
Do we have E(X(n)) <= E(Z(n))?
换句话说,当加上一个mean大于等于0的R.V.以后,first order statistic是不是一定
变大了呢?
B****n
发帖数: 11290
2
X(n)+Y(n)=Z(n)

【在 f**********g 的大作中提到】
: Let X be a random variable with mean >=0
: Y is a random variable with mean >=0, independent of X.
: Z=X+Y
: Let X(n) and Z(n) be the first order statistics of X and Z respectively,
: Do we have E(X(n)) <= E(Z(n))?
: 换句话说,当加上一个mean大于等于0的R.V.以后,first order statistic是不是一定
: 变大了呢?

hs
发帖数: 1549
3
我觉得结论正确: Z(n)>=X(n)+Y_m
其中m表示X_m=X(n)
不等式取期望,由于X,Y独立,E(Y_m)=E(Y)>=0

【在 f**********g 的大作中提到】
: Let X be a random variable with mean >=0
: Y is a random variable with mean >=0, independent of X.
: Z=X+Y
: Let X(n) and Z(n) be the first order statistics of X and Z respectively,
: Do we have E(X(n)) <= E(Z(n))?
: 换句话说,当加上一个mean大于等于0的R.V.以后,first order statistic是不是一定
: 变大了呢?

s******h
发帖数: 539
4
1. X(n) + Y(n) ? = Z(n)

Not necessarily true.
Suppose X1, X2, Y1, Y2 are iid Bernoulli(1/2) r.v.'s.
Zi = Xi + Yi (i = 1, 2), then Zi IID~ Binomial(2,1/2).
Their corresponding order statistics are denoted by
X(i), Y(i), Z(i) (i = 1, ..., n).
The first order statistic's Z(1), X(1), Y(1) have the following
properties:

P(Z(1) = 0) = 1 - P( Z1 != 0, Z2 != 0)
= 1 - [1 - P(Z1 = 0)]*[1 - P(Z2 = 0)]
= 1 - [1 - 1/4] * [1 - 1/4]
f**********g
发帖数: 107
5
感谢大家回复。可惜没解决问题。首先X(n)+Y(n)不等于Z(n),Z(n)甚至不等于X(i)+Y(
j) for some i, j. Z(n)也不等于Xi+Yj for some i, j。道理很简单,如果Z(n)=Xi+
Yj for some i, j,并且Z(n-1)=Xk+Yl for some k, l,那么with positive
probability Xk+Yl>=Xi+Yj, contradicting the definition of order stat。Z(n)实
际上是reconstructed from the ordered sample space。
hs
发帖数: 1549
6

Y(
~~~~~~~~~~~~~~~~~~~~~~~~~Z(n)是Z_1...Z_n之一啊
Z(n)也不等于Xi+Yj for some i, j。道理很简单,如果Z(n)=Xi+

【在 f**********g 的大作中提到】
: 感谢大家回复。可惜没解决问题。首先X(n)+Y(n)不等于Z(n),Z(n)甚至不等于X(i)+Y(
: j) for some i, j. Z(n)也不等于Xi+Yj for some i, j。道理很简单,如果Z(n)=Xi+
: Yj for some i, j,并且Z(n-1)=Xk+Yl for some k, l,那么with positive
: probability Xk+Yl>=Xi+Yj, contradicting the definition of order stat。Z(n)实
: 际上是reconstructed from the ordered sample space。

hs
发帖数: 1549
7
讨论一下最后一行:
max(EX(1), EY(1)) <= E(X(1)+Y(1))
X(1),Y(1)并不一定是均值非负的

【在 s******h 的大作中提到】
: 1. X(n) + Y(n) ? = Z(n)
:
: Not necessarily true.
: Suppose X1, X2, Y1, Y2 are iid Bernoulli(1/2) r.v.'s.
: Zi = Xi + Yi (i = 1, 2), then Zi IID~ Binomial(2,1/2).
: Their corresponding order statistics are denoted by
: X(i), Y(i), Z(i) (i = 1, ..., n).
: The first order statistic's Z(1), X(1), Y(1) have the following
: properties:
:

f**********g
发帖数: 107
8
Z(n)是Z_1...Z_n之一啊
~~~~~~~~~~~~
Z(n)绝对不是Z_1...Z_n之一。Z(n)是在ordered sample space上重新construct出来的。
换句话说Z(n)不等于Z_i,for some i。

【在 hs 的大作中提到】
: 讨论一下最后一行:
: max(EX(1), EY(1)) <= E(X(1)+Y(1))
: X(1),Y(1)并不一定是均值非负的

f**********g
发帖数: 107
9
max(EX(1), EY(1)) <= EZ(1)的结论好像和问题并不相关啊。
这和E(X(n)), E(Z(n))有什么关系呢?
hs
发帖数: 1549
10
好吧,我承认我没看懂,谢谢了

的。

【在 f**********g 的大作中提到】
: Z(n)是Z_1...Z_n之一啊
: ~~~~~~~~~~~~
: Z(n)绝对不是Z_1...Z_n之一。Z(n)是在ordered sample space上重新construct出来的。
: 换句话说Z(n)不等于Z_i,for some i。

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进入Mathematics版参与讨论
s******h
发帖数: 539
11
You are right, I thought the condition was Y_i, X_i are all nonnegative r.v.
's and lost following the condition E X1 >=0, E Y1 >=0.

【在 hs 的大作中提到】
: 讨论一下最后一行:
: max(EX(1), EY(1)) <= E(X(1)+Y(1))
: X(1),Y(1)并不一定是均值非负的

s******h
发帖数: 539
12
In general, it's not true that E Z(i) >= E(X(i)), for all i= 1,...,n
Take Xi ~IID N(0,1), Yi ~IID N(0,1), then Zi ~ IID N(0,2)
E Z(i) = \sqrt{2} * E{ Z(i)/\sqrt{2}}
= \sqrt{2} * E{ X(i) }

Thus E { Z(i) } - E{ X(i) }
= (\sqrt{2} - 1) * E X(i)
And we know that E X(i) < = 0 for i < = n/2
E X(i) > = 0 for i > = n/2
###################################################################
But in general, if Zi >= Xi ( for all i=1, ..., n), it's true that
Z(i) > = X
hs
发帖数: 1549
13
窃以为最早那篇文中的Z(n), X(n)都是最大的那一个。。。

【在 s******h 的大作中提到】
: In general, it's not true that E Z(i) >= E(X(i)), for all i= 1,...,n
: Take Xi ~IID N(0,1), Yi ~IID N(0,1), then Zi ~ IID N(0,2)
: E Z(i) = \sqrt{2} * E{ Z(i)/\sqrt{2}}
: = \sqrt{2} * E{ X(i) }
:
: Thus E { Z(i) } - E{ X(i) }
: = (\sqrt{2} - 1) * E X(i)
: And we know that E X(i) < = 0 for i < = n/2
: E X(i) > = 0 for i > = n/2
: ###################################################################

s******h
发帖数: 539
14
Then the result should be true,I'm just saying that it's not true for all
order statistics, it's always good to know the extra going-on things in a
problem.
In order to make sure that E Z(n) >= E X(n) :
1. Independence assumption may not be ignored,
because if Xi = - Yi ~ IID N(0,1), then Z(n) = 0, a.s., and E X(n) > 0.
2. Z(n) > = Xi + Yi, for all i =1, ..., n, by definition of Z(n) (largest
value)

Thus
E{ Z(n) | Y1, ..., Yn } > = E Xi + Yi a.s. for all i=1, ..., n

Then
E{

【在 hs 的大作中提到】
: 窃以为最早那篇文中的Z(n), X(n)都是最大的那一个。。。
hs
发帖数: 1549
15
我觉得你这个很对
一点补充: E Xi 不是随机变量而是一个非负数了,所以min_{1 <=i <=n } E Xi直接可
以认为>=0就是了

【在 s******h 的大作中提到】
: Then the result should be true,I'm just saying that it's not true for all
: order statistics, it's always good to know the extra going-on things in a
: problem.
: In order to make sure that E Z(n) >= E X(n) :
: 1. Independence assumption may not be ignored,
: because if Xi = - Yi ~ IID N(0,1), then Z(n) = 0, a.s., and E X(n) > 0.
: 2. Z(n) > = Xi + Yi, for all i =1, ..., n, by definition of Z(n) (largest
: value)
:
: Thus

s******h
发帖数: 539
16
说得很对。

【在 hs 的大作中提到】
: 我觉得你这个很对
: 一点补充: E Xi 不是随机变量而是一个非负数了,所以min_{1 <=i <=n } E Xi直接可
: 以认为>=0就是了

f**********g
发帖数: 107
17
Good analysis.However, I have the a question about your analysis.
I agree with you until the following part:
"Then
E{ Z(n) | Y1, ..., Yn } > = min_{1 <=i <=n } E Xi + Y(n) a.s."
But how did you get the next part:
"Thus
E{ Z(n) } >= E{Y(n)} + min_{1 <=i <=n} E Xi >= E{ Y(n) }"
I understand that you want to use the result that E(E(X|Y))=E(X). However,
notice that the outer expectation on the left hand side is taken with
respect to Y. Therefore, in your analysis, E(Z(n))=E(E(Z(n)|Y1,...,Yn)), the
o

【在 s******h 的大作中提到】
: 说得很对。
H****h
发帖数: 1037
18
很好。

Then the result should be true,I'm just saying that it's not true for all
order statistics, it's always good to know the extra going-on things in a
problem.
In order to make sure that E Z(n) >= E X(n) :
1. Independence assumption may not be ignored,
because if Xi = - Yi ~ IID N(0,1), then Z(n) = 0, a.s., and E X(n) > 0.
2. Z(n) > = Xi + Yi, for all i =1, ..., n, by definition of Z(n) (largest
value)

Thus
E{ Z(n) | Y1, ..., Yn } > = E Xi + Yi a.s. for all i=1, ..., n

Then

【在 s******h 的大作中提到】
: Then the result should be true,I'm just saying that it's not true for all
: order statistics, it's always good to know the extra going-on things in a
: problem.
: In order to make sure that E Z(n) >= E X(n) :
: 1. Independence assumption may not be ignored,
: because if Xi = - Yi ~ IID N(0,1), then Z(n) = 0, a.s., and E X(n) > 0.
: 2. Z(n) > = Xi + Yi, for all i =1, ..., n, by definition of Z(n) (largest
: value)
:
: Thus

s******h
发帖数: 539
19
E{ Y(n) } can also use the pdf of (Y1, ..., Yn), and Y(n) is a function of (
Y1, ..., Yn). It's the change of variable theorem that you can find in most
Real Analysis books that guarantees this.
It's like if Xi IID~ N(0,1), say T = X1^2 + X2^2 + ... + Xn^2
E(T) can be calculated by both the normal density and chi-sqaure density.

the

【在 f**********g 的大作中提到】
: Good analysis.However, I have the a question about your analysis.
: I agree with you until the following part:
: "Then
: E{ Z(n) | Y1, ..., Yn } > = min_{1 <=i <=n } E Xi + Y(n) a.s."
: But how did you get the next part:
: "Thus
: E{ Z(n) } >= E{Y(n)} + min_{1 <=i <=n} E Xi >= E{ Y(n) }"
: I understand that you want to use the result that E(E(X|Y))=E(X). However,
: notice that the outer expectation on the left hand side is taken with
: respect to Y. Therefore, in your analysis, E(Z(n))=E(E(Z(n)|Y1,...,Yn)), the

f**********g
发帖数: 107
20
Thanks for the elaboration. Yes, this makes sense. Right now I cannot find
any error in your logic, but I still have a feeling that some part is not
that solid. I will look into this and get back to you sometime next week.

(
most

【在 s******h 的大作中提到】
: E{ Y(n) } can also use the pdf of (Y1, ..., Yn), and Y(n) is a function of (
: Y1, ..., Yn). It's the change of variable theorem that you can find in most
: Real Analysis books that guarantees this.
: It's like if Xi IID~ N(0,1), say T = X1^2 + X2^2 + ... + Xn^2
: E(T) can be calculated by both the normal density and chi-sqaure density.
:
: the

1 (共1页)
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