b*****e
发帖数: 88 | | B****n
发帖数: 11290 | 2 在一些regular的條件下 這兩個應該一樣巴 兩邊同時對a微分 只要微分和積分可以交
換
【在 b*****e 的大作中提到】
: Please see the link
: http://math.stackexchange.com/questions/827318/please-help-comp
: Thanks.
| b*****e
发帖数: 88 | 3 请问，你能说详细点吗
给定上面那个等式，because a+t < a+b(a)+t，所以看起来有这个不等关系呀。
就是不知道如何能证明.
can any 大牛 help?
Thanks.
【在 B****n 的大作中提到】
: 在一些regular的條件下 這兩個應該一樣巴 兩邊同時對a微分 只要微分和積分可以交
: 換
| x********y
发帖数: 189 | 4 两个distribution functions的support是[0,c]???
这个本身就有问题，distribution function的support只可能是[constant,+infinity).
另外，你的first order stochastic dominance看起来给的也不对，并不是多任意的t
，而是对某些t （如果我理解正确的话），你所给的和我所说的差很多。
你先把这些写清楚了我再给你看怎么做一下。。。 | b*****e
发帖数: 88 | 5 distribution function's support could be [constant,constant], for example,
uniform distribution on [a, b].
first order stochastic dominance requires "less than or equal to" and "less
than" for some t. Strictly speaking, the condition given is a little bit
stronger than FOSD.
To be more specific, let's assume two cumulative distribution functions, G1(
t) and G2(t), on [0,c] are two strictly increasing functions. Then, let F1(t
)=G1(t)*G2(t) and F2(t)=G1(t), which leads to F1(t)
Thanks.
infinity).
t
【在 x********y 的大作中提到】
: 两个distribution functions的support是[0,c]???
: 这个本身就有问题，distribution function的support只可能是[constant,+infinity).
: 另外，你的first order stochastic dominance看起来给的也不对，并不是多任意的t
: ，而是对某些t （如果我理解正确的话），你所给的和我所说的差很多。
: 你先把这些写清楚了我再给你看怎么做一下。。。
| x********y
发帖数: 189 | 6 Again, how can a cdf's support be [constant,constant]???
The definition of a cfd is F(x)=P(X less or equal x), i.e. F(infinity)=1!!!
Anyway, I can guess what you meant, your "support" is actually the non-
constant part.
The result you want to claim is not right! A counterexample would be
F1(t)=delta_0(t)
F2(t)=1/2 delta_0(t)+ 1/2 delta_c(t)
Wisely choose a small c in order to make your b<0 and b+c<0. | b*****e
发帖数: 88 | 7 Yes, you are right on "support".
I am not sure I got your counter example. what kind of function is delta_0(t
)? Also, please notice that b(a) must be positive from the equality
condition, since integration of 1/(a+t)*d(F1(t)-F2(t)) is negative (use
integration by parts, and 1/(a+t) is decreasing in t).
Could you elaborate more? Thanks.
【在 x********y 的大作中提到】
: Again, how can a cdf's support be [constant,constant]???
: The definition of a cfd is F(x)=P(X less or equal x), i.e. F(infinity)=1!!!
: Anyway, I can guess what you meant, your "support" is actually the non-
: constant part.
: The result you want to claim is not right! A counterexample would be
: F1(t)=delta_0(t)
: F2(t)=1/2 delta_0(t)+ 1/2 delta_c(t)
: Wisely choose a small c in order to make your b<0 and b+c<0.
| x********y
发帖数: 189 | 8 delta_0(t)=1 if t=0, and =0 if t not equal to 0.
I don't understand why you need to consider the integration of 1/(a+t)*d(F1(
t)-F2(t))???
The thing is that you never know if d(F1(t)-F2(t))>0...
Just try what I mentioned to you, and work it out. I can't write everything
down! Put some effort yourself, ok? | B********e
发帖数: 10014 | 9 that's not what he wants.
any cdf F here satisfies F(0)=0, F(c)=1.
I think the answer is positive.
F1(
everything
【在 x********y 的大作中提到】
: delta_0(t)=1 if t=0, and =0 if t not equal to 0.
: I don't understand why you need to consider the integration of 1/(a+t)*d(F1(
: t)-F2(t))???
: The thing is that you never know if d(F1(t)-F2(t))>0...
: Just try what I mentioned to you, and work it out. I can't write everything
: down! Put some effort yourself, ok?
| B********e
发帖数: 10014 | 10 数值试了试，貌似什么情况都可以发生。
examples: take c=1, a=10, denote A1,A2 the integrals of the square functions
corresponding to F1 and F2.
1.
F1=t, F2=t+1/2t*(1-t),
b=0.0826, then
A1=0.00909091>0.00909024=A2;
2.
F1=t-1/2t(1-t),
F2=t,
b=0.08392,
A1=0.00894654<0.008947=A2.
也试了a=20, 100,一样结果。
感觉a＝100已经足够大了吧？lz不妨验算一下，看看是否解决问题。
【在 B********e 的大作中提到】
: that's not what he wants.
: any cdf F here satisfies F(0)=0, F(c)=1.
: I think the answer is positive.
:
: F1(
: everything
| | | b*****e
发帖数: 88 | 11 Thanks, you are right.
I have been able to show that if b>2*c, then the inequality holds. Also
notice that as 'a' goes to infinity, b(a) has to go to infinity as well. So,
as 'a' is large, the inequality holds.
functions
【在 B********e 的大作中提到】
: 数值试了试，貌似什么情况都可以发生。
: examples: take c=1, a=10, denote A1,A2 the integrals of the square functions
: corresponding to F1 and F2.
: 1.
: F1=t, F2=t+1/2t*(1-t),
: b=0.0826, then
: A1=0.00909091>0.00909024=A2;
: 2.
: F1=t-1/2t(1-t),
: F2=t,
| B********e
发帖数: 10014 | 12 sorry, i don't see b goes to infinity as a does so.
So,
【在 b*****e 的大作中提到】
: Thanks, you are right.
: I have been able to show that if b>2*c, then the inequality holds. Also
: notice that as 'a' goes to infinity, b(a) has to go to infinity as well. So,
: as 'a' is large, the inequality holds.
:
: functions
| b*****e
发帖数: 88 | 13 if b does not go to infinity (so goes to a finite number), then 1/(a+t) is
almost equal to 1/(a+b+t) as a is extremely large. Then, since F1(t)
the first equality cannot hold (i.e., the integration of 1/(a+t)*dF1(t) <
the integration of 1/(a+b+t)*dF2(t)), right?
【在 B********e 的大作中提到】
: sorry, i don't see b goes to infinity as a does so.
:
: So,
| D******g
发帖数: 125 | 14 xxxy的意思是density function 换成distribution function就是
F_1= 0 when x
F_2=0 when x<0, F_2= 1/2 when 0<=x
然后很容易验证楼主的不等式不对
【在 B********e 的大作中提到】
: that's not what he wants.
: any cdf F here satisfies F(0)=0, F(c)=1.
: I think the answer is positive.
:
: F1(
: everything
| y***s
发帖数: 23 | 15 The answer seems NO.
Let X_2 be uniform on (0,1), and X_1=X_2+1. Thus F_1(t)=t-1 on (1,2) and 0
otherwise.
F_2(t)=t on (0,1) and 0 otherwise. And let c=2.
\int_0^c 1/(a+t) dF_1(t) =log (a+2)/(a+1)
\int_0^c 1(a+b+c) dF_2(t) =log (a+b+1)/(a+b);
Hence the first equation will entail b=1.
But
\int_0^c 1/(a+t)^2 dF_1(t) =1/(a+1) -1/(a+2)
equals
\int_0^c 1/(a+b+t)^2 dF_2(t) =1/(a+1) -1/(a+2). | B********e
发帖数: 10014 | 16 no.
let c=1, F1=t^2, F2=t. You can show b->1/6, as a->\infty.
,
【在 b*****e 的大作中提到】
: if b does not go to infinity (so goes to a finite number), then 1/(a+t) is
: almost equal to 1/(a+b+t) as a is extremely large. Then, since F1(t): the first equality cannot hold (i.e., the integration of 1/(a+t)*dF1(t) <
: the integration of 1/(a+b+t)*dF2(t)), right?
| b*****e
发帖数: 88 | 17 in this case, as it can be shown, b=((1+a)*(1+1/a)^(2*a)-a*e^2)/(e^2-(1+1/a)
^(2*a)) > 0.
when a=10000, b=0.166647431218232 (about 1/6), but when a=10^7, b=1.
181533089297586e+005
I used matlab to do the calculation. It seems matlab has some problem with
the calculation since I could get negative b by trying some a values (which
is wrong since b has to be positive). to give an example, we know (1+1/a)^(a
) goes to e as a goes to infinity, but matlab would give 1 as a=10^16.
【在 B********e 的大作中提到】
: no.
: let c=1, F1=t^2, F2=t. You can show b->1/6, as a->\infty.
:
: ,
| B********e
发帖数: 10014 | 18 that's truncation error
http://www.wolframalpha.com/input/?i=limit+of+%28%281%2Ba%29*%2
a)
which
(a
【在 b*****e 的大作中提到】
: in this case, as it can be shown, b=((1+a)*(1+1/a)^(2*a)-a*e^2)/(e^2-(1+1/a)
: ^(2*a)) > 0.
: when a=10000, b=0.166647431218232 (about 1/6), but when a=10^7, b=1.
: 181533089297586e+005
: I used matlab to do the calculation. It seems matlab has some problem with
: the calculation since I could get negative b by trying some a values (which
: is wrong since b has to be positive). to give an example, we know (1+1/a)^(a
: ) goes to e as a goes to infinity, but matlab would give 1 as a=10^16.
| b*****e
发帖数: 88 | 19 Thanks!
【在 B********e 的大作中提到】
: that's truncation error
: http://www.wolframalpha.com/input/?i=limit+of+%28%281%2Ba%29*%2
:
: a)
: which
: (a
|
|
