z*3
8 楼
借此楼问个问题
Prove that if F(x) is a cumulative distribution function then F(x) is right
continuous, that is F(x + h) ↓ F(x) as h ↓ 0, for all x ∈ ℜ
Prove that if F(x) is a cumulative distribution function then F(x) is right
continuous, that is F(x + h) ↓ F(x) as h ↓ 0, for all x ∈ ℜ
M*e
14 楼
我不知道你们用的什么书,有些词可能不一样。
你们应该在引进measure之后证明/quote了一些定理,其中一条是说
Measure has continuity from above.
Let \mu be the measure. If you have events A_i with A_i \downarrow A, then \
mu(A_i) \downarrow \mu(A).
In this case, let's consider A_i = (-\infty, x+1/i], then we have
A_i \downward (-\infty, x]. According to the theorem, we have
\lim_{i\to\infty}F(x+1/i)=\lim_{i\to \infty} P((-\infty,x+1/i]) = P((-\infty
, x]) = F(x)
Now we only have to argue it is true when we replace 1/i by positive h.
【在 z*****3 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 教授能不能解释下和题干中h有关的那部分是什么意思
你们应该在引进measure之后证明/quote了一些定理,其中一条是说
Measure has continuity from above.
Let \mu be the measure. If you have events A_i with A_i \downarrow A, then \
mu(A_i) \downarrow \mu(A).
In this case, let's consider A_i = (-\infty, x+1/i], then we have
A_i \downward (-\infty, x]. According to the theorem, we have
\lim_{i\to\infty}F(x+1/i)=\lim_{i\to \infty} P((-\infty,x+1/i]) = P((-\infty
, x]) = F(x)
Now we only have to argue it is true when we replace 1/i by positive h.
【在 z*****3 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 教授能不能解释下和题干中h有关的那部分是什么意思
M*e
15 楼
我第一个回的那个帖子你可以不看了,就看这两个吧。
Continue:
To prove
\lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
>0. Therefore
\lim_{h\to 0^+} F(x+h) >= F(x).
On the other hand, for any integer n>0, you can alway find a g>0, such that
g<1/n, and hence for such g and n
F(x+g) <= F(x+1/n), which means
\lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
Since this is true for any n, we have
\lim_{h\to 0^+}F(x+h) <= lim_{n\to\infty} F(x+1/n) = F(x)
Combining these two inequalities
\lim_{h\to 0^+} F(x+h) = F(x).
The monotone part is trivial from the nonnegativity of the measure.
\
infty
【在 M*****e 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 我不知道你们用的什么书,有些词可能不一样。
: 你们应该在引进measure之后证明/quote了一些定理,其中一条是说
: Measure has continuity from above.
: Let \mu be the measure. If you have events A_i with A_i \downarrow A, then \
: mu(A_i) \downarrow \mu(A).
: In this case, let's consider A_i = (-\infty, x+1/i], then we have
: A_i \downward (-\infty, x]. According to the theorem, we have
: \lim_{i\to\infty}F(x+1/i)=\lim_{i\to \infty} P((-\infty,x+1/i]) = P((-\infty
: , x]) = F(x)
: Now we only have to argue it is true when we replace 1/i by positive h.
Continue:
To prove
\lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
>0. Therefore
\lim_{h\to 0^+} F(x+h) >= F(x).
On the other hand, for any integer n>0, you can alway find a g>0, such that
g<1/n, and hence for such g and n
F(x+g) <= F(x+1/n), which means
\lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
Since this is true for any n, we have
\lim_{h\to 0^+}F(x+h) <= lim_{n\to\infty} F(x+1/n) = F(x)
Combining these two inequalities
\lim_{h\to 0^+} F(x+h) = F(x).
The monotone part is trivial from the nonnegativity of the measure.
\
infty
【在 M*****e 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 我不知道你们用的什么书,有些词可能不一样。
: 你们应该在引进measure之后证明/quote了一些定理,其中一条是说
: Measure has continuity from above.
: Let \mu be the measure. If you have events A_i with A_i \downarrow A, then \
: mu(A_i) \downarrow \mu(A).
: In this case, let's consider A_i = (-\infty, x+1/i], then we have
: A_i \downward (-\infty, x]. According to the theorem, we have
: \lim_{i\to\infty}F(x+1/i)=\lim_{i\to \infty} P((-\infty,x+1/i]) = P((-\infty
: , x]) = F(x)
: Now we only have to argue it is true when we replace 1/i by positive h.
M*e
17 楼
静芬跑路了?把妹去了?
h
that
【在 M*****e 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 我第一个回的那个帖子你可以不看了,就看这两个吧。
: Continue:
: To prove
: \lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
: >0. Therefore
: \lim_{h\to 0^+} F(x+h) >= F(x).
: On the other hand, for any integer n>0, you can alway find a g>0, such that
: g<1/n, and hence for such g and n
: F(x+g) <= F(x+1/n), which means
: \lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
h
that
【在 M*****e 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 我第一个回的那个帖子你可以不看了,就看这两个吧。
: Continue:
: To prove
: \lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
: >0. Therefore
: \lim_{h\to 0^+} F(x+h) >= F(x).
: On the other hand, for any integer n>0, you can alway find a g>0, such that
: g<1/n, and hence for such g and n
: F(x+g) <= F(x+1/n), which means
: \lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
m*g
18 楼
也赐我一死吧
我和静芬同归于尽
我和静芬同归于尽
z*3
19 楼
第一个方法里lim_{hto 0^+} P((x,x+h]) = P(emptyset) 不懂
第二个方法里lim_{hto 0^+}F(x+h) <= F(x+g)不懂
that
【在 M*****e 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 我第一个回的那个帖子你可以不看了,就看这两个吧。
: Continue:
: To prove
: \lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
: >0. Therefore
: \lim_{h\to 0^+} F(x+h) >= F(x).
: On the other hand, for any integer n>0, you can alway find a g>0, such that
: g<1/n, and hence for such g and n
: F(x+g) <= F(x+1/n), which means
: \lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
第二个方法里lim_{hto 0^+}F(x+h) <= F(x+g)不懂
that
【在 M*****e 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 我第一个回的那个帖子你可以不看了,就看这两个吧。
: Continue:
: To prove
: \lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
: >0. Therefore
: \lim_{h\to 0^+} F(x+h) >= F(x).
: On the other hand, for any integer n>0, you can alway find a g>0, such that
: g<1/n, and hence for such g and n
: F(x+g) <= F(x+1/n), which means
: \lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
M*e
23 楼
不要管第一个了
第二个方法里这一步我确实偷了懒,你眼睛真尖。其实在证明lim存在之前就用它是不
对的。
Since for any f > h > 0, F(x+f) >= F(x+h) >= F(x), the limit is monotone
decreasing and bounded from below, according to some theorem in mathematical
analysis, the limit exists and is the greatest lower bound of the set,
which means it's always less than or equal to any particular value.
我不知道你们证明要多严谨……不加这个似乎也没关系吧。
【在 z*****3 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 第一个方法里lim_{hto 0^+} P((x,x+h]) = P(emptyset) 不懂
: 第二个方法里lim_{hto 0^+}F(x+h) <= F(x+g)不懂
:
: that
第二个方法里这一步我确实偷了懒,你眼睛真尖。其实在证明lim存在之前就用它是不
对的。
Since for any f > h > 0, F(x+f) >= F(x+h) >= F(x), the limit is monotone
decreasing and bounded from below, according to some theorem in mathematical
analysis, the limit exists and is the greatest lower bound of the set,
which means it's always less than or equal to any particular value.
我不知道你们证明要多严谨……不加这个似乎也没关系吧。
【在 z*****3 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 第一个方法里lim_{hto 0^+} P((x,x+h]) = P(emptyset) 不懂
: 第二个方法里lim_{hto 0^+}F(x+h) <= F(x+g)不懂
:
: that
z*3
25 楼
可是我喜欢第一个方法,简单快捷,宛如刀一样插入要害,只不过刀把暂时没找到;
根据这个statement,你是认为g>h了?(就是你原论证里面的那个g),怎么看出来的?
既然h是个变量,它可以大也可以小呀
mathematical
【在 M*****e 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 不要管第一个了
: 第二个方法里这一步我确实偷了懒,你眼睛真尖。其实在证明lim存在之前就用它是不
: 对的。
: Since for any f > h > 0, F(x+f) >= F(x+h) >= F(x), the limit is monotone
: decreasing and bounded from below, according to some theorem in mathematical
: analysis, the limit exists and is the greatest lower bound of the set,
: which means it's always less than or equal to any particular value.
: 我不知道你们证明要多严谨……不加这个似乎也没关系吧。
根据这个statement,你是认为g>h了?(就是你原论证里面的那个g),怎么看出来的?
既然h是个变量,它可以大也可以小呀
mathematical
【在 M*****e 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 不要管第一个了
: 第二个方法里这一步我确实偷了懒,你眼睛真尖。其实在证明lim存在之前就用它是不
: 对的。
: Since for any f > h > 0, F(x+f) >= F(x+h) >= F(x), the limit is monotone
: decreasing and bounded from below, according to some theorem in mathematical
: analysis, the limit exists and is the greatest lower bound of the set,
: which means it's always less than or equal to any particular value.
: 我不知道你们证明要多严谨……不加这个似乎也没关系吧。
M*e
26 楼
两个方法是一样的
第一个方法我把关键步骤用了~~~~此处展开代替
第二个方法我把所有步骤都展开了
我没有假设g>h. h是个变量,g是某个给定值。
The limit, which is the greatest lower bound, which is a lower bound, is
less than or equal to any particular value, which in this case is F(x+g).
【在 z*****3 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 可是我喜欢第一个方法,简单快捷,宛如刀一样插入要害,只不过刀把暂时没找到;
: 根据这个statement,你是认为g>h了?(就是你原论证里面的那个g),怎么看出来的?
: 既然h是个变量,它可以大也可以小呀
:
: mathematical
第一个方法我把关键步骤用了~~~~此处展开代替
第二个方法我把所有步骤都展开了
我没有假设g>h. h是个变量,g是某个给定值。
The limit, which is the greatest lower bound, which is a lower bound, is
less than or equal to any particular value, which in this case is F(x+g).
【在 z*****3 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 可是我喜欢第一个方法,简单快捷,宛如刀一样插入要害,只不过刀把暂时没找到;
: 根据这个statement,你是认为g>h了?(就是你原论证里面的那个g),怎么看出来的?
: 既然h是个变量,它可以大也可以小呀
:
: mathematical
z*3
27 楼
我看一些其他人的证明,
A cdf is always continuous from the right; that is, F(x) = F(x
+) at every point x.
Proof: Let y1 > y2 > · · · be a sequence of numbers that are decreasing
such that limn→∞ yn = x.
Then the event {X ≤ x} is the intersection of all the events {X ≤ yn} for
n = 1, 2, . . .. Hence, by the continuity property of the probability
function,F(x) = P(X ≤ x) = limn→∞P(X ≤ yn) = F(x+).
中间似乎少了点什么
【在 M*****e 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 两个方法是一样的
: 第一个方法我把关键步骤用了~~~~此处展开代替
: 第二个方法我把所有步骤都展开了
: 我没有假设g>h. h是个变量,g是某个给定值。
: The limit, which is the greatest lower bound, which is a lower bound, is
: less than or equal to any particular value, which in this case is F(x+g).
A cdf is always continuous from the right; that is, F(x) = F(x
+) at every point x.
Proof: Let y1 > y2 > · · · be a sequence of numbers that are decreasing
such that limn→∞ yn = x.
Then the event {X ≤ x} is the intersection of all the events {X ≤ yn} for
n = 1, 2, . . .. Hence, by the continuity property of the probability
function,F(x) = P(X ≤ x) = limn→∞P(X ≤ yn) = F(x+).
中间似乎少了点什么
【在 M*****e 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 两个方法是一样的
: 第一个方法我把关键步骤用了~~~~此处展开代替
: 第二个方法我把所有步骤都展开了
: 我没有假设g>h. h是个变量,g是某个给定值。
: The limit, which is the greatest lower bound, which is a lower bound, is
: less than or equal to any particular value, which in this case is F(x+g).
M*e
30 楼
对啊,他们的证明比我的简单多了。
我的证明用了特殊的sequence 1/i, 但是同样的逻辑可以对付任意的sequence y_i上,
就像他们做的那样。我从单个sequence跳到 h 要写好几行,但是如果你已经证明了对
于任意的sequence都成立,就已经够了。
http://en.wikipedia.org/wiki/Continuous_function#Definition_in_
for
【在 z*****3 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 我看一些其他人的证明,
: A cdf is always continuous from the right; that is, F(x) = F(x
: +) at every point x.
: Proof: Let y1 > y2 > · · · be a sequence of numbers that are decreasing
: such that limn→∞ yn = x.
: Then the event {X ≤ x} is the intersection of all the events {X ≤ yn} for
: n = 1, 2, . . .. Hence, by the continuity property of the probability
: function,F(x) = P(X ≤ x) = limn→∞P(X ≤ yn) = F(x+).
: 中间似乎少了点什么
我的证明用了特殊的sequence 1/i, 但是同样的逻辑可以对付任意的sequence y_i上,
就像他们做的那样。我从单个sequence跳到 h 要写好几行,但是如果你已经证明了对
于任意的sequence都成立,就已经够了。
http://en.wikipedia.org/wiki/Continuous_function#Definition_in_
for
【在 z*****3 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 我看一些其他人的证明,
: A cdf is always continuous from the right; that is, F(x) = F(x
: +) at every point x.
: Proof: Let y1 > y2 > · · · be a sequence of numbers that are decreasing
: such that limn→∞ yn = x.
: Then the event {X ≤ x} is the intersection of all the events {X ≤ yn} for
: n = 1, 2, . . .. Hence, by the continuity property of the probability
: function,F(x) = P(X ≤ x) = limn→∞P(X ≤ yn) = F(x+).
: 中间似乎少了点什么
z*3
33 楼
我可不可以理解成P(x)就是P(x+h)的交集?
【在 M*****e 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 对啊,他们的证明比我的简单多了。
: 我的证明用了特殊的sequence 1/i, 但是同样的逻辑可以对付任意的sequence y_i上,
: 就像他们做的那样。我从单个sequence跳到 h 要写好几行,但是如果你已经证明了对
: 于任意的sequence都成立,就已经够了。
: http://en.wikipedia.org/wiki/Continuous_function#Definition_in_
:
: for
【在 M*****e 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 对啊,他们的证明比我的简单多了。
: 我的证明用了特殊的sequence 1/i, 但是同样的逻辑可以对付任意的sequence y_i上,
: 就像他们做的那样。我从单个sequence跳到 h 要写好几行,但是如果你已经证明了对
: 于任意的sequence都成立,就已经够了。
: http://en.wikipedia.org/wiki/Continuous_function#Definition_in_
:
: for
m*n
38 楼
万能的Joke版。。
that
【在 M*****e 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 我第一个回的那个帖子你可以不看了,就看这两个吧。
: Continue:
: To prove
: \lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
: >0. Therefore
: \lim_{h\to 0^+} F(x+h) >= F(x).
: On the other hand, for any integer n>0, you can alway find a g>0, such that
: g<1/n, and hence for such g and n
: F(x+g) <= F(x+1/n), which means
: \lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
that
【在 M*****e 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 我第一个回的那个帖子你可以不看了,就看这两个吧。
: Continue:
: To prove
: \lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
: >0. Therefore
: \lim_{h\to 0^+} F(x+h) >= F(x).
: On the other hand, for any integer n>0, you can alway find a g>0, such that
: g<1/n, and hence for such g and n
: F(x+g) <= F(x+1/n), which means
: \lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
M*e
39 楼
你可以这样理解,但不能这么写。
对于任意y_1 > y_2..... >0, with lim_{i\to \infty} y_i = 0
(-\infty, x]是 所有(-\infty, x+y_i]的交集
但是你不能把 y_i直接换成h, 因为在mearuable space的定义里面
the intersection of countably many events is an event
but the intersection of arbitrarily many events is not necessarily so
所以你的证明里引用的那个continuity from above property也是只适用于countably
many events的。
靠,老子中英文夹杂真恶心。
【在 z*****3 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 我可不可以理解成P(x)就是P(x+h)的交集?
对于任意y_1 > y_2..... >0, with lim_{i\to \infty} y_i = 0
(-\infty, x]是 所有(-\infty, x+y_i]的交集
但是你不能把 y_i直接换成h, 因为在mearuable space的定义里面
the intersection of countably many events is an event
but the intersection of arbitrarily many events is not necessarily so
所以你的证明里引用的那个continuity from above property也是只适用于countably
many events的。
靠,老子中英文夹杂真恶心。
【在 z*****3 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 我可不可以理解成P(x)就是P(x+h)的交集?
m*g
40 楼
为什么xiaopo还不回答我的问题
S*u
41 楼
哆哆嗦嗦路过,吓尿了 好高深
h
that
【在 M*****e 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 我第一个回的那个帖子你可以不看了,就看这两个吧。
: Continue:
: To prove
: \lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
: >0. Therefore
: \lim_{h\to 0^+} F(x+h) >= F(x).
: On the other hand, for any integer n>0, you can alway find a g>0, such that
: g<1/n, and hence for such g and n
: F(x+g) <= F(x+1/n), which means
: \lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
h
that
【在 M*****e 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 我第一个回的那个帖子你可以不看了,就看这两个吧。
: Continue:
: To prove
: \lim_{h\to 0^+} F(x+h) = F(x), first use the fact that F(x+h)>=F(x) since h
: >0. Therefore
: \lim_{h\to 0^+} F(x+h) >= F(x).
: On the other hand, for any integer n>0, you can alway find a g>0, such that
: g<1/n, and hence for such g and n
: F(x+g) <= F(x+1/n), which means
: \lim_{h\to 0^+}F(x+h) <= F(x+g)<=F(x+1/n).
b*p
43 楼
太高深鸟,开始以为进错了版!
p*g
67 楼
我考, 我还以为进错版了
x*o
68 楼
看到标题提到我,兴奋地冲进来,看到楼歪成这样,就软了
ps: 把妹不是随便掰俩笑话,一定要你比她更黄才行
ps: 把妹不是随便掰俩笑话,一定要你比她更黄才行
相关阅读