n*****g
发帖数: 112 | 1 If X's 95% confidence interval (CI) = (Lx, Hx)
then can I say the 95% CI of X^2 = (Lx^2, Hx^2)?
i.e., just taking the square of the upper and lower limits?
If so, does it apply to all the monotonic functions?
Thank you! | n*****n
发帖数: 3123 | 2 yes, if lx and ux are >=0 | n*****g
发帖数: 112 | 3 thanks!
【在 n*****n 的大作中提到】
: yes, if lx and ux are >=0
| j******o
发帖数: 127 | 4 I don't think so. You can try a simulation. | T*******I
发帖数: 5138 | 5 You definitely cannot say that. Do you really think a statistical measure is
a mathematical function? They are totally different.
A statistical measure is from a real and uncertain sampling data; and a mathematical function is from a theoretical and certain assumption, just like you assumed in your statement.
There is apparently not a certain assumption between the two things you mentioned, thus you cannot say or do that.
【在 n*****g 的大作中提到】
: If X's 95% confidence interval (CI) = (Lx, Hx)
: then can I say the 95% CI of X^2 = (Lx^2, Hx^2)?
: i.e., just taking the square of the upper and lower limits?
: If so, does it apply to all the monotonic functions?
: Thank you!
| n*****g
发帖数: 112 | 6 thanks. I kind of feel one cannot do that, but what is the correct way to
get the CI without doing simulation or bootstrap?
for example, sqrt(x) is normally distributed, so we can get the 95% CI of
the mean(sqrt(x)). Then how to get the 95% CI of mean(x)?
is
mathematical function is from a theoretical and certain assumption, just
like you assumed in your statement.
mentioned, thus you cannot say or do that.
【在 T*******I 的大作中提到】
: You definitely cannot say that. Do you really think a statistical measure is
: a mathematical function? They are totally different.
: A statistical measure is from a real and uncertain sampling data; and a mathematical function is from a theoretical and certain assumption, just like you assumed in your statement.
: There is apparently not a certain assumption between the two things you mentioned, thus you cannot say or do that.
| T*******I
发帖数: 5138 | 7 Suppose you have two variables X and Y, and for each y_i you have
y_i = sqrt(x_i), then you will have
y_i^2 = x_i
or
Y^2 = X
thus, the mean(Y^2) as well as the 95% CI of the Y^2 are what you want.
Obviously you will have that
the [mean(X)]^2 is not equal to the mean(Y^2). For example, you are given:
X Y
1 1
2 4
3 9
4 16
5 25
the mean(X) is 3, and the mean(Y) is 11.
【在 n*****g 的大作中提到】
: thanks. I kind of feel one cannot do that, but what is the correct way to
: get the CI without doing simulation or bootstrap?
: for example, sqrt(x) is normally distributed, so we can get the 95% CI of
: the mean(sqrt(x)). Then how to get the 95% CI of mean(x)?
:
: is
: mathematical function is from a theoretical and certain assumption, just
: like you assumed in your statement.
: mentioned, thus you cannot say or do that.
| j******o
发帖数: 127 | 8 I guess you need to derive the distribution of the square of your random
variable. For example, the square of standard normal RV will be Chi-squared
one.
【在 n*****g 的大作中提到】
: thanks. I kind of feel one cannot do that, but what is the correct way to
: get the CI without doing simulation or bootstrap?
: for example, sqrt(x) is normally distributed, so we can get the 95% CI of
: the mean(sqrt(x)). Then how to get the 95% CI of mean(x)?
:
: is
: mathematical function is from a theoretical and certain assumption, just
: like you assumed in your statement.
: mentioned, thus you cannot say or do that.
| g******n
发帖数: 339 | 9 The question is: a CI for what?
When people say the confidence interval of x, in general x is a unknown
parameter, say the population mean, not a random variable. 95% CI of Y^2: A
confidence interval of a random variable? It is possible but rarely in the
general context, and in these cases they may have to called by other names
such as the prediction intervals etc.
If a 95% CI of a parameter x =(L,U), then the 95% CI for the parameter x^2=(
L^2, U^2), provided that the transformation x->x^2 is o
【在 T*******I 的大作中提到】
: Suppose you have two variables X and Y, and for each y_i you have
: y_i = sqrt(x_i), then you will have
: y_i^2 = x_i
: or
: Y^2 = X
: thus, the mean(Y^2) as well as the 95% CI of the Y^2 are what you want.
: Obviously you will have that
: the [mean(X)]^2 is not equal to the mean(Y^2). For example, you are given:
: X Y
: 1 1
| o***o
发帖数: 43 | 10 Again, I can not follow your logic. Why we need the mean of Y^2?
I think the statement of LZ can be true under some conditions.
For example, P(Lx
.
Is there anything wrong here?
Besides, a lot of confidence intervals are constructed in a way that they are
invariant to monotone transformations.
【在 T*******I 的大作中提到】
: Suppose you have two variables X and Y, and for each y_i you have
: y_i = sqrt(x_i), then you will have
: y_i^2 = x_i
: or
: Y^2 = X
: thus, the mean(Y^2) as well as the 95% CI of the Y^2 are what you want.
: Obviously you will have that
: the [mean(X)]^2 is not equal to the mean(Y^2). For example, you are given:
: X Y
: 1 1
| | | e****t
发帖数: 766 | 11 It is really a good catch and explantion, thanks !
A
=(
type
log-
【在 g******n 的大作中提到】
: The question is: a CI for what?
: When people say the confidence interval of x, in general x is a unknown
: parameter, say the population mean, not a random variable. 95% CI of Y^2: A
: confidence interval of a random variable? It is possible but rarely in the
: general context, and in these cases they may have to called by other names
: such as the prediction intervals etc.
: If a 95% CI of a parameter x =(L,U), then the 95% CI for the parameter x^2=(
: L^2, U^2), provided that the transformation x->x^2 is o
| T*******I
发帖数: 5138 | 12 你指出的问题非常好。
我在我的那个回答里省略了关于Y^2的总体期望的说法，因而不太严谨。我以为这是一
个不用特别说明的事情，因为任何统计量都是对相应的总体参数的估计，因此，根据本
题讨论中的上下文可知，我所说的关于随机变量Y^2的可信区间时，当然指的是用样本
的mean(Y^2)来估计总体的mu(Y^2)时的可信区间了。
LZ的问题如下：
If X's 95% confidence interval (CI) = (Lx, Hx)
then can I say the 95% CI of X^2 = (Lx^2, Hx^2)?
显然，这样的数学假设和算法是没有根据的。我在上文中给出的实例简单地推翻了这样
的假设。
A
=(
type
log-
【在 g******n 的大作中提到】
: The question is: a CI for what?
: When people say the confidence interval of x, in general x is a unknown
: parameter, say the population mean, not a random variable. 95% CI of Y^2: A
: confidence interval of a random variable? It is possible but rarely in the
: general context, and in these cases they may have to called by other names
: such as the prediction intervals etc.
: If a 95% CI of a parameter x =(L,U), then the 95% CI for the parameter x^2=(
: L^2, U^2), provided that the transformation x->x^2 is o
| s********s
发帖数: 13 | 13 TNEGIETNI和graceman貌似是在讨论2个不同的问题。
graceman讨论的是: 假设我们要estimate一个distribution的某个参数t, which is a
constant but unknown from frequentist's point of view.那么我们显然可以用观察
到的data来construct confidence interval at level alpha. Let's say这个
interval for t是 [f(x), g(x)], 这里x是observed data. 那么对于t^2, which is
also constant but unknown, 它的CI就是[f(x)^2, g(x)^2]. As simple as this.本
质上来讲t和t^2是一个parameter.因为它们用的相同的statistics.
TNEGIETNI讨论的情况是estimate 2个没有直接关系的parameter.或许可以换个例子来
考虑。比如说x follows N(0,1), y = x^2, 那么y follo
【在 T*******I 的大作中提到】
: 你指出的问题非常好。
: 我在我的那个回答里省略了关于Y^2的总体期望的说法，因而不太严谨。我以为这是一
: 个不用特别说明的事情，因为任何统计量都是对相应的总体参数的估计，因此，根据本
: 题讨论中的上下文可知，我所说的关于随机变量Y^2的可信区间时，当然指的是用样本
: 的mean(Y^2)来估计总体的mu(Y^2)时的可信区间了。
: LZ的问题如下：
: If X's 95% confidence interval (CI) = (Lx, Hx)
: then can I say the 95% CI of X^2 = (Lx^2, Hx^2)?
: 显然，这样的数学假设和算法是没有根据的。我在上文中给出的实例简单地推翻了这样
: 的假设。
| T*******I
发帖数: 5138 | 14 我认为你错了。t和t^2是两个不同的随机变量，它们的变异性不一样，因而CI的估计也
不一样，且我们没有任何样本基础可以假定[CI(t)]^2 = CI(t^2)。从上面我给出的X和
Y（Y=X^2）的实例中不难看出，[mean(X)]^2 不等于mean(Y)，因而，我们也就不能用
[CI(mean(X))]^2来作为CI(mean(Y))，两者之间不存在这样的函数关系。
所谓的频率主义就是经验主义，是一种从经验中抽象出理性的哲学思想。这正是统计学
理论和方法的基石。所谓的贝叶斯主义是一种主观先验主义，是一种从先验理性到客观
经验的哲学思想。LZ提出的问题正是基于这样的一种先验的数学理性，而这个理性根本
不存在，因为根据算术均数和可信区间的算法定义，他所期望的等式根本不成立。
【在 s********s 的大作中提到】
: TNEGIETNI和graceman貌似是在讨论2个不同的问题。
: graceman讨论的是: 假设我们要estimate一个distribution的某个参数t, which is a
: constant but unknown from frequentist's point of view.那么我们显然可以用观察
: 到的data来construct confidence interval at level alpha. Let's say这个
: interval for t是 [f(x), g(x)], 这里x是observed data. 那么对于t^2, which is
: also constant but unknown, 它的CI就是[f(x)^2, g(x)^2]. As simple as this.本
: 质上来讲t和t^2是一个parameter.因为它们用的相同的statistics.
: TNEGIETNI讨论的情况是estimate 2个没有直接关系的parameter.或许可以换个例子来
: 考虑。比如说x follows N(0,1), y = x^2, 那么y follo
| s********s
发帖数: 13 | 15 那么假定X服从N(u,1), 且有samples: x1, x2,... xn. 怎么construct u^2的
confidence interval?
【在 T*******I 的大作中提到】
: 我认为你错了。t和t^2是两个不同的随机变量，它们的变异性不一样，因而CI的估计也
: 不一样，且我们没有任何样本基础可以假定[CI(t)]^2 = CI(t^2)。从上面我给出的X和
: Y（Y=X^2）的实例中不难看出，[mean(X)]^2 不等于mean(Y)，因而，我们也就不能用
: [CI(mean(X))]^2来作为CI(mean(Y))，两者之间不存在这样的函数关系。
: 所谓的频率主义就是经验主义，是一种从经验中抽象出理性的哲学思想。这正是统计学
: 理论和方法的基石。所谓的贝叶斯主义是一种主观先验主义，是一种从先验理性到客观
: 经验的哲学思想。LZ提出的问题正是基于这样的一种先验的数学理性，而这个理性根本
: 不存在，因为根据算术均数和可信区间的算法定义，他所期望的等式根本不成立。
| s********s
发帖数: 13 | 16 I think from frequentists' point of view, parameters are unknown numbers
without any randomness, while the Baysian views parameters as random
variables, so they talk about the prior and posterior distribution of
parameters.
所以t和t^2不是随机变量，from frequentists' point of view。
【在 T*******I 的大作中提到】
: 我认为你错了。t和t^2是两个不同的随机变量，它们的变异性不一样，因而CI的估计也
: 不一样，且我们没有任何样本基础可以假定[CI(t)]^2 = CI(t^2)。从上面我给出的X和
: Y（Y=X^2）的实例中不难看出，[mean(X)]^2 不等于mean(Y)，因而，我们也就不能用
: [CI(mean(X))]^2来作为CI(mean(Y))，两者之间不存在这样的函数关系。
: 所谓的频率主义就是经验主义，是一种从经验中抽象出理性的哲学思想。这正是统计学
: 理论和方法的基石。所谓的贝叶斯主义是一种主观先验主义，是一种从先验理性到客观
: 经验的哲学思想。LZ提出的问题正是基于这样的一种先验的数学理性，而这个理性根本
: 不存在，因为根据算术均数和可信区间的算法定义，他所期望的等式根本不成立。
| T*******I
发帖数: 5138 | 17 鉴于Graceman和SeekDreams的very valuable comments， 我想还是请LZ澄清一下他/她
的问题中的X和X^2的含义及其相互关系，即X究竟是一个随机变量，还是关于一个总体
参数（在我看来，一个总体参数也具有随机性，而不是确定性）的期望估计？
其实，无论X是一个随机变量，还是关于一个总体参数的期望估计，在讨论X^2的CI时，
都不能简单地采用数学函数的方式将其定义在CI(X)上。事实上，LZ的问题可以归结为
如下陈述：
如果我在X和X^2之间定义一个函数关系（这里是平方关系），那么，我可否在CI(X)和
CI(X^2)之间定义这个同样的函数关系？
我的回答是不可以，理由参见我在前面的评论。我认为jackdiao的观点和建议不失可取
之处，一个简单的simulation应该可以从统计上证实这个假设是否成立。除此以外，没
有别的办法。在这个问题上，LZ采用的那个数学理性（在我看来那个数学理性在统计学
上是一个非理性）帮不上忙。
【在 n*****g 的大作中提到】
: If X's 95% confidence interval (CI) = (Lx, Hx)
: then can I say the 95% CI of X^2 = (Lx^2, Hx^2)?
: i.e., just taking the square of the upper and lower limits?
: If so, does it apply to all the monotonic functions?
: Thank you!
| n*****n
发帖数: 3123 | 18 没有怎么看前面的讨论，
其实很简单的问题
Pr(lx=0
so (lx^2, ux^2) 是 x^2的95% CI
请注意，一般意义上x是常量参数，这样区间估计才有意义。这是frequentist里才有
interval estimate. lx和ux是关于random variable. | T*******I
发帖数: 5138 | 19 “请注意，一般意义上x是常量参数，这样区间估计才有意义。”
——这是一个误解。
【在 n*****n 的大作中提到】
: 没有怎么看前面的讨论，
: 其实很简单的问题
: Pr(lx=0
: so (lx^2, ux^2) 是 x^2的95% CI
: 请注意，一般意义上x是常量参数，这样区间估计才有意义。这是frequentist里才有
: interval estimate. lx和ux是关于random variable.
| n*****n
发帖数: 3123 | 20 你先去看看interval estimate的书再来吧
【在 T*******I 的大作中提到】
: “请注意，一般意义上x是常量参数，这样区间估计才有意义。”
: ——这是一个误解。
| | | D******n
发帖数: 2836 | 21 Ya, i already pointed this out, CI is hard to deal with when it is not a
regular one, even hard to define it, if we look at how CI is derived for
sample mean of normal population. It is really ad hoc, u cant apply that
,for example, to CI derivation of Binomial. This is all because the fucked
up definition of CI by frequentist' point of view.
On the other hand Bayesian counterpart of CI, creditable region, is pretty
generic, u can apply that methodology to many distributions.
【在 s********s 的大作中提到】
: I think from frequentists' point of view, parameters are unknown numbers
: without any randomness, while the Baysian views parameters as random
: variables, so they talk about the prior and posterior distribution of
: parameters.
: 所以t和t^2不是随机变量，from frequentists' point of view。
| T*******I
发帖数: 5138 | 22 我想我是懂得CI的含义的。但我也有一套概念系统用来说明总体参数的随机性，而用样
本估计的关于总体随机参数的CI依然有意义。说到底，任何样本统计量都是一个随机测
量或随机可变的量，因为样本是随机的。
【在 n*****n 的大作中提到】
: 你先去看看interval estimate的书再来吧
| T*******I
发帖数: 5138 | 23 Bayesian方法之所以容易habdle，是因为假设在你的脑海里，对于你来说可控；而频率
主义或经验主义的方法不在你可假设的能力之下，因而超出了你的可控性。这就是统计
学之所以困难的原因。令人遗憾的是，统计学本来就不是像数学那样从一个公理体系推
到另一个公理体系，而是从一个样本经验事实到一个理性（但非公理性的）结论。这里，
经验事实并非构成一个公理系统。
【在 D******n 的大作中提到】
: Ya, i already pointed this out, CI is hard to deal with when it is not a
: regular one, even hard to define it, if we look at how CI is derived for
: sample mean of normal population. It is really ad hoc, u cant apply that
: ,for example, to CI derivation of Binomial. This is all because the fucked
: up definition of CI by frequentist' point of view.
: On the other hand Bayesian counterpart of CI, creditable region, is pretty
: generic, u can apply that methodology to many distributions.
| n*****g
发帖数: 112 | | T*******I
发帖数: 5138 | 25 我想这个simulation在SAS中用MACRO很容易实现。我就做过类似的simulation， 且至
今还保存着1000多个simualted的随机样本，只需修改一下程序就可以出结果。
【在 j******o 的大作中提到】
: I don't think so. You can try a simulation.
|
|
