首页 | 学习中心 | 注册 登录
返回列表 发帖
EastCoastJ

Rank: 2

UID
223223 
帖子
183 
主题
163 
注册时间
2011-7-11 
最后登录
2013-8-21 
1# 跳转到 » 倒序看帖
打印
tT
发表于 2011-7-26 01:20 | 只看该作者

Mean Absolute Deviation and Standard Deviation

positive, absolute, standard, average, median
For a normal distribution , if the mean and the median values are zero , and the average of positive values and the absolute value of average of negative values is 1% , what is the standard deviation?



Edited 1 time(s). Last edit at Wednesday, June 15, 2011 at 03:38PM by janakisri.
收藏 分享

d2rockstar

Rank: 2

UID
223341 
帖子
249 
主题
31 
注册时间
2011-7-11 
最后登录
2013-8-23 
2#
发表于 2011-7-26 01:25 | 只看该作者
This is only true for E(X) = 0. The formula for E(x) = m is much more complicated.

x ~ normal(0, var(x))
1% = E(|x|) = ((2*var(x))/pi)^(1/2)

so

stdev(x) = (pi/2)^.5 = 1.2533...%

TOP

bchadwick

UID
222250 
帖子
515 
主题
123 
注册时间
2011-7-2 
最后登录
2016-4-19 
3#
发表于 2011-7-26 01:30 | 只看该作者
Thanks MathMan , you're really good

TOP

WarrenB1

Rank: 2

UID
223415 
帖子
346 
主题
18 
注册时间
2011-7-11 
最后登录
2014-7-31 
4#
发表于 2011-7-26 01:35 | 只看该作者
What? Really?
So for the set {1, -1, 1, -1, 1, -1 ... } the standard deviation is 1.2533?

TOP

needhelp1700

Rank: 2

UID
223424 
帖子
319 
主题
22 
注册时间
2011-7-11 
最后登录
2014-8-7 
5#
发表于 2011-7-26 01:41 | 只看该作者
Is {1, -1, 1, -1, 1, -1 ... } normally distributed?

seemorr Wrote:
-------------------------------------------------------
> What? Really?
> So for the set {1, -1, 1, -1, 1, -1 ... } the
> standard deviation is 1.2533?

TOP

joehogue

Rank: 2

UID
222272 
帖子
301 
主题
72 
注册时间
2011-7-2 
最后登录
2014-6-28 
6#
发表于 2011-7-26 01:46 | 只看该作者
MathMan is implying that it is not !

It seems to be alternating 1s and -1s ( kind of like a barbell )

TOP

anish

Rank: 2

UID
223378 
帖子
231 
主题
35 
注册时间
2011-7-11 
最后登录
2014-8-6 
7#
发表于 2011-7-26 01:51 | 只看该作者
MathMan's derivation is correct. If you use

E(|x|)=(2/(sqrt(2pi)*sigma)*intergral from 0 to infinity of x*exp(-x^2/2sigma^2)dx, then change variable (for example, y = x^2/2sigma^2), you will get his formula.

Since the question was about normal distribution with zero mean, the answer is correct. If distributional assumption is different, the answer would be different as well. If you'd like to come up with a range of values for a wide range of symmetric distributions, you can try to minimize and maximize sqrt((1/n)*sum(x_k^2)) given 1/n*(sum(x_k^2))=1% and all x_k>=0. That might a fun problem to think about.

TOP

chandsingh

Rank: 2

UID
223451 
帖子
355 
主题
11 
注册时间
2011-7-11 
最后登录
2014-8-7 
8#
发表于 2011-7-26 01:57 | 只看该作者
Oh I guess {1, -1, 1, -1 ...} isn't normally distributed ...



Edited 1 time(s). Last edit at Tuesday, June 21, 2011 at 04:37PM by seemorr.

TOP

transferpricing

Rank: 2

UID
222278 
帖子
295 
主题
68 
注册时间
2011-7-2 
最后登录
2016-4-19 
9#
发表于 2011-7-26 02:02 | 只看该作者
Not true that "{1, -1, 1, -1, 1, -1 ...} is clearly normally distributed"

If your pattern contains only 1's and -1s , how can it be normally distributed? Every normal distribution chart I've seen is shaped like a bell ( hence bell shaped curve).

The pattern of alternating 1's and -1s results in a frequency diagram with two vertical , equal height lines , one at +1 and one at -1

TOP

Kapie

UID
223384 
帖子
638 
主题
62 
注册时间
2011-7-11 
最后登录
2014-8-6 
10#
发表于 2011-7-26 02:07 | 只看该作者
Variance bound is [E(|x|)^2 – E(x)^2, inf).

Regarding the variance bounds for this type of problem, the following is true regardless of symmetry.

Var(|x|) = E(|x|^2) – E(|x|)^2 = E(x^2) – E(|x|)^2
Var(x) = E(x^2) – E(x)^2

So, Var(|x|) + E(|x|)^2 = Var(x) + E(x)^2

In this case, E(|x|) = 1 and E(x) = 0. So,

Var(|x|) + 1 = Var(x)

Since we know Var(|x|) >= 0 then the minimum value of Var(x) is 0 + 1 = 1. Seemorr already produced a distribution where Var(|x|) = 0 is possible. (Generally the minimum is E(|x|)^2 – E(x)^2, note that E(|x|)^2 >= E(x)^2)

So what is the max….well, assume a case where the distribution of |x| is lognormal. It should be clear that when we set E(|x|) = 1 (or any constant), Var(|x|) is unbounded above.

This is far from mathematical proof but it should also be obvious from above that combinations of seemorr’s distribution and a lognormal distribution for |x| should be able to produce a set where E(|x|) = 1, E(x) = 0, and Var(x) is equal to whatever we want.

TOP

返回列表