如果和是两个只能具有两个可能状态的随机变量,我如何证明表示独立性?这种违背了我回想起并不意味着独立...
提示说从和开始作为可能的状态,并从那里开始进行概括。我可以这样做并显示,但这并不意味着独立?
我猜这有点困惑如何数学上做到这一点。
如果和是两个只能具有两个可能状态的随机变量,我如何证明表示独立性?这种违背了我回想起并不意味着独立...
提示说从和开始作为可能的状态,并从那里开始进行概括。我可以这样做并显示,但这并不意味着独立?
我猜这有点困惑如何数学上做到这一点。
Answers:
对于二进制变量,其期望值等于它们等于1的概率。因此,
如果两个协方差为零,则意味着,这意味着
使用关于独立事件的基本规则(即,如果和是独立的,则它们的补码是独立的,等等),看到所有其他联合概率也相乘是很简单的,这意味着联合质量函数是因式分解的两个随机变量是独立的。
相关性和协方差都可以测量两个给定变量之间的线性关联,并且没有义务检测其他任何形式的关联。
因此,这两个变量可能以其他几种非线性方式关联,并且协方差(因此,相关性)无法与独立情况区分开。
作为一个非常说教,人工和非现实的例子,可以考虑,使得P (X = X )= 1 / 3为X = - 1 ,0 ,1,并且还考虑ÿ = X 2。请注意,它们不仅是关联的,而且是另一个的功能。但是,它们的协方差为0,因为它们的关联与协方差可以检测到的关联正交。
编辑
确实,正如@whuber所指出的那样,以上原始答案实际上是关于如果两个变量都不一定是二分法的话该断言不是普遍正确的评论。我的错!
因此,让我们数学起来。(与Barney Stinson的“ Suit up!”相当)
如果两个和ÿ是二分,则可以假定,不失一般性,这两个假设只的值0和1具有任意的概率p,q和- [R由下式给出 P (X = 1 )= p ∈ [ 0 ,1 ] P (Ý = 1 )= q ∈ [ 0 ,1 ] P (X = 1 ,ÿ 表征完全的联合分布X和ÿ。接受@DilipSarwate的提示,请注意,这三个值足以确定(X,Y)的联合分布,因为 P (X = 0 ,Y = 1 )
Notice that might be equal to the product , which would render and independent, since
Yes, might be equal to , BUT it can be different, as long as it respects the boundaries above.
Well, from the above joint distribution, we would have
Now, notice then that and are independent if and only if . Indeed, if and are independent, then , which is to say . Therefore, ; and, on the other hand, if , then , which is to say . Therefore, and are independent.
About the without loss of generality clause above, if and were distributed otherwise, let's say, for and ,
Also, we would have
=D
IN GENERAL:
The criterion for independence is . Or
This is nicely explained by Macro here, and in the Wikipedia entry for independence.
, yet
Great example: , and Covariance is zero (and , which is the criterion for orthogonality), yet they are dependent. Credit goes to this post.
IN PARTICULAR (OP problem):
These are Bernoulli rv's, and with probability of success , and .
This is equivalent to the condition for independence in Eq.
:
: by LOTUS.
As pointed out below, the argument is incomplete without what Dilip Sarwate had pointed out in his comments shortly after the OP appeared. After searching around, I found this proof of the missing part here:
If events and are independent, then events and are independent, and events and are also independent.
Proof By definition,
and are independent
But , so , which yields:
Repeat the argument for the events and this time starting from the statement that and are independent and taking the complement of
Similarly. and are independent events.
So, we have shown already that