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Math Help - How to get the covariance of x and |x|----cov(x,|x|)?

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    How to get the covariance of x and |x|----cov(x,|x|)?

    How to get the covariance of x and |x|----cov(x,|x|)? under the condition of continuous X or discrete X, thanks!
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    Hello,

    cov(X,|X|)=E[X*|X|]-E[X]E[|X|]...

    But then it can be simplified or not, depending on what your exercise really is. Please state the context...
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    thanks for your quick reply, The context is that the E(X) is the given, but E(|x|) is unknown, when X, as a random variable, can be located in both the positive interval[0,inf) and negative interval(-inf,0), what's the results of Cov(X,|x|)? I encounter a problem that is X is a random variable, so it can jump between positive and negative case, so whether we can simply consider the covariance in the two case? if not, what's the result? Thanks!
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    thanks for your quick reply, The context is that the E(X) has been given, but E(|x|) is unknown. When X, as a random variable, can be located in both the positive interval[0,inf) and negative interval(-inf,0), what's the results of Cov(X,|x|)? I encounter a problem that is X is a random variable, so it can jump between positive and negative case, so whether we can simply consider the covariance in positive case x>0 and negative case x<0 separately? if not, what's the result? Thanks!
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    Hello,

    You can't consider it if X>0 or not. You have to condition by it.

    Okay, let's set sgn(X)=-1 if X<0 and sgn(X)=1 if X\geq 0. In other words sgn(X)=\bold{1}_{X\geq 0}-\bold{1}_{X<0} (it's a random variable following a simple discrete distribution, let's say p is the probability that it equals 1)

    Then cov(|X|,X)=E[sgn(X)X^2]-E[sgn(X)X]E[X]

    Now let's consider a conditional covariance. (sorry, I love using conditional expectations )

    cov(|X|,X \mid sgn(X))=E[sgn(X)X^2 \mid sgn(X)] - E[sgn(X)X \mid sgn(X)]E[X \mid sgn(X)].
    Since sgn(X) is indeed sgn(X)-measurable, we can pull it out from the expectations :
    cov(|X|,X \mid sgn(X))= sgn(X) E[X^2 \mid  sgn(X)]-sgn(X) (E[X \mid sgn(X)])^2 = sgn(X) Var(X \mid  sgn(X))
    So cov(|X|,X)=E[sgn(X) Var[X \mid sgn(X)]], which is an expectation with respect to sgn(X)'s distribution (meaning that sgn(X) is the sole remaining random variable in the expectation).
    So this gives pVar(X)-(1-p)Var(X)=(2p-1)Var(X), where p is, as said before, P(sgn(X)=1)=P(X\geq 0)

    That's the furthest one can go with the provided information. And it doesn't matter whether X is discrete or continuous.
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    thank you so much for your very smart and creative answer, but I still have one more question, which is that VAR[X|sgn(x)=-1]=VAR(X)? and also VAR[X|sgn(x)=1]=VAR(X)?
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    Quote Originally Posted by lyb66 View Post
    thank you so much for your very smart and creative answer, but I still have one more question, which is that VAR[X|sgn(x)=-1]=VAR(X)? and also VAR[X|sgn(x)=1]=VAR(X)?
    Yes it's equal. So we can say that Var[X|sgn(X)]=Var[X], which lets us simplify :
    E[sgn(X) Var[X|sgn(X)]=E[sgn(X) Var[X]]=Var[X] E[sgn(X)] , since Var[X] is a constant !

    If you understand my method, I have some doubts about your thread belonging to the pre-university subforum
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    acctually, http://latex.codecogs.com/png.latex?Var[X|sgn(X)]=Var[X] this equation is not hold when I use integration to solve cov(x,|x|), the two final results are not consistent...please check it again to avoid future mistake...
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