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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- May 31st 2011, 03:07 PMlyb66How 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!

- May 31st 2011, 04:14 PMMoo
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... - May 31st 2011, 05:40 PMlyb66
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!

- May 31st 2011, 05:50 PMlyb66
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!

- June 1st 2011, 03:28 AMMoo
Hello,

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

Okay, let's set if and if . In other words (it's a random variable following a simple discrete distribution, let's say p is the probability that it equals 1)

Then

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

.

Since sgn(X) is indeed sgn(X)-measurable, we can pull it out from the expectations :

So , 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 , where p is, as said before,

That's the furthest one can go with the provided information. And it doesn't matter whether X is discrete or continuous. - June 5th 2011, 09:53 AMlyb66
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)?

- June 12th 2011, 01:36 AMMoo
- June 13th 2011, 10:40 AMlyb66
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...