What Does Expected Return Mean in Coefficient of Variation?

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The discussion revolves around understanding the concept of expected return in relation to the coefficient of variation. The coefficient of variation is calculated using the formula: standard deviation divided by expected return. In the given example, the mean value of the distances run by three individuals is considered as the expected return. Participants confirm that the expected return can indeed be interpreted as the mean value in this context. The formula for the coefficient of variation is affirmed as correct.
sotellme
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Please give me some hints on resolving this problem. Thanks!

For example there are 3 persons who run 2, 2 and 2 metres. n=3 and the mean of the metres is 2. Standard deviation is 2 (let us pretend so). The formula for Coefficient of Variation is ; Standard deviation/ Expected return.

I wonder what is Expected return in this case? Is that the mean value? Besides is the formula for Coefficient of Variation right? I have got it from this link http://www.investopedia.com/terms/c/coefficientofvariation.asp


Thanks
 
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sotellme said:
The formula for Coefficient of Variation is ; Standard deviation/ Expected return.
Yes.
sotellme said:
what is Expected return in this case? Is that the mean value?
Yes.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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