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Let X and Y be two random variables and let (x1,y1),…,(xn,yn) be an i.i.d. sample of n observations from the joint distribution of X and Y. You know that the sample average of $\bar X_n$ is 4, each X and Y have a sample variance equal to 2 and that the sample correlation coefficient of X and Y is -0.25. Compute the following quantities.

$$ \frac1n\sum_1^n (x_i-\bar X_n) $$

I tried to solve this problem but I'm stuck. I started by splitting up the problem so that I had $$ \frac1n(\sum_1^nx_i - \sum_1^n\bar X_n) $$ I know that $\frac1n \sum_1^nx_i$ is $\bar X$. So, now I have $$ \bar X - \frac1n \sum_1^n\bar X_n $$

I just want to know if $\bar X$ is the same as $\bar X_n$. I just don't know where to go from this point. I could really use a hint! Thank you!

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    Are you making some unstated distinction between $x_i$ and $X_i$? Not just being fussy, some authors do make a distinction. Similarly, is $\bar X$ different than $\bar X_n$? If you want $\sum(X_i - \bar X),$ then why mention the $y_i$s? $\sum(X_i - \bar X) = 0.$ – 2017-02-16
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    @BruceET I don't really know. The information above is all that I was given. But I don't think there is a distinction in both. – 2017-02-16
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    $\sum X_i = n\bar X$ and $\sum \bar X = \bar X + \bar X + \cdots + \bar X = n\bar X.$ (Sum with $\cdots$ has $n$ identical terms.) – 2017-02-16
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    Maybe we need some clarification. If $x_i$'s are *observed* values and $\bar X_n$ is a random variable then it does not make sense to "compute" the first expression (it is a r.v., not a quantity). – 2017-02-16

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