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I was playing around with non-central second moments, and noticed that $E[x(y-\mu_y)] = E[(x-\mu_x + \mu_x)(y-\mu_y)] = COV[x,y] + E[\mu_x(y-\mu_y) = COV[x,y] + 0$.

I find this very surprising. It appears that as long as either variable is a central moment, then you can move the other variable arbitrarily without changing the value of the cross-moment.

I was trying to intuit this geometrically by making MATLAB surface plots of the integrand $x(y-\mu_y)$ with some correlated multivariate normal pdf to understand why shifting x arbitrarily doesn't change the integral, and it... seems very puzzling to me.

Any thoughts?

Thanks.

user49404
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  • This result follows immediately from my geometric characterization of covariance at https://stats.stackexchange.com/a/18200/919. Indeed, there's nothing special about the fact you have subtracted the mean of $y$ from $y:$ the covariance remains the same when you shift $y$ by *any* constant amount. – whuber Sep 07 '19 at 18:03
  • @whuber I'm still trying to parse your interesting linked answer, but I don't think its true the covariance is constant if you shift y by any constant amount. In general, $E[x(y-c)] = E[(x-\mu_x+\mu_x)(y-\mu_y+\mu_y-c)] = COV[x,y] + 0 + 0 + \mu_x (\mu_y - c)$. It was only because above I have $c=\mu_y$ that translating x doesn't change things. – user49404 Sep 07 '19 at 18:29
  • You seem to have a different idea of what covariance is than the standard one: your formulas fail to subtract the product of expectations. – whuber Sep 08 '19 at 14:15
  • @whuber $COV[x,y] = E[(x-\mu_x)(y-\mu_y)] = E[XY]-E[X]E[Y]$. So we can use either characterization. In my comment above, to make this more explicit, I have $E[(x-\mu_x+\mu_x)(y-\mu_y+\mu_y-c)]$ = $E[(x-\mu_x)(y-\mu_y) + \mu_x (y-\mu_y) + (x-\mu_x)(\mu_y-c) + \mu_x (\mu_y-c)]$ = $COV[x,y] + 0 + 0 + \mu_x (\mu_y-c).$ – user49404 Sep 08 '19 at 16:08
  • That's correct, but not relevant: you are asking about the *covariance* rather than the expectation of the product, so the relevant calculation subtracts the product of the means. In your example the means of $X$ and $Y-c$ are $\mu_x$ and $\mu_y-c,$ respectively; and when you subtract their product from your result you find that the covariance of $X$ and $Y-c$ equals the covariance of $X$ and $Y.$ – whuber Sep 09 '19 at 13:27

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