Is it correct that $\text{Cov}(X,Y) = \text{E}((X-\text{E}(X))Y)$? We have that $\text{Cov}(X,Y) = \text{E}\left( (X-\mu_x)(Y-\mu_y)\right) = \text{E}\left(XY -\mu_yX - \mu_xY + \mu_x\mu_y  \right) = E(XY) - \mu_y\mu_x - \mu_x\mu_y + \mu_x\mu_y = E(XY) - \mu_x\mu_y$. 
We also have that $\text{E}\left( (X-\mu_x)Y\right) = \text{E}\left(XY - \mu_xY \right) = \text{E}(XY) - \mu_x\text{E}(Y) = \text{E}(XY) -\mu_x\mu_y = \text{Cov}(X,Y)$
I have never seen $\text{E}\left( (X-\mu_x)Y\right)$ being used to compute or define the covariance of $X$ and $Y$, but it is correct, or? 
 A: Interesting; I wasn't aware of this fact, but I don't see any errors in your derivation. (it's not a concept I've ever dealt much with)
At first glance, it looks like the usual definition has pedagogical advantages, making obvious two key properties:


*

*Covariance is symmetric: Cov(X,Y) = Cov(Y,X)

*Covariance depends only on how much the random variables differ from their means


As for computation, I think the alternative you suggest only has niche uses: I imagine that for nearly every purpose, at least one of $E((X - \mu_x)(Y - \mu_y))$ or $E(XY) - \mu_x \mu_y$ is more computationally convenient than $E((X - \mu_x) Y)$: the former due to smaller numbers and better numerical stability, and the latter due to ease of tabulation and simpler formulas.
A: It is correct and it is fairly frequently seen in statistics textbooks.  In the case of a finite sample, one sees it written as
$$
\frac1n \sum_{i=1}^n (x_i-\bar x)y_i = \frac1n \sum_{i=1}^n x_i (y_i-\bar y) = \frac1n \sum_{i=1}^n (x_i-\bar x)(y_i - \bar y).
$$
Or sometimes with Bessel's correction, so that $n-1$ appears instead of $n$.  It can be somewhat more computationally efficient to use one of the first two forms above than the third one.
