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What is the intuition behind this expected covariance formula? Why we do not use the first one (first line) and we use the last one. E[X] and E[Y] are means and easy to find. Why we derive the last equation. I do not get the idea.

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Mas A
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    In many cases, finding the mean $XY$ is easier than finding the mean of $(X-\mathbb EX)(Y-\mathbb EY)$. E.g. if $X$ and $Y$ are iid Bernoulli. – jlammy Dec 24 '21 at 20:29
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    You can use both of them. If you find the first one easier, then you can use that. I think in the first line you can find much easier the intuition behind the definition of covariance, but the last line is much easier to calculate. (But it is just mine opinion.) – Kapes Mate Dec 24 '21 at 20:30

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This is not about "expected" covariance, but simply about covariance.

In some contexts it is a very bad idea to use this formula. For example, suppose \begin{align} & \operatorname E(X) = 1\,000\,000, \\[2pt] & \operatorname E(Y) = 2\,000\,000, \\[2pt] & \operatorname{sd}(X) = 0.03, \\[2pt] & \operatorname{sd}(Y) = 0.02, \\[2pt] & \operatorname{corr}(X,Y) = 0.9. \end{align} Then we have $\operatorname{cov}(X,Y) = 0.03\times0.02\times0.9 = 0.00054$ and so $$\operatorname E(XY) = \operatorname{cov}(X,Y) + \operatorname E(X)\operatorname E(Y) = 0.00054 + 2\,000\,000\,000\,000.00054. $$ So what happens when you try to use this formula then? Watch: \begin{align} \operatorname{cov}(X,Y) & = \operatorname E(XY) - \operatorname E(X) \operatorname E(Y) \\[4pt] & = \underbrace{2\,000\,000\,000\,000}_\text{rounded} {} - 2\,000\,000\,000\,000 \\[15pt] & = 0. \quad \text{So all of the desired information was lost in rounding.} \end{align}

But if you have something like $\operatorname E(X)=2$ and $\operatorname{sd}(X)=3$ and $\operatorname E(Y)=4$ and $\operatorname{sd}(Y)=8,$ then sometimes doing the arithmetic is a bit quicker with this formula, so it gets called a shortcut.

Michael Hardy
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