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What is square absolute value of covariance $X$ and $Y$? $\operatorname{cov}(X,Y)=E[XY]-E[X]E[Y]$ Why we put $\operatorname{cov}(X,Y)$ inside absolute value? $|\operatorname{cov}(X,Y)|^2=|E[XY]-E[X]E[Y]|^2=$?

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Welcome to Math.SE! In its present form your question does not contain enough information for us to fully understand what is going on. Please edit the question and include an explanation of where you saw the covariance appearing inside of absolute values. – user53153 Dec 29 '12 at 7:21

Are you looking for something like this?

Let $$\varrho(X,Y) := \frac{\text{cov}(X,Y)}{\sqrt{\text{var} X} \cdot \sqrt{\text{var} Y}}$$

the correlation coefficient. Then one can easily show (using Cauchy-Schwarz inequality) that $\varrho(X,Y) \in [-1,1]$. Thus

$$1 \geq |\varrho(X,Y)|^2 = \frac{\text{cov}(X,Y)^2}{\text{var} X \cdot \text{var} Y}$$

i.e. $$\text{cov}(X,Y)^2 \leq \text{var} X \cdot \text{var} Y$$ where $\text{var} X := \mathbb{E}((X-\mathbb{E}X)^2)$ denotes the variation of $X$.

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