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I learnt in probability theorem class that correlation coefficient is $$ \rho=\frac{\sigma_{XY}}{\sigma_X \sigma_Y}=\frac{E\left[(X-\mu_X)(Y-\mu_Y)\right]}{\sigma_X \sigma_Y} $$


However, my communication professor taught me that the correlation of two random variables is equal to expectation of X times Y. $$ \mbox{Corr}(X, Y)=E(XY)=\int\int_{\mathbb R^2}xy\cdot f(x,y) dxdy $$


"Correlation between two random variables" and "Correlation coefficient of two random variables" are different????

How can I have to understand? Do Both mean that the linear similarity of two variables?

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In general, we have $$\text{Corr}(X,Y) = \frac{E[(X-\mu_X)(Y-\mu_Y)]}{\sigma_X\sigma_Y} = \frac{E[XY]-E[X]E[Y]}{\sigma_X\sigma_Y}.$$ If your instructor or you did not make some kind of mistake, then this suggests that in your course, the random variables will be $X$ and $Y$ such that $$E[X]\times E[Y] = 0$$ and $$\sigma_X\times \sigma_Y = 1$$ so that $$ \text{Corr}(X,Y) = \frac{E[XY]-0}{1} =E(XY)=\int\int_{\mathbb R^2}xy\cdot f(x,y) dxdy.$$ Unless that is actually true, this seems very suspicious.

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  • $\begingroup$ My professor gives us examples but all signals consist of sinusoidal functions. Thank you for good intuitions. $\endgroup$ – Danny_Kim Jan 7 '16 at 6:15
  • $\begingroup$ @Danny_Kim Sure, no problem. You usually post good questions. $\endgroup$ – Em. Jan 7 '16 at 6:20
  • $\begingroup$ Thank you for encouragement. $\endgroup$ – Danny_Kim Jan 7 '16 at 7:17
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The integral you've written is $E(XY)$, but:

  1. $Cov(XY)=E((X-\mu_X)(Y-\mu_Y))=E(XY)-E(X)E(Y)$ is the covariance of $X$ and $Y$
  2. $Corr(XY)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$ is the correlation between $X$ and $Y$.

In general, if it exists, $Cov(X,Y)$ can take any real value. On the other hand, the denominator in the expression for $Corr(X,Y)$ guarantees that correlation always takes a value between $-1$ and $1$.

These both measure how two variables change together - we can think of Correlation as a non-dimensionalised version of covariance.

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  • $\begingroup$ However, my professor used the function $E(XY)$ without minus $E(X)E(Y)$. and he determined the random process is stationary or not by using only the result of $E(XY)$. $\endgroup$ – Danny_Kim Jan 7 '16 at 5:56

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