Suppose $\binom{X}{Y}$ follows a bivariate density $f=0.5f_{1}+0.5f_{2}$ where f$_{1}$ and $f_{2}$ are densities of $N(\mu_1, \Sigma)$ and $N(\mu_2, \Sigma)$. $\mu_1 = \binom{1}{1} , \mu_2= \binom{-1}{-1}$ and $\Sigma= I_2$. How do I check if X and Y are correlated?
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Normality of the variables is irrelevant, because the correlation coefficient depends only on the second central moments, which can explicitly be computed. (See https://stats.stackexchange.com/a/16609/919 for general formulas.) Even if by "correlated" you mean "not independent," computing the correlation coefficient is useful because a nonzero value implies lack of independence. By sketching a contour plot of the density (again, regardless of Normality) you will be led to expect non-independence. See the "spherical" plot at https://stats.stackexchange.com/a/326678/919 for an example. – whuber Nov 03 '19 at 13:47
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They appear to be correlated because if you know one of them, e.g. $X=1$, then the other will likely be in the same (first) quadrant. This is the intuition. Mathematically, the steps are simple. Since the means are $0$, we just need $E[XY]$: $$E[XY]=0.5\int xy f_1dxdy+0.5\int xy f_2dxdy$$ These two integrals represent $E[XY]$ when $X$ and $Y$ are distributed according to $f_1$ and $f_2$ respectively (call these situations as $Z=0$ and $Z=1$). $$\int xy f_1dxdy=E[XY|Z=0]=\underbrace{\operatorname{cov}(X,Y|Z=0)}_0+\underbrace{E[X|Z=0]}_1\underbrace{E[Y|Z=0]}_1=1$$ Similarly, $$\int xy f_2dxdy=\underbrace{E[X|Z=0]}_{-1}\underbrace{E[Y|Z=0]}_{-1}=1$$ which yields $E[XY]=2\neq 0$, so they're correlated.

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Thank. Got the mathematical part. But didn't get the intuition that X and Y will likely be in the same quadrant. – Harry Nov 03 '19 at 08:41
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Because the probability masses are around $(1,1)$ and $(-1,-1)$. Not around $(1,-1)$ or $(-1,1)$. Try to think visually (as peaks). – gunes Nov 03 '19 at 10:08