I am reading a statistics paper. It said generally uncorrelatedness does not imply independence. But if the case is two binary variables, then the uncorrelatedness can imply independence. How to show this statement?
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$\begingroup$ Could you link the paper? $\endgroup$– Samuel Adrian AntzAug 5 at 19:56
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$\begingroup$ "Uncorrelatedness implying independence: Although uncorrelatedness usually does not imply independence, it is well known that it does for two binary variables." $\endgroup$– JonathenAug 5 at 19:59
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$\begingroup$ The paper is Zhang, K. (2019). BET on independence $\endgroup$– JonathenAug 5 at 19:59
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$\begingroup$ It’s also true for normally distributed random variables $\endgroup$– BeyAug 10 at 13:26
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1 Answer
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Let $X,Y$ be two binary random variables with $P(X=1)=p_X,P(Y=1)=p_Y$.
Note that zero correlation implies zero covariance and we get:
$$Cov[X,Y] = E[XY]-E[X]E[Y] = 0 \implies E[XY]=E[X]E[Y]=p_Xp_Y$$
For brevity let's define the events $X_i:= \{X=i\}$ and $Y_i:=\{Y=i\}$
Define $Z=XY \in \{0,1\}$
$$E[XY]= P(Z=1) = P(X_1,Y_1)=p_Xp_Y \implies X_1 \perp Y_1$$
In general, $A\perp B \implies A^c \perp B, \;A \perp B^c, \;A^c \perp B^c$ (see here) therefore,
$$X_1 \perp Y_1 \implies X_1^c \perp Y_1, \;X_1 \perp Y_1^c, \;X_1^c \perp Y_1^c$$
This happens to exhaust the sample space of the joint experiment $(X,Y)$ so we conclude:
$$X_i \perp Y_j\;\;\forall i,j \in \{0,1\} \implies X \perp Y$$
