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?

  • $\begingroup$ Could you link the paper? $\endgroup$ Aug 5 at 19:56
  • $\begingroup$ "Uncorrelatedness implying independence: Although uncorrelatedness usually does not imply independence, it is well known that it does for two binary variables." $\endgroup$
    – Jonathen
    Aug 5 at 19:59
  • $\begingroup$ The paper is Zhang, K. (2019). BET on independence $\endgroup$
    – Jonathen
    Aug 5 at 19:59
  • $\begingroup$ It’s also true for normally distributed random variables $\endgroup$
    – Bey
    Aug 10 at 13:26

1 Answer 1


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:

$$X_i \perp Y_j\;\;\forall i,j \in \{0,1\} \implies X \perp Y$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.