# Example that two distributions are both dependent and independent

Let $X_1,X_2$ be random vectors such that $F_{X_1}=F_{X_2}$. ($F$ denotes CDFs)

Let $Y_1,Y_2$ be random vectors such thar $F_{Y_1}=F_{Y_2}$.

If $X_1,Y_1$ are independent, then are $X_2,Y_2$ necessarily independent? I guess this is obviously false, but since I am new to probability theory, I am not sure how to construct such one. What would be a counterexample?

• $F_1$ and $F_2$ are joint distribution functions of what variable? – quallenjäger Apr 19 '18 at 15:14
• Actually, it need not be specified, because joint distributions can be completely characterized without invoking random vectors. But, here, let’s say $F_1=F_{X_1}$ and $F_2=F_{X_2}$, as given in my post – Rubertos Apr 19 '18 at 15:16
• I don’t see an issue here, but then I will edit my question with that terminology – Rubertos Apr 19 '18 at 15:23
• Done :) ${}{}{}$ – Rubertos Apr 19 '18 at 15:24
• Sorry, it was my fault. I didn't understand your question first. Now I see it makes sense. I agree with you. You can somehow construct r.v.s such that $F_{X_2}*F_{Y_2} \neq F_{X_2,Y_2}$. Then you can construct your $X_1$ and $Y_1$ with $F_{X_1}=F_{X_2}$ and $F_{Y_1}=F_{Y_2}$according to the distribution $F_{X_1,Y_1}=F_{X_2}*F_{Y_2}$. – quallenjäger Apr 19 '18 at 15:28

Distributions are not dependent or independent; random variables are. Dependence or independence is determined by the joint distribution, which his determined by the joint c.d.f. $F_{X_1,Y_1}.$
It follows that $f_{X_1} = f_{X_2}$ and $f_{Y_1} = f_{Y_2},$ but $X_1$ and $Y_1$ are independent but $X_2$ and $Y_2$ are not.
• Why is $f_{X_1}=f_{X_2}$? How do I deduce that? What are $X_1,X_2,Y_1,Y_2$ here? – Rubertos Apr 19 '18 at 15:52
• @Rubertos : $$f_{X_1}(0) = \Pr(X_1=0) = \Pr( (X_1,Y_1)=(0,0) \text{ or } (X_1,Y_1) = (0,1)) = \frac 1 4 + \frac1 4 = \frac 1 2$$ and $f_{X_1}(1)$ is found similarly, and also $f_{X_2}. \qquad$ – Michael Hardy Apr 19 '18 at 15:56