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?

  • $\begingroup$ $F_1$ and $F_2$ are joint distribution functions of what variable? $\endgroup$ – quallenjäger Apr 19 '18 at 15:14
  • $\begingroup$ 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 $\endgroup$ – Rubertos Apr 19 '18 at 15:16
  • $\begingroup$ I don’t see an issue here, but then I will edit my question with that terminology $\endgroup$ – Rubertos Apr 19 '18 at 15:23
  • $\begingroup$ Done :) ${}{}{}$ $\endgroup$ – Rubertos Apr 19 '18 at 15:24
  • $\begingroup$ 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 $F_{X_1,Y_1}=F_{X_2}*F_{Y_2}$. $\endgroup$ – 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. $F_{X_1,Y_1}.$

Consider these probability mass functions \begin{align} & \begin{array}{ll} f_{X_1,Y_1}(0,0)= 1/4, & f_{X_1,Y_1}(0,1) = 1/4 \\ f_{X_1,Y_1}(1,0) = 1/4, & f_{X_1,Y_1}(1,1) = 1/4 \end{array} \\[12pt] \hline & \begin{array}{ll} f_{X_2,Y_2}(0,0)= 1/2, & f_{X_2,Y_2}(0,1) = 0 \\ f_{X_2,Y_2}(1,0) = 0, & f_{X_2,Y_2}(1,1) = 1/2 \end{array} \end{align}

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.

  • $\begingroup$ Why is $f_{X_1}=f_{X_2}$? How do I deduce that? What are $X_1,X_2,Y_1,Y_2$ here? $\endgroup$ – Rubertos Apr 19 '18 at 15:52
  • $\begingroup$ @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 $f_{X_2}. \qquad$ $\endgroup$ – Michael Hardy Apr 19 '18 at 15:56

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