0
$\begingroup$

Context: Suppose I have two pairs of sequences of random variables $X_n, \tilde{X}_n$ and $Y_n, \tilde{Y}_n$, where $X_n$ and $Y_n$ are not necessarily independent for any $n$, but $\tilde{X}_n$ and $\tilde{Y}_n$ are independent for any $n$.

Moreover I know that dependence of $X_n$ and $Y_n$ is getting weaker when $n \to \infty$ in the following sense:

$$ \text{sup}_{A, B} |P(\{X_n \in A\} \& \{Y_n \in B\}) - P(X_n \in A) P(Y_n \in B)| \to 0, n \to \infty. $$

Question: I have the following convergences for two sets $A$ and $B$:

  1. $|P(X_n \in A) - P(\tilde{X_n} \in A)| \to 0, n \to \infty$;
  2. $|P(Y_n \in B) - P(\tilde{Y_n} \in B)| \to 0, n \to \infty$;

Do this convergences imply the following convergence?

$$ |P \left( \{ X_n \in A \} \& \{Y_n \in B \} \right) - P \left( \{ \tilde{X}_n \in A \} \& \{\tilde{Y}_n \in B \} \right)| \to 0 $$

I was trying to find any method on how to prove it... Any support will be very appreciated.

Improve this question
$\endgroup$
10
  • $\begingroup$ It all depends on how $X_n$ and $Y_n$ depend on each other... $\endgroup$
    – tsnao
    Commented Mar 21 at 13:32
  • $\begingroup$ @tsnao, thank you for the comment, I corrected the question, there was a typo: $X_n$ and $Y_n$ could be dependent, but $\tilde{X}_n$ and $\tilde{Y}_n$ are independent for any $n$. $\endgroup$
    – Grigori
    Commented Mar 21 at 13:35
  • $\begingroup$ This is hopeless without information on the joint distribution of $X_n,Y_n$. For example, suppose that $P(X_n=1)=P(X_n=0)=1/2$, $Y_n=X_n$, and the tilde variables are independent with the same distribution. Then $P(X_n=Y_n=1)=1/2$ while $P(\tilde{X_n}=\tilde{Y_n}=1)=1/4$. $\endgroup$
    – Christian Remling
    Commented Mar 21 at 14:34
  • $\begingroup$ @ChristianRemling, but in this case there won't be convergence of $|P(X_n \in A) - P(\tilde{X}_n \in A)| \to 0$. $\endgroup$
    – Grigori
    Commented Mar 21 at 14:46
  • 3
    $\begingroup$ Now this is a simple fact about sequences of numbers. $\endgroup$
    – Iosif Pinelis
    Commented Mar 21 at 16:53

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .