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Assume we have two sequences $X_1,X_2\ldots$ and $Y_1,Y_2,\ldots$ which are all independent random variables. Assume the partial sums $\sum\limits_{k=1}^N X_k$ and $\sum\limits_{k=1}^N Y_k$ do converge in probability to some RVs $X$ and $Y$. Are then $X$ and $Y$ also independent?

Is their a proof of this that purely relies on the definition of independence of $\sigma$-Algebra's without using characteristic functions ect.

MackeyTopology
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    I would suspect that if $X_k$ converges in probability to $X$ and the $X_k$ are independent, then $X$ is a constant ae. – copper.hat Jun 23 '23 at 18:25
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    @copper.hat To my understanding, the definition of "convergence in probability" is completely unaffected by whether the $X_k$ are independent or dependent. The limit being taken is over the numerical values of probabilities which are computed separately for each $k$. – Greg Martin Jun 23 '23 at 18:34
  • @GregMartin I don't get your point. I am not suggesting that convergence in probability is affected by anything, only that if $X_k$ are independent and converge in probability to $X$ then $X$ is essentially a constant. – copper.hat Jun 23 '23 at 18:45
  • @copper.hat I'm sorry. I've formulated the question wrong. Now the edit should be correct. – MackeyTopology Jun 23 '23 at 18:54
  • @GregMartin I edited the question it was formulated wrong. – MackeyTopology Jun 23 '23 at 19:09
  • It’s sure that they are independent, thank to this answer https://math.stackexchange.com/a/377909/195378 – NN2 Jun 23 '23 at 22:08
  • I don’t know whether there exists or not a proof without using characteristic function. But why don’t you want use characteristic function? – NN2 Jun 23 '23 at 22:09
  • @NN2 I thought of maybe generalizing it to more general spaces and then things could get more difficult. I think with youre comment before where you provided the link you suggest to use levys contunuity theorem. I thought the same thing. However I don't know if this holds also in Hilbert spaces when we use characteristic functionals. Or in more general vector spaces we can't even define what a characteristic functional is yet. – MackeyTopology Jun 24 '23 at 10:30
  • @NN2 See also the discussion here: https://math.stackexchange.com/questions/4724252/levys-continuity-theorem-fail-in-general-hilbert-spaces/4724708#4724708 – MackeyTopology Jun 25 '23 at 12:43

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