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it is well-known that if $X_n \to Y$ and $X_n \to X$ in probability, then $X=Y$ almost surely.

Does the same result also hold for the weaker assumption that $X_n \to Y$ and $X_n \to X$ in distribution?

Maybe for a bit more context: I have shown that a sequence of random variables converges almost surely and using characteristic functions I have also calculated the limit in distribution. Now using the aforementioned statement, it would be easy to show that the calculated limit holds for the almost sure convergence.

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  • $\begingroup$ Obviously, if the convergence is in distribution, the limit is not unique... notice that if $X_n\to X$ in distribution, $(X_n)$ and $X$ can leave in different spaces... so obviously, the limit can't be unique $\endgroup$
    – Surb
    Commented Feb 2, 2022 at 17:07
  • $\begingroup$ @Surb Thanks, actually as soon as I posted the question, I realized that it wasn't the smartest :D What I really needed was that X and Y have the same distribution, which can be proven using Portmanteau's theorem. $\endgroup$
    – max_121
    Commented Feb 2, 2022 at 17:10
  • $\begingroup$ When you are dealing with weak covergence, you really are dealing with convergence on the space of measures (as a dual of some topological space) and the random variables attached to this convergence are meaningless. To exemplify, consider the two wildly different experiments of throwing a fair coin or guessing correctly your next offspring (assuming 50-50 chance). These two "experiments" both yield the same distribution Bernoulli with parameter 0.5, yet, the underlying space (i.e. the random variables) are not associated with each other. $\endgroup$
    – William M.
    Commented Feb 2, 2022 at 21:38
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    $\begingroup$ What is true is that if $X_n \to X$ weakly and $X_n \to X'$ weakly, then $X = X'$ in distribution (that is to say, the distributions of both $X$ and $X'$ are the same). $\endgroup$
    – William M.
    Commented Feb 2, 2022 at 21:42

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If your limit $X$ is symmetric then $X_n$ also converges to $-X$ which is of course different from $X$.

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