# Limit of convergence in distribution is unique almost surely

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.

• 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
– Surb
Commented Feb 2, 2022 at 17:07
• @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. Commented Feb 2, 2022 at 17:10
• 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. Commented Feb 2, 2022 at 21:38
• 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). Commented Feb 2, 2022 at 21:42

## 1 Answer

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