Let $X_n, X,Y_n,Y$ be real valued random variables. I know that if $X_N\stackrel{(d)}{\rightarrow}X$ and $X_N\stackrel{(d)}{\rightarrow}Y$ then $X$ and $Y$ have the same distribution. Now I asked myself does this imply that $X=Y$ almost surely?
I think that this do not work in general because ...
If $\{ X_n \}_{n\in \mathbb N}$ and $\{ Y_n \}_{n\in \mathbb N}$ are independent and have the same distribution, then you get your counter example.
Well, you could, however the limit would be $X=0$ almost surely and then $Y=0$ almost surely, therefore $X=Y$ almost surely. You want to take a more interesting example that doesn't converge to a constant (try starting by making it random).
@P.Quinton So you mean if I take $F=B(\Bbb{R})$ to be the Borel sigma algebra. Then I can take $X_n:F\rightarrow \Bbb{R}$ by saying that $X_n(A)=n\cdot\lambda(A)$?
Because I would like to take the explicit example instead
@LeanderTilstedKristensen @P.Quinton If I take my $X_n$ as above then I know that it converges in distribution but not almost surely as we discussed above. Can I then immediately say that if I take $Y$ to be independent of $X$ but with the same distribution so that $X_n->Y$ in distribution that then Y!=X a.s. since we don't have convergence a.s.?
Yes, you are correct that this choice of $Y$ will give convergence that X_n -> Y in distribution. To prove that X and Y are not equal almost surely, it is not a sufficient argument, that we do not have almost sure convergence. You can argue for it directly by proving that P(X=Y) != 1. This can be showed using inpendence, for instance we have that P(X< 1/2 < Y) = 1/4, when X and Y are independent and uniformly distributed on (0,1)
Discussion between aprozz and P. Quinton
Discussion between aprozz and P. Quinton