Suppose that I have $\{Y_1^N\}_N$ and $\{Y_2^N\}_N$ two sequences of random variables defined in the same probability space $(\Omega, F, P)$. Suppose that $P(Y_1^N\leq Y_2^N)=1$ and that $Y_1^N\xrightarrow[N\to +\infty]{D}Y_1^*$ and $Y_2^N\xrightarrow[N\to +\infty]{D}Y_2^*$ and that the last two limits are in distribution. Can I conclude that
$$P(Y_1^*\leq Y_2^*)=1$$
Thank you very much
The problem comes from the fact that the limit is only in distribution. We could imagine the case where $Y_1^N=Y_2^N=Y_1^*=U$, where $U$ has a uniform distribution in the unit interval and $Y_2^*=1-U$. Then it is true that $Y_1^N\leqslant Y_2^N$ almost surely, $Y_1^N\to Y_1^*$ and $Y_2^N\to Y_2^*$ in distribution but we do not have that $Y_1^*\leqslant Y_2^*$ with probability one (only with probability $1/2$).
However, if the sequence of vectors $\left(\left(Y_1^N,Y_2^N\right)\right)_{N\geqslant 1}$ converges to the vector $\left(Y_1^*,Y_2^*\right)$, then it is true that $Y_1^*\leqslant Y_2^*$ almost surely. This follows from an application of portmanteau theorem to the closed set $C:=\{(x_1,x_2)\in\mathbb R^2\mid x_1\leqslant x_2\}$.
This is what failed in the previous example: it is true that $\left(\left(Y_1^N,Y_2^N\right)\right)_{N\geqslant 1}$ converges to $(U,U)$ but not to $(U,1-U)$.