# Random variables $X_n\leq Y_n$ with common weak limit $L$

Let $$X_n,Y_n$$ be sequences of random variables defined on some common probability space. Suppose $$X_n\leq Y_n$$ (a.s.) and $$X_n\Rightarrow L$$, $$Y_n\Rightarrow L$$ as $$n\rightarrow\infty$$, for some distribution $$L$$. Is it true that $$Y_n-X_n\Rightarrow 0$$?

If we have joint convergence $$(X_n,Y_n)\Rightarrow (L_1,L_2)$$ with each $$L_i$$ having marginal distribution $$L$$, the claim follows from an application of the Skorohod representation theorem. I'm not sure about the claim without this condition.

• It is also true if $X_n,Y_n$ are uniformly integrable. Can you reduce to this case by a truncation argument? – pre-kidney Sep 2 '19 at 20:59

Assume that for some $$\epsilon,\eta> 0$$, $$P(Y_n-X_n > \epsilon) > \eta$$ hold for infinitely many $$n$$.
Since $$X_n$$ and $$Y_n$$ converge in distribution, there are $$\epsilon >0,\eta > 0, M> 0$$ such that $$P(|Y_n| < M,|X_n| < M, Y_n-X_n >\epsilon) > \eta$$ holds for infinitely many $$n$$ (the complement of the event above is $$A_n$$).
Consider $$f$$ piecewise affine, $$f(x)= x$$ for $$|x| \leq M$$, $$f(x \leq -M)=f(-M)$$, $$f(x \geq M)=f(M)$$, $$\alpha > 0$$.
Now, $$f(Y_n)-f(X_n)$$ is a nonnegative random variable, with expected value $$o(1)$$, and is greater than $$\epsilon$$ with probability $$\eta$$: a contradiction.