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.

  • $\begingroup$ It is also true if $X_n,Y_n$ are uniformly integrable. Can you reduce to this case by a truncation argument? $\endgroup$ – 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.


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