# Compute limit of $\mathbb P(Y \le X_n)$ using limiting information on the sequence of random variables $X_n$

Let $$Y$$ be a symmetric random variable, $$(X_n)_n$$ be a sequence of nonnegative random variables, and set $$p_n = \mathbb P(Y \le X_n)$$. It is known from Slutsky's theorem that, if $$c$$ is a constant such that $$X_n \to c$$ in probability, then $$p_n \to F_Y(c)$$, where $$Y$$ is the CDF of $$Y$$.

Question. Can convergence of $$X_n$$ in probability to a constant $$c$$, be replace by some other notion of convergence, say $$X_n \to X$$ (in some sense), for some random variable $$X$$, such that we can still compute the limit of $$p_n$$ only via the data $$X$$ and $$Y$$ ?

• Of course by convergence in distribution, since the limit r,v. here is constant. Sep 22, 2020 at 12:57
• It won't work. $X$ is not a constant. I've updated the text to make this point more explicit. Sep 22, 2020 at 12:59

Even if $$X_n\to c$$ in probability for some real constant $$c$$, it is not necessary that $$P(Y\le X_n)\to P(Y\le c)$$ -- you also need to require that $$P(Y=c)=0$$.
More generally, if the limit of $$X_n$$ is not a constant, then you need to assume the convergence, not just of the distribution of $$X_n$$, but of the joint distribution of $$X_n$$ and $$Y$$. In particular, if $$(X_n,Y)$$ converges to $$(X,Y)$$ in distribution and $$P(Y=X)=0$$, then, by the Portmanteau theorem, $$P(Y\le X_n)\to P(Y\le X)$$.