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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 $Y$ ?

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

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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, $P(Y\le X_n)\to P(Y\le X)$.

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  • $\begingroup$ Makes sense. Thanks! $\endgroup$
    – dohmatob
    Sep 22, 2020 at 13:22

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