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Consider two continuous functions $f,g: \mathbb{R}^{2} \rightarrow \mathbb{R}$ with $f(x,\cdot), g(x,\cdot) \in \mathcal{L}^2(\mathbb{R},\mathcal{B},\lambda)$ for all $x \in \mathbb{R}$ and a sequence of real random variables $(X_n)_{n \in \mathbb{N}}$ and another real random variable $X$. Then $T_n :=f(X_n,\cdot)$ and $V :=g(X,\cdot)$ can be interpreted as $\mathcal{L}^2$-valued random variables. Denote the distribution of $T_n$ by $\mathbb{P}_n$ and the distribution of $V$ by $\mathbb{P}$ (then $\mathbb{P}_n$ and $\mathbb{P}$ are defined on the Borel-$\sigma$-field on $\mathcal{L}^2$).

Now if we know that $(\mathbb{P}_n)_{n \in \mathbb{N}}$ is tight and that $(f(X_n,t_1), \ldots ,f(X_n,t_k)) \longrightarrow (g(X,t_1), \ldots ,g(X,t_k))$ in distribution (as $\mathbb{R}^k$-valued random variables) for all $t_1,\ldots,t_k \in \mathbb{R},k \in \mathbb{N}$, can we conclude that $\mathbb{P}_n$ is converging weakly to $\mathbb{P}$ (in the Levy-Prokhorov metric)?

I know that similar results hold for the spaces $C[0,1]$, $D[0,1]$ and the Frechet space $C[0,\infty)$, but I just don't know how to deal with this one.

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  • $\begingroup$ And how do you define convergence in distribution for $L^2$-valued random variables? (Is it just the convergence of $\mathbb P_n$ in the Levy-Prokhorov metric or something weaker or stronger than that?) $\endgroup$
    – fedja
    Commented Jan 12, 2020 at 15:12
  • $\begingroup$ Yes, I mean the convergence in the Levy-Prokhorov metric. $\endgroup$
    – esner1994
    Commented Jan 12, 2020 at 15:20
  • $\begingroup$ Can you give references to those "similar results [that] hold for the spaces $C[0,1]$, $D[0,1]$ and the Frechet space $C(\mathbb{R})$"? $\endgroup$
    – Iosif Pinelis
    Commented Jan 12, 2020 at 18:50
  • $\begingroup$ $C[0,1]$: Billingsley, "Convergence of probability measures", Theorem 8.1 $D[0,1]$: Billingsley, "Convergence of probability measures", Theorem 15.1 $C[0,\infty)$ Whitt, "Weak Convergence of Probability Measures on the Function Space $C[ 0, \infty)$", Theorem 3 Sorry, I thought the last one was about $C(\mathbb{R})$, but it probably holds as well. I corrected it. $\endgroup$
    – esner1994
    Commented Jan 12, 2020 at 19:38
  • $\begingroup$ The difficulty here is that we have to deal with $L^2(\mathbb R)$, and $\mathbb R$ is not compact. $\endgroup$
    – Iosif Pinelis
    Commented Jan 12, 2020 at 20:38

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