Convergence of random measure

Suppose that $$S$$ is a separable metric space or Polish. Let $$μ_{n},n∈N$$ be a random probability measures and let μ be a deterministic probability measure on $$S$$. That is to say, that the $$μ_{n}$$ are measurable maps from a probability space $$(Ω,T,P)$$ to the space of $$M_1(S)$$ equipped with the Borel-σ-algebra generated by the topology of weak convergence. Assume that the expected measures $$ν_{n}:=Eμ_{n}$$, defined via duality as $$∫f dν_{n}:=E∫f dμ_{n}$$ for all $$f∈C_{b}(S)$$, converge weakly to μ, i.e. that for all $$f∈C_{b}(S)$$ we have the convergence of $$E∫fdμ_{n}$$ to $$∫fdμ$$.

1. For all ϵ>0 the sequence $$P(d_{BL}(μ_{n},μ)>ϵ)$$ converges to zero, where $$d_{BL}$$ is the bounded Lipschitz metric $$d_{BL}(μ,ν)=\sup\Big\{\Big\vert∫f dμ−∫f dν\Big|: f:S\to\mathbb{R} \; \mbox{is 1-Lipschitz and \Vert f\Vert_\infty\leq 1}\Big\}$$ which completely metrizes the topology of weak convergence)

2. For all $$f∈C_{b}(S)$$ it holds that $$∫f dμ_{n}$$ converges in probability to $$∫f dμ$$.

please i want to know if 1) implies 2) or if 2) implies 1) or if 1) is equivalent to 2)

and in the case where $$μ_{n}$$ is not necessary a probability measure but just a random measure, do the same results remain true?

In your setting, the two are equivalent. Let's first show that your assumptions imply that the $$\mu_n$$, viewed as random variables in $$M_1(S)$$, are tight. Since $$S$$ is Polish and the $$\nu_n$$ converge weakly to a limit, we know from Prokhorov that the sequence $$\nu_n$$ is tight so, for every $$\varepsilon > 0$$ there exists a compact set $$K_\varepsilon \subset S$$ such that $$\nu_n(K_\varepsilon) > 1-\varepsilon$$, uniformly in $$n$$. A simple application of Markov's inequality then shows that $$P(\mu_n(K_\varepsilon) > 1-\sqrt \varepsilon) > 1-\sqrt \varepsilon$$.

Define now $$\mathcal{K}_\delta = \{\mu \in M_1(S)\,:\, \mu(K_{2^{-2m}\delta^2}) > 1- 2^{-m} \delta \quad \forall m \in \mathbb{N}\}.$$ The set $$\mathcal{K}_\delta$$ is tight and therefore precompact in $$M_1(S)$$. Furthermore, we know from above that $$P(\mu_n(K_{2^{-2m}\delta^2}) > 1-2^{-m} \delta) > 1-2^{-m} \delta$$ so that, summing all of these probabilities, one has $$P(\mu_n \in \mathcal{K}_\delta) > 1-\delta$$. Now that we know that the $$\mu_n$$ are tight, we know that there are (possibly random) accumulation points $$\hat \mu$$.

The claim now is that both of your conditions are equivalent to the statement that $$\hat \mu = \mu$$ almost surely. It is clear that this statement implies your second claim by continuity of the maps $$\eta \mapsto \int f\,d\eta$$. It also implies the first claim since $$d_{BL}$$ is a metric for $$M_1(S)$$ and convergence in law to a deterministic element is the same as convergence in probability. The fact that your claims both imply that $$\hat \mu$$ cannot be anything other than $$\mu$$ is pretty much immediate.

I haven't checked the details, but I guess that if the $$\mu_n$$ are allowed to be random positive measures, then the claim still holds. (Testing with $$f=1$$ gives an a priori bound on the mass of the measure and identifies the mass of the limit.)

The equivalence of 1. and 2. is essentially a special case of  van der Vaart/Wellner (1996), Weak Convergence and Empirical Processes - With Applications to Statistics. (Unfortunately this book mainly covers nonseparable metric spaces, so it may be hard to read.)

That 1. implies 2. even if $$\mu$$ is arbitrary, is an easy consequence of , Lemma 1.9.2 (ii), rewritten to the separable case: $$\mu_n \to \mu$$ in probability iff every subsequence $$(\mu_{n'})$$ has a further subsequence $$(\mu_{n''})$$ with $$\mu_{n''} \to \mu$$ almost surely.

The metric space $$\mathbb{D}$$ in  here are $$(M_1(S),d_{BL})$$ resp. $$\mathbb{R}$$. These spaces are separable.

That 2. implies 1., now for constant $$\mu$$, is a direct consequnce of , Lemma 1.10.2 (iii).