Given a sequence (mn) of random probability measures on a metric space S, consider the conditions: (i) mn/m (weakly) a.s. for some random probability measure m on S; (ii) mn(f) converges a.s. for all f2Cb(S). Then, (i) implies (ii), while the converse is not true, even if S is separable. For (i) and (ii) to be equivalent, it is enough that S is Radon (i.e. each probability on the Borel sets of S is tight) or that the sequence (Pmn) is tight, where Pmn($)ZE(mn($)). In particular, (i)5(ii) in case S is Polish. The latter result is still available if a.s. convergence is weakened into convergence in probability. In case SZTN with T Radon, a.s. convergence of mn(f), for those f2Cb(S) which are finite products of elements of Cb(T), is sufficient for (i). In case SZRd and the limit m is given in advance, a.s. convergence of characteristic functions is enough for mn/m (weakly) a.s.
Almost sure weak convergence of random probability measures
Pietro Rigo
2006
Abstract
Given a sequence (mn) of random probability measures on a metric space S, consider the conditions: (i) mn/m (weakly) a.s. for some random probability measure m on S; (ii) mn(f) converges a.s. for all f2Cb(S). Then, (i) implies (ii), while the converse is not true, even if S is separable. For (i) and (ii) to be equivalent, it is enough that S is Radon (i.e. each probability on the Borel sets of S is tight) or that the sequence (Pmn) is tight, where Pmn($)ZE(mn($)). In particular, (i)5(ii) in case S is Polish. The latter result is still available if a.s. convergence is weakened into convergence in probability. In case SZTN with T Radon, a.s. convergence of mn(f), for those f2Cb(S) which are finite products of elements of Cb(T), is sufficient for (i). In case SZRd and the limit m is given in advance, a.s. convergence of characteristic functions is enough for mn/m (weakly) a.s.I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione.
