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Let $(\Omega,\mathcal{F},P)$ be a probability space and let $(E,\mathcal{E})$ be a separable Hilbert space ($E$) with Borel $\sigma$-algebra $\mathcal{E}$. For concreteness let us set $E=L^{2}[a,b]$ with the usual inner product.

For a sequence of stochastic processes $\{X_{N}(t)\}_{N=1}^{\infty}$ and $X(t)$ whose sample paths belong to $L^{2}[a,b]$, suppose I have shown that

$a)$ Convergence of finite-dimensional distributions, i.e. that

$$(X_{N}(t_{1}),\ldots,X_{N}(t_{k})) \to (X(t_{1}),\ldots,X(t_{k})), \qquad N \to \infty$$

in distribution, for all $(t_{1},\ldots,t_{k}) \in [a,b]^{k}$, any $k \in \mathbb{N}$ and

$b)$ Tightness of the family of random elements $\{X_{N}(t)\}_{N=1}^{\infty}$ in $L^{2}[a,b]$, i.e. that for all $\epsilon>0$ there is a compact $K \subset L^{2}[a,b]$ such that

$$\mu_{N}(K) > 1-\epsilon$$

uniformly in $N$, where $\mu_{N}$ is the probability measure on $L^{2}[a,b]$ induced by the $X_{N}$.

$\textbf{My question:}$ Are conditions $a)$ and $b)$ enough to conclude that for all bounded continuous functionals $f : L^{2}[a,b] \to \mathbb{R}$, one has the weak convergence

$$\mathbb{E}(f(X_{N})) \to \mathbb{E}(f(X)), \qquad N \to \infty.$$

From what I can tell in the literature, this type of result is usually applied to spaces like $C[0,1]$ (continuous functions on $[0,1]$) (see e.g. Billingsley's book "convergence of probability measures"). But for more general function spaces (e.g. $L^{2}[a,b]$) I can't find a good reference. Is it still true here? Are there interesting examples of function spaces where this claim is false?

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I don't think so. Let $f(X)=\langle X,1\rangle$.

Now define the stochastic process $X_n^\omega(t)$ as follows:

Let $Z_n(\omega)$ be a sequence of iid random variables uniformly distributed in $[a,(a+b)/2]$ and let $Y_n$ be a sequence of independent random variables taking the value 1 with probability $1/n$ and 0 otherwise.

Let $X_n=2^nY_n\mathbf 1_{[Z_n(\omega),Z_n(\omega)+2^{-n}]}$, so that in particular, $X_n$ is the 0 function with probability $1-1/n$. Given $\epsilon>0$, the values taken by $X_1,\ldots,X_{1/\epsilon}$ lie in a compact subset of $L^2[a,b]$ and so the sequence of measures $\mu_n$ is tight.

Also, for any fixed $t_1,\ldots,t_k$, almost surely $X_n(t_i)=0$ for all sufficiently large $n$.

But, of course, by Borel-Cantelli II, $f(X_n(\omega))=1$ infinitely often and $f(X_n(\omega))=0$ infinitely often for almost every $\omega$.

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  • $\begingroup$ Just to remark that if we include the additional assumption that the stochastic process $X_{N}(t)$ is measurable and has bounded variance uniformly in $N$ then the conclusion I gave in the original post is actually correct. It is a special case ($p=2$) of the result given in the abstract of this paper. $\endgroup$ – Nigel Jul 1 '13 at 17:56
  • $\begingroup$ But we can conclude that $\operatorname Ef(X_n)\to\operatorname Ef(X)$ for all bounded, uniformly continuous $f$, right? According to Billingsley’s book, if $\{P_n\}$ is tight and the fdds of $P_n$ converge weakly to those of $P$, $P_n\Rightarrow_nP$. $\endgroup$ – Cm7F7Bb Mar 21 '17 at 14:51

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