2
$\begingroup$

Let $\Omega$ be a subset of $\mathbb{R}_+^d$. $X$ should be the canonical process, i.e. $X_k(\omega)=\omega_k$, where $k\in \{1,\dots,d\}$ and $\omega_k$ is just the $k$-th component of $\omega\in\mathbb{R}^d$. The filtration should be generated by $X$.

Now suppose we are given a sequence of probability measures $\{\mu_n\}$ on $\Omega$. I would like to know if this sequence is tight. Tightness means that for every $\epsilon > 0$, $\exists K$ compact such that $\sup_n\mu_n(\Omega\backslash K)<\epsilon$. In this particular case it exists a $M$ such that $\sup_n\mu_n(\Omega\backslash [-M,M]^d)<\epsilon$.

Somehow tightness of $\{\mu_n\}$ should follow from the following: We assume that $$\sup_nE_{\mu_n}[X_d^p]<\infty\tag{1}$$ where $E_{\mu_n}[\cdot]$ denotes the expectation with respect to $\mu_n$ and $p$ is big enough, i.e. for sure bigger than $1$, maybe even bigger than $2$. Then tightness should be implied from

$$\lim_{L\to\infty}\sup_nE_{\mu_n}[X_k\mathbf1_{X_k> L}]=0 \tag{2}$$ for all $k=1,\dots,d-1$

My first question is, why is this true? $(2)$ is somehow related to uniform integrability of $X_i$. But instead of taking the supremum over the r.v. you take it over the measures. How does this imply tightness and why is $(1)$ needed (if it is)?

Then it should also follow that by the integrability of $X_k$ und any $\mu_n$ $(2)$ is implied if

$$\lim_{N\to\infty}\lim_{L\to\infty}\sup_{n\ge N}E_{\mu_n}E[X_k\mathbf1_{X_k> L}]=0 \tag{3}$$ for all $k=1,\dots,d-1$

Does $(2)$ also imply tightness in a more general setup, i.e. not discrete time parameters?

$\endgroup$
1
$\begingroup$

(2) implies tightness because we have $X_k\chi_{|X_k|\gt L}\geqslant L\chi_{|X_k|\gt L}$, which gives $$\lim_{L\to \infty}L\sup_n\mu_n\left(|X_k|\gt L\right)=0.$$ So for a fixed $\varepsilon$, take $L$ such that $\sup_n\mu_n\left(|X_k|\gt L\right)\lt\varepsilon$; then $$\mu_n\{\Omega\setminus [-L,L]^d\}\leqslant \sum_{k=1}^d\mu_n\{|X_k|>L\}.$$ If (1) is satisfied for some $p\gt 1$, then (2) is satisfied (as a consequence of Hölder's inequality).

(3) implies (2): if (2) does not hold, there is $\delta\gt 0$ such that for each $L$, $\sup_n\mathbb E_{\mu_n}[|X_k|\chi_{|X_k|\gt L}]\geqslant 2\delta$. For each integer $L$, pick $n_L\gt n_{L-1}$ such that $E_{\mu_{n_L}}[|X_k|\chi_{|X_k|\gt L}]\gt \delta$. then (3) is not satisfied.

$\endgroup$
  • $\begingroup$ Some question: 1. Why is $(2)$ satisfied if $(1)$ is true for a $p>1$? 2. Why can we remove "for all $k\dots$"? 3. What about $(3)$ implies $(2)$? $\endgroup$ – user20869 Sep 21 '13 at 8:20
  • $\begingroup$ @hulik See edit. $\endgroup$ – Davide Giraudo Sep 21 '13 at 8:27
  • $\begingroup$ Thanks a lot for your help. Am I right, that we just need the integrability condition $\sup_n E_{\mu_n}[X_k^p]$ for one $k\in\{1,\dots,d\}$ then the whole thing works. We do not need it as above for the case $k=d$. Is this right? $\endgroup$ – user20869 Sep 21 '13 at 12:13
  • $\begingroup$ Yes, it is right. $\endgroup$ – Davide Giraudo Sep 21 '13 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy