2
$\begingroup$

Assume we have a Levy process $(X_t)_{t\geq 0}$ with a finite second moment for all $t>0$. For simplicity, say $\operatorname{Var}\left[X_1\right]=1$. Let $\tilde{X}_t:=X_t-t\cdot E\left[X_1\right]$. Define $\bar{X}_t(u):=\tilde{X}_{tu}$. Can we say, that for fixed $u\in[0,1]$

$$ \frac{\bar{X}_{t}(u)}{\sqrt{t}}\xrightarrow{d} W_{u} $$ as $t \rightarrow \infty$, where $W$ is a standard Wiener process?

More generally, do we have the convergence $$ \left(\frac{\bar{X}_{t}(u)}{\sqrt{t}}\right)_{u\in[0,1]}\xrightarrow{d} (W_{u})_{u\in[0,1]}\text{?} $$

$\endgroup$
2
$\begingroup$

Let $Y_t(u):=\frac{\bar{X}_{t}(u)}{\sqrt{t}}$ and $W(u):=W_u$. The convergence (as $t\to\infty$) in distribution (in the Skorokhod space $D[0,1]$) of $Y_t$ to $W$ can be proved quite similarly to the way it was done e.g. in the proof of Theorem 16.14 in Foundations of Modern Probability by Kallenberg.

Alternatively and a bit more directly, one may use e.g. Theorem 15.6 in Convergence of Probability Measures by Billingsley (1968). Indeed, the convergence of the finite-dimensional distributions of $Y_t$ to those of $W$ follows by the convergence of the one-dimensional distributions and the independence of the increments of the processes $Y_t$. In view of the condition $Var\;X_1=1$, condition (15.21) in Billingsley holds (with $Y_t$ in place of $X_n$) for $\gamma=2$, $\alpha=1$, and $F(u)\equiv u$ -- cf. the condition sufficient for (15.21) displayed right after Theorem 15.6 in Billingsley.

$\endgroup$
  • $\begingroup$ thanks for you support. I try to reamphasize your second statement on Theorem 15.6.. Why doesn't from $(Y_t(s_1),\ldots ,Y_{t}(s_k))\xrightarrow{d} (W(s_1),\ldots,W(s_k))$ for $0\leq s_1< \ldots<s_k\leq 1$ (are times of continuity of W, which is fullfilled anyways) as $t\rightarrow \infty$ (this is assumption (15.20) in Billingsley), where $W=(W(s))_{s\leq 1}$ follow directly $(Y_t(s))_{s\leq 1}\xrightarrow{d} (W(s))_{s\leq 1}$? I mean the process is completely determined by its finite dimensional distribution. Why do we need more than (15.20) ? Best regards ziT $\endgroup$ – ziT Jul 19 '16 at 12:34
  • $\begingroup$ You did not specify the meaning of the symbol $\xrightarrow{d}$. Usually, it means the convergence in distribution, in an appropriate space. In this case, concerning convergence of Levy processes, the natural space is the Skorokhod one, $D[0,1]$. Whereas a probability distribution in $D[0,1]$ is indeed uniquely determined by the finite-dimensional distributions, the questions of convergence and even the existence of a random function in $D[0,1]$ with given finite-dimensional distributions (see Theorem 15.7 in Billingsley) are quite different matters. $\endgroup$ – Iosif Pinelis Jul 19 '16 at 13:42
  • $\begingroup$ [Previous comment continued:] Condition (15.21) in Billingsley is needed to establish the tightness, which is necessary for the convergence. The convergence in distribution in $D[0,1]$ is much more than that of the finite-dimensional distributions; recall the Portmanteau theorem (Theorem 2.1 in Billingsley). In particular, the convergence in distribution of $Y_t$ to $W$ implies that $P(Y_t(u)<g(u)\ \forall u\in[0,1])\to P(W(u)<g(u)\ \forall u\in[0,1])$ as $t\to\infty$, for any (say) continuous function $g$. $\endgroup$ – Iosif Pinelis Jul 19 '16 at 13:42
  • $\begingroup$ Ok thanks. I prefer your 2. statement of billingsley then. Because bY the convergence of $Y_t(1)$ to $W(1)$ the convergence to the Brownian law on the skorohodspace on $[0,1]$ follows directly with your remarks, if i am right. $\endgroup$ – ziT Jul 19 '16 at 15:16

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.