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FINAL EDIT: There is one main question left: According to the answer, we have choosen $\theta=1$ , where we could choose $0<\theta<\infty$ as we like. His this sufficient, if we regarde the convergence to $\theta$ as a finite positive random variable?

This post seems long, but its almost everything proofed except the last step. The unknown part is marked especially.

Assumptions

Given a Levy-Process $U_{t}$ with with $E(U_t)=0$ (then $U_t$ is a martingale). Let $U_t$ have finite variance and $Var(X_1)=\sigma^{2}$ and the limit theorem holds: \begin{align} F_t:=\sqrt{t}\left(\frac{U_t}{t}-E(U_1) \right)=\frac{U_t}{\sqrt{t}}\xrightarrow{d}\mathcal{N}(0,\sigma^{2})\quad as \,\,t\rightarrow \infty.\tag1 \end{align} Let $K_t$ a non-decreasing positive ($K_{t}>0$ a.s.) process with cadlag-paths with the property that $K_{t}\rightarrow \infty$ almost sureley, as $t\rightarrow \infty$.

I want to show that \begin{align} F_{K_t}:=\frac{U_{K_t}}{\sqrt{K_{t}}} \xrightarrow{d}\mathcal{N}(0,\sigma^{2})\quad as \,\,t\rightarrow \infty. \tag2 \end{align} For this one requires a positive non-random cadlag-function $a(t)$ with $a(t)\rightarrow \infty$ as $t\rightarrow \infty$ such that \begin{align} \frac{K_{t}}{a(t)}\rightarrow \theta\quad P\, a.s. \tag3 \end{align} holds. Where $\theta$ is a positive finite random-variable. Then the convergence in distribution of $F_{t}\xrightarrow{d} \mathcal{N}(0,\sigma^{2})$ implies the convergence in distribution of $F_{K_t}\xrightarrow{d} \mathcal{N}(0,\sigma^{2})$.

The suggestion how it has to be proofed is given in the post below: Here he takes $\theta=1$. So that we have $K_{t}\in ((1-\epsilon)a(t),(1+\epsilon)a(t))$ a.s. as $t\rightarrow \infty$.

Is this legit, concerning that we have generally $\theta$ a positive ($\theta>0$) finite random variable? For small $m$ we have $$ P(U_{K_t}<x\sqrt{K_t})\leq P\left(K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))\right)+P\left(U_{a_t}<x\sqrt{(1+\epsilon)a(t)}+m\cdot \sqrt{\epsilon a(t))}\right)+ P\left(\sup_{s\in ((1-\epsilon)a(t),(1+\epsilon)a(t))}|U_{s}-U_{a(t)}|>m\cdot \sqrt{\epsilon a(t))}\right) $$ The first term converges to 0 due to (3). The second term converges to $\Phi(x+m)$ (Why?) by the central limit theorem (1).

Due to the martingale maximale inequality the third term is bounded by $$\frac{1}{(m\cdot \sqrt{\epsilon a(t))})^{2}}$$ and tends to zero as $a(t)\rightarrow \infty$.

Why should this proof (2)? So far we have only that the distribution $P(U_{K_{t}}/\sqrt{K_t}\leq x)$ is bounded by $\Phi(x+m)$ then. A lower bound converging to $\Phi(x)$ is necessary i guess?

Idea:

$$ P(U_{K_t}<x\sqrt{K_t})\geq P(U_{K_t}<x\sqrt{K_t},|U_{a(t)}-U_{K_t}|<m\sqrt{\epsilon a(t)},K_{t}\in ((1-\epsilon)a(t),(1+\epsilon)a(t))\\ \geq P(U_{a(t)}<x\sqrt{(1-\epsilon)a(t)}-m\sqrt{\epsilon a(t)}) \\ \rightarrow \Phi(x-m) $$

And we have it sandwiched, converging to $\Phi(x)$ right?

Btw: How to come to this inequality is due to $$ P(U_{k_t}<x \sqrt{K_t})\\ \leq P[U_{k_t}<x \sqrt{K_t},K_{t}\in((1-\epsilon) a(t),(1+\epsilon) a(t))]+P[U_{K_t}<x\sqrt{K_t},K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))] \\ \leq P[K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))]+ P[U_{k_t}<x \sqrt{K_t},K_{t}\in((1-\epsilon) a(t),(1+\epsilon) a(t))] \\ \leq P[K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))] \\ +P(U_{K_{t}}<x\sqrt{(1+ \epsilon)a(t)},|U_{K_t}-U_{a(t)}|\leq m\sqrt{\epsilon a(t)},|U_{K_t}-U_{a(t)}|> m\sqrt{\epsilon a(t)}] \\ \leq P[U_{a(t)}<x\sqrt{(1+\epsilon)a(t)}+m \sqrt{\epsilon a(t)}] + P\left(\sup_{s\in ((1-\epsilon)a(t),(1+\epsilon)a(t))}|U_{s}-U_{a(t)}|>m\cdot \sqrt{\epsilon a(t))}\right)+P\left(K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))\right) $$

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For square integrable Levy processes this should be easy because you can control the difference between $U_{K_t}$ and $U_{a(t)} $ with the martingale maximal inequality. Subtract out the mean and so suppose one has a square integrable independent increments process. Then e.g. $$P(U_{K_t} < x \sqrt{K_t }) \le I_t+J_t+L_t$$ with $$I_t=P\left(K_t \notin ((1-\epsilon)a(t) , (1+\epsilon a(t))\right)$$ $$J_t= P\left(U_{a(t)} < x\sqrt{(1 + \epsilon) a(t)} + m \sqrt{\epsilon a(t)} \right)$$ and $$L_t= P\left(\sup_{s \in ((1-\epsilon)a(t) , (1+\epsilon a(t))} |U_s - U_{a(t)}| > m\sqrt{\epsilon a(t)} \right) $$ $m$ should be small. Use convergence is probability on $I_t$, CLT on $J_t$, and maximal inequality on $L_t$ (which will tell you that the sup can't be worse than $\frac 1 {(m \sqrt{\epsilon a(t)})^2}$). Don't take every term literally as I am tex-ing on the fly, but that is the idea.

The assumption that $\frac{K_t}{a(t)}$ converges to a constant shows that $I_t$ is small. $J_t$ converges to $\Phi(x+m)$ by the CLT applied to $U_t$ or $U_{a(t)}$ as you prefer, and $L_t$ is used as mentioned above. Btw, I call this Anscombe's theorem.

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  • $\begingroup$ Hey thanks for the answer. I guess your suggestion tends to the right direction. But it's a bit blurry for me. Could you be a little bit more precisely? And why follows from this inequality $P(U_{K_{t}}<x\sqrt{K_{t}})$ the desired result? And where we use the Assumption that $K_{t}/a_{t}\rightarrow \theta$ a.s.? Thanks so far. I really appreciate your help. Best regards $\endgroup$ – GuildY123 Mar 7 '16 at 16:59
  • $\begingroup$ Have edited to address yr questions. It won't let me fix those tex mistakes but I think its readable. $\endgroup$ – user83457 Mar 8 '16 at 11:24
  • $\begingroup$ you showed this for $\theta=1$ however we have convergence to a positive finite random variable. What is left to proof there? $\endgroup$ – GuildY123 Jun 2 '16 at 9:26

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