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We say that a sequence of processes $X^n$ converges to a process $X$ uniformly on compacts in probability if for all $\epsilon >0, t>0$ $$P[\sup_{s\le t}|X^n_s-X_s|>\epsilon]\to 0 $$ for $n\to\infty$. We suppose that the processes $X^n$, $X$ are nice such that the supremum is measurable, e.g. left or right continuous. To get a feeling for the definition I wanted to prove several things. I stuck on three of these and it would be appreciated if someone could help me. I will first state the three questions and then explain my attempts so far.

  1. I want to prove that $X^n\to X$ in ucp if and only $d(X^n,X)\to 0$, where $$d(X,Y):=\sum_{m=1}^\infty 2^{-m}E[1\wedge \sup_{s\le m}|X_s-Y_s|] $$
  2. If $X^n\to X$ then we can pass to a subsequence which converges a.s. uniformly on compacts.
  3. For a left-continuous process with right limits $Y$ we define $T_n:=\inf\{t:|Y_t|> n\}$. By the Début theorem $T_n$ is a stopping time (we assume the usual assumption on the filtration). Let $Y^n:=Y^{T_n}\mathbf1_{T_n>0}$, where $X^T:=(X)_{t\wedge T}$. Why does $Y^n \to Y$ in ucp?

My thoughts:

  1. Suppose $X^n\to X$ in ucp. Let $\delta>0$, we have to prove that there is a $N\in\mathbb{N}$ such that for all $n\ge N$ we have $$d(X^n,X)<\delta $$ i.e. $$\big(\sum_{m=1}^\infty 2^{-m}E[1\wedge \sup_{s\le m}|X^n_s-X_s|] \big)<\delta$$ Of course we want to manipulate the expectation in such a way to apply ucp convergence. Let $A:=\{\sup_{s\le m}|X^n_s-X_s|\le 1\}$, I started $$E[1\wedge \sup_{s\le m}|X^n_s-X_s|]=E[(1\wedge \sup_{s\le m}|X^n_s-X_s|)\mathbf1_A]+E[(1\wedge \sup_{s\le m}|X^n_s-X_s|)\mathbf1_{A^c}]$$ The first term can be bounded by $$E[(1\wedge \sup_{s\le m}|X^n_s-X_s|)\mathbf1_A]\le P[A]=P[\sup_{s\le m}|X^n_s-X_s|\le 1] $$ The second one is equal $$E[(1\wedge \sup_{s\le m}|X^n_s-X_s|)\mathbf1_{A^c}]=P[\sup_{s\le m}|X^n_s-X_s|> 1]$$ The second one is nice, since it is of the form to apply ucp convergence. However, I can make this probability as small as I want but it will still depend on $m$. Moreover it is unclear how I get rid of the first term. For the converse direction I have no idea so far.
  2. By convergence in ucp we can pass to a subsequence again denoted by $X^n$ such that $d(X^n,X)<2^{-n}$. To apply Borel-Cantelli and establish the result we must prove for every $\epsilon >0,t>0$ that $$\sum_{n=1}^\infty P[\sup_{s\le t}|X^n_s-X_s|>\epsilon]<\infty $$ Of course the idea is to bound $P[\sup_{s\le t}|X^n_s-X_s|>\epsilon]<D(X^n,X)$. By $1.$ there is a $m$ such that $P[\sup_{s\le t}|X^n_s-X_s|>1]\le P[\sup_{s\le m}|X^n_s-X_s|>1]$. in fact ever $m>t$ does the job. For $\epsilon >1 $ I also have $$P[\sup_{s\le t}|X^n_s-X_s|>1]\ge P[\sup_{s\le t}|X^n_s-X_s|>\epsilon]$$. But how should I deal the case $\epsilon <1$? Moreover I have still to think about the $2^{-m}$.

  3. Intuitively this is clear but I have trouble to write it down formally. I tried to apply Markov's inequality without success. What I need to prove is that for every $\epsilon,t,\delta>0$ there is a $N\in\mathbb{N}$ such that for all $n\ge N$ we have $$ P[\sup_{s\le t}|Y^n_s-Y_s|>\epsilon]\le \delta$$

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up vote 3 down vote accepted
  1. First of all, we choose $M \in \mathbb{N}$ sufficiently large such that $\sum_{m \geq M} 2^{-m} < \delta/3$. Moreover, we note that $$\begin{align*} \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \right) &\leq \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \cdot 1_A \right)+ \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \cdot 1_{A^c} \right) \\ &\leq \frac{\delta}{3} + \mathbb{P} \left( \sup_{s \leq m} |X_s^n-X_s| > \frac{\delta}{3} \right) \\ &\leq \frac{\delta}{3} + \mathbb{P} \left( \sup_{s \leq M} |X_s^n-X_s| > \frac{\delta}{3} \right)\end{align*}$$ for any $m \leq M$ and $$A := \left\{\sup_{s \leq m} |X_s^n-X_s| \leq \frac{\delta}{3} \right\}.$$ By the UCP-convergence, we may choose $N \in \mathbb{N}$ sufficiently large such that $$ \mathbb{P} \left( \sup_{s \leq M} |X_s^n-X_s| > \frac{\delta}{3} \right) < \frac{\delta}{3}$$ for all $n \geq N$. Plugging these estimates into the definition of the metric yields $d(X^n,X) \to 0$ as $n \to \infty$. For the converse, we apply Markov's inequality: $$\begin{align*} \mathbb{P} \left( \sup_{s \leq t} |X_s^n-X_s| > \varepsilon \right) &= \mathbb{P} \left( 1 \wedge \sup_{s \leq t} |X_s^n-X_s| > \varepsilon \right) \\ &\leq \frac{1}{\varepsilon} \mathbb{E} \left( 1 \wedge \sup_{s \leq t} |X_s^n-X_s| > \varepsilon \right)\\ &\leq \frac{1}{\varepsilon} \mathbb{E} \left( 1 \wedge \sup_{s \leq M} |X_s^n-X_s| > \varepsilon \right)\end{align*}$$ for any $t \leq M$, $0<\varepsilon<1$. Hence, $$\mathbb{P} \left( \sup_{s \leq t} |X_s^n-X_s| > \varepsilon \right) \leq \frac{2^M}{\varepsilon} d(X^n,X) \stackrel{n \to \infty}{\to} 0 \tag{1}.$$
  2. If $d(X^n,X) < 2^{-n}$ and $t \leq m$, then $(1)$ shows $$\mathbb{P} \left( \sup_{s \leq t} |X_s^n-X_s|>\varepsilon \right) \leq \frac{2^m}{\varepsilon} d(X^n,X) = \frac{2^{m-n}}{\varepsilon}.$$ Thus, $$\sum_{n =1}^{\infty} \mathbb{P} \left( \sup_{s \leq t} |X_s^n-X_s|>\varepsilon \right) \leq \frac{2^m}{\varepsilon} \sum_{n=1}^{\infty} 2^{-n} < \infty.$$
  3. We have $$\begin{align*} \mathbb{P} \left( \sup_{s \leq t} |Y_s^n-Y_s| > \varepsilon \right) &= \underbrace{\mathbb{P} \left( \sup_{s \leq t} |Y_s^n-Y_s| > \varepsilon, t < T_n \right)}_{0} + \mathbb{P} \left( \sup_{s \leq t} |Y_s^n-Y_s| > \varepsilon,t \geq T_n \right) \\ &\leq \mathbb{P} \left(T_n \leq t \right) \end{align*}$$ Now by the definition of $T_n$, $$\mathbb{P}(T_n \leq t) \leq \mathbb{P} \left( \sup_{s \leq t} |Y_s| \geq n \right).$$ Since $Y$ has cadlag paths, its path are bounded on compact sets, and therefore the latter probability converges to $0$ as $n \to \infty$.
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Great answer! Thank you so much. Just one small question: For question 1. the second direction you assmued $\epsilon <1$. What about $\epsilon \ge 1$? Or why can we assume w.l.o.g. that for convergence in probability it is enough to consider $\epsilon<1$? – math Mar 11 '14 at 16:07
@math Yes, it is enough to consider $\varepsilon <1$. This follows from the fact that $$\mathbb{P}\left( \sup_{s \leq t} |X_s^n-X_s|>\varepsilon \right) \leq \mathbb{P}\left( \sup_{s \leq t} |X_s^n-X_s|> \varepsilon_0 \right)$$ for any $\varepsilon>\varepsilon_0$. In particular, for $\varepsilon \geq 1$, we can choose arbitrary $\varepsilon_0<1$, and then the right-hand side converges to $0$; hence, also the left-hand side. – saz Mar 11 '14 at 16:23

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