Almost Sure

22 December 09

U.C.P. and Semimartingale Convergence

A mode of convergence on the space of processes which occurs often in the study of stochastic calculus, is that of uniform convergence on compacts in probability or ucp convergence for short.

First, a sequence of (non-random) functions ${f_n\colon{\mathbb R}_+\rightarrow{\mathbb R}}$ converges uniformly on compacts to a limit ${f}$ if it converges uniformly on each bounded interval ${[0,t]}$. That is,

 $\displaystyle \sup_{s\le t}\vert f_n(s)-f(s)\vert\rightarrow 0$ (1)

as ${n\rightarrow\infty}$.

If stochastic processes are used rather than deterministic functions, then convergence in probability can be used to arrive at the following definition.

Definition 1 A sequence of jointly measurable stochastic processes ${X^n}$ converges to the limit ${X}$ uniformly on compacts in probability if

$\displaystyle {\mathbb P}\left(\sup_{s\le t}\vert X^n_s-X_s\vert>K\right)\rightarrow 0$

as ${n\rightarrow\infty}$ for each ${t,K>0}$.

The notation ${X^n\xrightarrow{\rm ucp}X}$ is sometimes used, and ${X^n}$ is said to converge ucp to ${X}$. Note that this definition does not make sense for arbitrary stochastic processes, as the supremum is over the uncountable index set ${[0,t]}$ and need not be measurable. However, for right or left continuous processes, the supremum can be restricted to the countable set of rational times, which will be measurable. In fact, for jointly measurable processes, it can be shown that the supremum is measurable with respect to the completion of the probability space, so ucp convergence makes sense. However, that is not needed for these notes.

For each time ${t\ge 0}$, the following pseudometric can be defined on the space of locally bounded deterministic processes

$\displaystyle D_t(X-Y)=\sup_{s\le t}|X_s-Y_s\vert.$

Then, ${X^n\rightarrow X}$ uniformly on compacts if ${D_t(X^n-X)\rightarrow 0}$ for each ${t\ge 0}$. This shows that uniform convergence on compacts is in fact given by the following metric

$\displaystyle D(X-Y)=\sum_{k=1}^\infty 2^{-k}\wedge D_k(X-Y).$

Similarly, a sequence of stochastic processes ${X^n}$ converges ucp to ${X}$ if ${D(X^n-X)\rightarrow 0}$ in probability. By bounded convergence, this is equivalent to ${{\mathbb E}[D(X^n-Y)]}$ tending to zero. So, ucp convergence is given by the following metric.

 $\displaystyle D^{\rm ucp}(X-Y)={\mathbb E}[D(X-Y)]=\sum_{k=1}^\infty{\mathbb E}\left[2^{-k}\wedge\sup_{s\le k}\vert X_s-Y_s\vert\right].$ (2)

If a sequence of processes converges ucp, then it is always possible to pass to a subsequence for which uniform convergence on compacts holds with probability one on the sample paths. This is a simple application of the Borel-Cantelli lemma, but allows properties of ucp convergence to be inferred from corresponding properties of uniform convergence on compacts.

Theorem 2 The space of cadlag (resp. continuous) adapted processes is complete under ucp convergence.

Furthermore, if ${X^n\xrightarrow{\rm ucp}X}$ then there is a subsequence whose sample paths almost surely converge to those of ${X}$ uniformly on compacts.

Proof: Let ${X^n}$ be a Cauchy sequence under ucp convergence, so that ${D^{\rm ucp}(X^m-X^n)\rightarrow 0}$ as ${m,n\rightarrow\infty}$. Then, there is a subsequence ${Y^n=X^{r_n}}$ satisfying ${D^{\rm ucp}(Y^m-Y^n)\le 2^{-n}}$ for all ${m\ge n}$. In this case,

$\displaystyle {\mathbb E}\left[\sum_{n=1}^\infty D(Y^{n+1}-Y^n)\right]=\sum_{k=1}^\infty D^{\rm ucp}(Y^{n+1}-Y^n)\le 1$

so that ${\sum_n D(Y^{n+1}-Y^n)}$ is almost surely finite. Restricting to a set of probability one if necessary, we suppose that this is always finite. Then, the sample paths of ${Y^n}$ are Cauchy convergent under uniform convergence on compacts, and there is a limit ${X}$. As ${X_t=\lim_nY^n_t}$, this will be measurable, so ${X}$ is a stochastic process and is adapted whenever ${Y^n}$ are adapted. If the processes ${Y^n}$ have left limits, then it is clear that ${\sup_{s\le t}\vert Y^n_{s-}-Y^m_{s-}\vert\le\sup_{s\le t}\vert Y^n_s-Y^m_s\vert}$ and, therefore, the left limits ${Y^n_-}$ are also Cauchy under uniform convergence on compacts.

As limits can be commuted with uniform convergence of sequences, if the processes ${Y^n}$ are cadlag then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\lim_{s\downarrow\downarrow t}X_s=\lim_{n\rightarrow\infty}\lim_{s\downarrow\downarrow t}Y^n_s=\lim_{n\rightarrow\infty}Y^n_t=X_t,\smallskip\\ &\displaystyle\lim_{s\uparrow\uparrow t}X_s = \lim_{n\rightarrow\infty}\lim_{s\uparrow\uparrow t}Y^n_s=\lim_{n\rightarrow\infty} Y^n_{t-}. \end{array}$

So, ${X}$ is cadlag and ${Y^n_-\rightarrow X_-}$ under uniform convergence on compacts. In particular, ${X}$ is continuous whenever ${Y^n}$ are.

To complete the proof, it just remains to show that the original sequence ${X^n}$ does indeed converge ucp to ${X}$. As ${D(Y^n-X)\rightarrow 0}$ it follows that this converges in probability, so ${Y^n\xrightarrow{\rm ucp}X}$. Then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rcl} \displaystyle\limsup_{n\rightarrow\infty}D^{\rm ucp}(X^n-X) &\displaystyle\le &\displaystyle \limsup_{m,n\rightarrow\infty}\left(D^{\rm ucp}(X^n-X^{r_m})+D^{\rm ucp}(Y^m-X)\right)\smallskip\\ &\displaystyle =&\displaystyle 0 \end{array}$

as required. ⬜

A consequence of Doob’s martingale inequalities is that ${L^1}$ convergence of martingales implies ucp convergence.

Lemma 3 Let ${X^n}$ be a sequence of cadlag martingales, and ${X}$ be a process such that ${{\mathbb E}\vert X^n_t-X_t\vert\rightarrow 0}$ as ${n\rightarrow\infty}$.

Then, ${X}$ is a martingale and has a cadlag version which is the ucp limit of ${X^n}$.

Proof: Clearly, by ${L^1}$-convergence, ${X}$ is a martingale. Furthermore, Doob’s martingale inequality shows that

$\displaystyle {\mathbb P}\left(\sup_{s\le t}\vert X^n_s-X^m_s\vert>K\right)\le K^{-1}{\mathbb E}\vert X^n_t-X^m_t\vert\rightarrow 0$

as ${m,n\rightarrow\infty}$ for all ${t,K>0}$. Consequently, the sequence is Cauchy under ucp convergence and has a cadlag limit ${X^n\xrightarrow{\rm ucp}Y}$. As ${Y_t=\lim_n X^n_t=X_t}$ (convergence in probability), this limit is a cadlag version of ${X}$. ⬜

In particular, the space of continuous martingales is complete under ${L^1}$ convergence.

Corollary 4 Let ${X^n}$ be a sequence of continuous martingales such that ${{\mathbb E}\vert X^n_t-X_t\vert\rightarrow 0}$ as ${n\rightarrow\infty}$ for all ${t}$ and some process ${X}$. Then, ${X}$ is a martingale and has a continuous version.

Proof: By the previous lemma, ${X}$ is a martingale and has a cadlag version ${Y}$ such that ${X^n\xrightarrow{\rm ucp}Y}$. Then, by Theorem 2, this limit is continuous. ⬜

The semimartingale topology

An even stronger topology than ucp convergence is the semimartingale topology. A sequence of cadlag and adapted processes ${X^n}$ converges to ${0}$ under this topology if, for every sequence ${\xi^n}$ of elementary predictable processes with ${\vert\xi^n\vert\le 1}$ and every ${t\ge 0}$, the limit

$\displaystyle \xi^n_0X^n_0+\int_0^t\xi^n\,dX^n\rightarrow 0$

holds, in probability. Then, a sequence of cadlag adapted processes ${X^n}$ converges to a limit ${X}$ under the semimartingale topology if ${X^n-X}$ converges to zero. As with ucp convergence, this can be described by a metric. For each ${t\ge 0}$ set

$\displaystyle D^{\rm sm}_{t}(X)\equiv \sup\left\{{\mathbb E}\left[\left\vert \xi_0X_0+\int_0^t\xi\,dX\right\vert\wedge 1\right]\colon \vert\xi\vert\le 1\text{ is elementary}\right\}.$

Then, ${X^n\rightarrow X}$ if ${D_t^{\rm sm}(X^n-X)\rightarrow 0}$. It follows that the semimartingale topology is defined by the metric

 $\displaystyle D^{\rm sm}(X-Y)\equiv\sum_{n=1}^\infty 2^{-n} D^{\rm sm}_n(X-Y).$ (3)

The semimartingale topology is indeed stronger than ucp convergence.

Theorem 5 If ${X^n,X}$ are cadlag adapted processes with ${X^n}$ converging to ${X}$ in the semimartingale topology then ${X^n\xrightarrow{\rm ucp}X}$.

Proof: If ${Y}$ is a cadlag adapted process and ${K>0}$, let ${\tau}$ be the stopping time

$\displaystyle \tau=\inf\left\{t\in{\mathbb R}_+\colon \vert Y_t\vert\ge K\right\}.$

Approximating ${\tau}$ from the right by the simple stopping times ${\tau_n=\min\{m/n\colon m/n\ge\tau\}}$ gives

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rcl} \displaystyle{\mathbb P}\left(\sup_{s\le t}\vert Y_s\vert> K\right)&\displaystyle\le&\displaystyle\limsup_{n\rightarrow\infty}{\mathbb P}\left(\vert Y_{\tau_n\wedge t}\vert> K\right)\smallskip\\ &\displaystyle\le&\displaystyle K^{-1}\limsup_{n\rightarrow\infty}{\mathbb E}\left[\vert Y_{\tau_n\wedge t}\vert\wedge 1\right] \end{array}$

for any real ${K\in(0,1]}$. However, the processes ${\xi^n_s=1_{\{s\le\tau_n\wedge t\}}}$ are elementary and ${\xi^n_0Y_0+\int_0^t\xi^n\,dY=Y_{\tau_n\wedge t}}$. So,

$\displaystyle {\mathbb P}\left(\sup_{s\le t}\vert Y_s\vert> K\right)\le K^{-1} D^{\rm sm}_t(Y).$

It follows that if ${X^n\rightarrow X}$ in the semimartingale topology then

$\displaystyle {\mathbb P}\left(\sup_{s\le t}\vert X^n_s-X_s\vert>K\right)\le K^{-1}D^{\rm sm}_t(X^n-X)\rightarrow 0$

as ${n\rightarrow\infty}$, and ucp convergence holds. ⬜

Lemma 3 above states that ${L^1}$ convergence of martingales implies ucp convergence. In fact, the stronger property of semimartingale convergence holds, although I do not prove that fact here.

Completeness of ucp convergence can be used to also prove completeness of the semimartingale topology.

Lemma 6 The space of cadlag and adapted processes is complete under the semimartingale topology.

Proof: Let ${X^n}$ be Cauchy under the semimartingale topology. By Theorem 5 this is also Cauchy under ucp convergence, so has a limit ${X^n\xrightarrow{\rm ucp}X}$ which is cadlag and adapted.

As ${X^n_t\rightarrow X_t}$ in probability for each time ${t}$, it follows that ${\int_0^t\xi\,dX^n\rightarrow\int_0^t\xi\,dX}$ in probability, for all elementary processes ${\vert\xi\vert\le 1}$. So, setting ${Y^n=X^n-X}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle{\mathbb E}\left[\left\vert \xi_0Y^n_0+\int_0^t\xi\,dY^n\right\vert\wedge 1\right]\smallskip\\ &\ \ \ \displaystyle=\lim_{m\rightarrow\infty}{\mathbb E}\left[\left\vert \xi_0(X^n_0-X^m_0)+\int_0^t\xi\,d(X^n-X^m)\right\vert\wedge 1\right]\smallskip\\ &\ \ \ \displaystyle\le\limsup_{m\rightarrow\infty}D^{\rm sm}_t(X^n-X^m). \end{array}$

Taking the supremum over all such elementary processes ${\xi}$ gives

$\displaystyle D^{\rm sm}_t(X^n-X)\le\limsup_{m\rightarrow\infty}D^{\rm sm}_t(X^n-X^m)$

which, by Cauchy convergence, goes to zero as ${n\rightarrow\infty}$. ⬜

Finally, the semimartingale topology is not a vector topology on the space of cadlag adapted processes. For that to be the case, we would need ${\lambda_nX\rightarrow 0}$ for all such processes ${X}$ and sequences of real numbers ${\lambda_n\rightarrow 0}$. However, this says that ${\lambda_n\int_0^t\xi^n\,dX\rightarrow 0}$ in probability for all elementary processes ${\vert\xi^n\vert\le 1}$. Equivalently, for each ${t\ge 0}$, the set

$\displaystyle \left\{\int_0^t\xi\,dX\colon \vert\xi\vert\le 1\text{ is elementary}\right\}$

is bounded in probability. The processes for which this holds are precisely the semimartingales, although I do not prove that here. On these processes, semimartingale convergence is indeed a vector topology.

1. Hi just to mention that in the second paragraph of the proof of lemma 6 I believe you mean :
“So, setting $Y^n= X^n- X$ ” and not “So, setting $Y^n= Y$

Regards

Comment by TheBridge — 11 August 11 @ 1:33 PM

2. […] can be shown that the space is complete under the above […]

Pingback by New Approach to Stochastic Integral « Jun's Blog — 16 September 11 @ 5:40 PM

3. Hi,

Would you be so kind to give a reference or sketch the poof of your statement:
” Lemma 3 above states that convergence of martingales implies ucp convergence. In fact, the stronger property of semimartingale convergence holds, although I do not prove that fact here. ”

Comment by Anonymous — 25 July 13 @ 7:17 AM

• It follows from equation (8) in my post “Martingales are integrators”. If $X^n$ is a sequence of martingales with $\mathbb{E}[\lvert X^n_t\rvert]\to0$ then this shows that for elementary processes $\xi^n$ bounded by 1, $\int_0^t\xi^n\,dX$ tends to 0 in probability.

Comment by George Lowther — 6 June 16 @ 3:28 AM

4. I had a question: also the space of càglàd processes is complete under the ucp topology? In other words, if we use the metric $D^{ucp}$ defined above in (2), do we obtain that the spec of càglàd processes is complete with this metric?

Comment by Almat — 19 May 15 @ 8:44 AM

• We do. This is because caglad processes are closed under pathwise uniform convergence.

Comment by George Lowther — 6 June 16 @ 3:21 AM

5. Why do you write $D(X-Y)$ instead of $D(X,Y)$ for the metric?

Comment by Anonymous — 17 December 16 @ 1:36 PM

• That was just to emphasise that it is translation invariant, so is a function of the difference, $X-Y$.

Comment by George Lowther — 18 December 16 @ 10:16 AM

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