# Question about a proof of Ansel-Stricker in a paper

I was working through a paper by M. De Donno, which proves the Ansel-Stricker lemma in a different way. The paper can be found here. I've chosen this paper instead of the original one by Ansel-Stricker, simply because my French is to bad. Reading this paper I have some question concerning the proof of Theorem 1.

1. Why is it needed that $\sum_n P(\tau_n< T)<\infty$. This looks like Borel-Cantelli, but I do not see where we use this.
2. How do I derive the bound for $(\Delta X^n_{\sigma_m})^-$?
3. Why is $M^n_{t\wedge \sigma_m}\ge \theta_m-1-(m-\theta_m)$ and why is this positive? Or why do we need this calculation?

hulik

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@ hulik : by the way the point (ii) of th 1 uses that $\eta_k$ converges stationarily to $T$. What does stationarily means precisely in this context ? – TheBridge Oct 1 '12 at 12:44
@TheBridge that's a good question! I was also unsure about that, since I've never heard this expression. I think it should mean stationary. If I look at Corollary 2 below of theorem 1, I guess it means monotone increasing – hulik Oct 1 '12 at 12:57
@ hulik : Iguess you are right, by the way I think I have figured out your point 1. – TheBridge Oct 1 '12 at 15:38

## 2 Answers

Suppose you have a sequence of martingales converging uniformly in probability to some process, $X$ say. The first theorem of the paper gives you a couple of conditions to check to ensure that the limit $X$ is a local martingale.

Understanding the theorem statement

1. What does stationarily mean? This word describes a sequence which is eventually stationary. That is, there is some random variable $N$ such that, for $n\geq N$, $\eta_n=T$.
2. What is the "right" definition of a local martingale indexed by a compact set of times? It seems as though it should be similar to the familiar definition, except that the reducing sequence $(\sigma_n)_{n\in\mathbb{N}}$ should converge stationarily to T, as this gives the useful property that $$\lim_{n\to\infty}X^{\sigma_n}_T = X_T.$$ This is a property that the familiar local martingales have, which doesn't necessarily hold if you drop the stationarity requirement, since $X$ isn't necessarily continuous at $T$.

Understanding the theorem proof

Here we define a sequence of stopping times $(\tau_n)_{n\in\mathbb{N}}$ is such a way that the system is well-bahaved up to time $\tau_n$, for each $n$, but so that we still have $\tau_n\uparrow T$ - note here that $X$ is cadlag and thus bounded on compacts.

I think the author passes to a subsequence here as then that sequence $(\tau_{n_k})_{k\in\mathbb{N}}$ converges stationarily to $T$, and thus, since $\eta$ converges stationarily to $T$, $\sigma$ does also, which would fit with the above definition of a local martingale.

$$(\Delta X_{\tau_n})^- = (X_{\sigma_n}-X_{\sigma_n-})^-\leq n - \theta_n,$$ since $\sigma_n\leq \tau_n$, and $\sigma_n\leq \eta_n$.

Now note that it's not necessarily true that $$M^n_{t\wedge\sigma_m}\geq X_{t\wedge\sigma_m} -1.$$

Why? Take, for example $t=T$, and suppose $X$ jumps far from $M^n$ at time $\sigma_m$. For this reason, we need to take into account a possible jump of $X$ at time $t\wedge \sigma_m$, which we've just shown is bounded by $m-\theta_m$. This gives the required inequality.

We can now apply Fatou's lemma as follows: $$\liminf_n ~~\mathbb{E}[ M^n_{t\wedge\sigma_m} - (2\theta_m-m-1)]\geq \mathbb{E} [X_{t\wedge\sigma_m} - (2\theta_m-m-1)],$$

and we can cancel $2\theta_m-m-1$ from both sides since it is integrable. Alternatively, as TheBridge mentions, we can learn this as an extension of Fatou's lemma.

I hope that helps!

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@ Ben Derrett : In my answer I have interpreted $()^-$ as the negative part of the classical decoposition of a function $f$ as $f(x)=f(x)^+-f(x)^-$. I think that you interpret it the other way around, but from point (iii) in th 1 in the article  I think that my interpretation makes more sense, don't you think ? – TheBridge Oct 2 '12 at 14:24
@TheBridge Thanks for drawing my attention to that - it's now fixed. I haven't received a down vote before, so I'd be very curious to know if anything else can be improved? – Ben Derrett Oct 2 '12 at 15:37
@ Ben Derrett : Actually I did the downvote, and I feel a bit sorry that I can't take it back. It was an epidermic reaction of mine due to the downvote that I received (just after your answer was posted) and that I interpreted coming from you. I found that a bit harsh as there were no comments explaining what I did wrong. I'll edit my answer accordingly to your comment when I find time. Best regards – TheBridge Oct 2 '12 at 20:11
@ Ben Derret: You can apply directly Fatou's Lemma extension that I found on wiki's page to $M^n_{t\wedge \sigma_n}$ to get to the result (you are basically redemonstrating it, though it is a quick thing to do). Best regards. – TheBridge Oct 2 '12 at 20:32
@TheBridge: No worries. – Ben Derrett Oct 3 '12 at 11:50
show 2 more comments

Hi for the first part of your 3rd question unless mistaken, I think that you have (using the elements in the proof of the paper):

$M^n_{\sigma_m\wedge t}\geq X_{\sigma_m\wedge t}-1 -(\Delta X_{\sigma_m\wedge t})^-\geq \theta_m -1 - (m-\theta_m)=2.\theta_m - m -1$

The need to substract the negative part of the jump of $X$ at $\sigma_n$ comes from $\tau_n$'s definition which implies that it is possible that $X$ jumps beyond $M^n_{\sigma_m\wedge t}+1$ at $\sigma_n$ (as pointed out by Ben Derrett).

For the second part, I think that it need not be positive, but the thing is that it is still integrable and I think that it is this hypothesis that is used here in a slight extension of Fatou's lemma which the authors abusively used without mentionning (check wiki's page of Fatou's Lemma).

Edit: Regarding 1 - The more I look at it, the more I think that the place where implicitly the Borel-Cantelli's lemma is used is when the authors define $\sigma_n$ where an $inf_{m>n}\tau_m$ is involved. So the fact that $\sigma_n$ is inferior to $T$ a.s. is coming from the fact that $inf_{m>n}\tau_m<T$ for some n>0 from Borel-Cantelli's Lemma as $P(limsup_n E_n)=0$ where $E_n=\{\tau_n<T\}$. Otherwise said it means that almost surely for a fixed $\omega$ there exists a $p>n$ such that $inf_{m>n}\tau_m(\omega)=\tau_p(\omega)$.

Edit two: Now point 2.Please check it carefully.

Suppose that we are on the event that $X$ has a strictly negative jump at $\sigma_n$, then we have $X_{\sigma_n-}<n$ (due to $\tau_n$'s definition and to the fact that $\sigma_n\leq \tau_n$), and on the other hand $X_{\sigma_n}>\theta_n$ by point (ii).So :

$\Delta X_{\sigma_n}^-=X_{\sigma_n-}-X_{\sigma_n}\leq n-\theta_n$

The case $\Delta X_{\sigma_n}^-=0$ doesn't matter.

Best regards

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1) Unfortunately, your first inequality doesn't hold quite as stated (hence the necessity of the $m-\theta_m$ term), since there may be a jump in $X$ at time $\sigma_m\wedge t$. 2) We define $\Delta X_{\sigma_n} = X_{\sigma_n}-X_{\sigma_n-}$. Hope that helps. :) – Ben Derrett Oct 2 '12 at 15:50
@ Ben Derrett : Hi you are right about 1) I misunderstood $\tau_n$ definition, I edit the first part of my answer accordingly. – TheBridge Oct 2 '12 at 20:16
Thank you so much for your help! I'm sorry for the late response, I had a busy week. Back to my question: There still some small points which are not entirely clear to me. regarding 1): What do you mean by $\sigma_n$ is inferior to $T$? 2) Why do I have $X_{\sigma_n-}< n$ From the definition of $\tau_n$ I have $X_{\tau_n}\ge n$ by right continuity, or am I wrong? 3) Sorry I really don't get the first inequality: $M_{\sigma_m\wedge t}^n\ge X_{\sigma_m\wedge t}-1-(\Delta X_{\sigma_m\wedge t})^-$. – hulik Oct 8 '12 at 13:32
Actually I even do not understand why $M^n_t\ge X_t-1$ for $n\ge m$ and $t<\sigma_m$ – hulik Oct 8 '12 at 13:35
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