Talk:Local martingale

F a filtration of itself?
The text currently says that "F = { ... } is a filtration of  F". Should those Fs be different? Which one is referred to in the rest of the article? LachlanA (talk) 22:01, 2 June 2008 (UTC)


 * F is a &sigma;-algebra. For each t, Ft is a &sigma;-subalgebra of F.  The collection of all such Ft's is denoted F&lowast;.  I like to think of the &ldquo;&lowast;&rdquo; as saying &ldquo;is a function of t, but we're not saying any particular value of t&rdquo;.  In other words, F and each Ft are &sigma;-algebras, but the filtration F&lowast; is a collection of &sigma;-algebras.  The rest of the article uses exactly this notation.  Sullivan.t.j (talk) 10:25, 3 June 2008 (UTC)

Martingale in a measurable space?
After the localization we still have a process with values in a measurable space, which in general is not a linear space. Then, what do we mean by a martingale? Boris Tsirelson (talk) 16:57, 6 October 2008 (UTC)

Having no answer, I delete these "values in a measurable space". Also I do some effort for making the article less technical, and remove the "too technical" flag. Boris Tsirelson (talk) 09:28, 8 October 2008 (UTC)

Readability/Compatibility
The "lowast" symbol renders as a rectangle in MSIE6, a typical rendering of nonexistent characters in Windows. It might make more sense to use TeX there...something along the lines of $$F_*$$, rather than F&lowast;. It shows up fine in Firefox 3.0.3. Just a thought. —Preceding unsigned comment added by 70.234.255.207 (talk) 16:26, 13 November 2008 (UTC)

Divergence of localisation times in Example 1
The times $$\tau_k$$ in Example 1 seem to be bounded above by 1. This seems to contradict the requirement that the localisation times diverge. Could someone who understands this please clarify it? LachlanA (talk) 06:27, 27 May 2009 (UTC)


 * They are not bounded above by 1, since sometimes they are infinite. Indeed, the stopped Wiener process in bounded a.s. If k exceeds the (random) maximum of the stopped Wiener process then $$\tau_k$$ is infinite by the convention that minimum of an empty set is plus infinity. But you are right in the sense that the article is not clear enough at this point. Boris Tsirelson (talk) 08:19, 27 May 2009 (UTC)

In the current definition of the stopping times of Example 1, shouldn't it be $$X_t$$ instead of $$W_t$$? —Preceding unsigned comment added by 62.141.176.1 (talk) 12:50, 16 November 2009 (UTC)


 * It was $$X_t$$ in my text; then it was changed to $$W_t$$ by 213.79.71.65 at 11:40, 2 November 2009. My first impression was that it became erroneous. However, then I have realized that it depends on details of definitions. If we construct $$X_t$$ from $$W_t$$ and then abandon $$W_t$$ and keep only the distribution of $$X_t$$, then of course we should use $$X_t$$. But if we treat $$X_t$$ as a function of $$W_t$$, living on the filtered probability space of $$W_t$$, then writing $$W_t$$ is acceptable. Anyway, the version with $$X_t$$ is easier to understand (I think so). Thus, feel free to restore $$X_t$$. Boris Tsirelson (talk) 15:46, 16 November 2009 (UTC)

Either there should be added a proof why $$E ( X_{\tau_k} ) = 0 $$ or the example should be removed it is very unclear. —Preceding unsigned comment added by 77.11.6.228 (talk) 19:29, 31 May 2010 (UTC)


 * More explanations are added. Boris Tsirelson (talk) 15:02, 1 June 2010 (UTC)
 * Here is an alternative explanation for why the process $$X_t$$ stopped at $$\tau_k$$ is a martingale. If $$X_t$$ hits $k$ then $$X_{t\wedge\tau_k}$$ is a stopped Wiener process. If $$X_t$$ doesn’t hit $$k$$, this means that as $$k\to\infty$$, $$X_t$$ remains $$<k$$. Therefore $$X_t$$ equals the stopped Wiener process $$W_T$$ a.s. and is a stopped martingale. Ofb (talk) 17:46, 30 May 2024 (UTC)

Why is the process in example 1 a local martingale?
Let's say one of my stopping times is t=2.5. If I take the conditional expectation E[X(2)|F(0.1)] (I mean the conditional expectation of the stopped process X(stopped in t=2.5) in time 2 given the Filtration in time 0.1) I do not get X(2) but -1, don't I ? Or do I not "know" at time 0.1 what is about to happen to the process after t=1? — Preceding unsigned comment added by 131.220.46.116 (talk) 18:47, 7 July 2011 (UTC)


 * No, this cannot happen. $$\tau_2$$ cannot be 2.5 since it never exceeds 2. $$\tau_3$$ cannot be 2.5 since the process X does not move after the instant 1. Either it reaches 3 before this instant, and then $$\tau_3<1,$$ or it never reaches 3, and then $$\tau_3=3.$$ Boris Tsirelson (talk) 19:28, 7 July 2011 (UTC)


 * Thanks for your answer! Let's say then that $$\tau_3=3.$$. If i stop the then take the conditional expectation E[X(2)|F(0.1)](notation as above, (sorry i dont knoiw much about editing)), I still don't get X(2) as result, do I? Or do I not "know" at time 0.1 what is about to happen to the process after t=1? — Preceding unsigned comment added by 131.220.74.16 (talk) 08:14, 8 July 2011 (UTC)


 * Well, $$\tau_3=3$$ means that the Brownian motion reaches -1 before reaching 3. Also the value of B at a time close to 0.1 is given. It is possible to condition this way, but somewhat untypical: you know something from the past, and also something from the future! But anyway, X(2)=-1 always, therefore the conditional expectation of X(2) is -1, no matter what is the condition. So, what do you really want to say by all that? Boris Tsirelson (talk) 10:24, 8 July 2011 (UTC)


 * Ah, yes, it seems, now I understand what do you really want to say by all that! Yes, it is a good question, and the answer should be instructive.


 * First, I reformulate the question, and I replace your 0.1 with just 0; 0.1 is an unneeded complication.


 * The stopped process $$X_t^{\tau_3} = X_{\min \{ t, \tau_3 \}}$$ must be a martingale. Its initial value is 0. Therefore its expected value at t=2 must be 0. However, its (random) value at t=2 is $$X_{\min \{ 2, \tau_3 \}} = X_2 = -1$$ whenever $$\tau_3=3.$$


 * However, $$\tau_3$$ need not be 3. There is a probability, denote it p3, that the Brownian motion hits 3 before hitting -1. In this case $$\tau_3<1,$$ $$X_{\tau_3}=3,$$ and $$X_{\min \{ 2, \tau_3 \}} = X_{\tau_3}=3.$$ In the other case, with probability 1-p3, the Brownian motion hits -1 before hitting 3; then $$\tau_3=3$$ and indeed, $$X_{\min \{ 2, \tau_3 \}} = -1.$$ Thus, $$ \mathbb{E} X_{\min \{ 2, \tau_3 \}} = 3 p_3 - (1-p_3) = 4p_3 - 1.$$ Can it be 0? Sure, it is, provided that p3 = 1/4. And this is really the case!


 * Boris Tsirelson (talk) 11:43, 8 July 2011 (UTC)


 * Hi, I just wanted to thank you for your explanations in july! It was a great help in preparing for my finance exam! — Preceding unsigned comment added by 178.6.194.50 (talk) 13:12, 13 November 2011 (UTC)


 * You are welcome. Have your finance! :-) Boris Tsirelson (talk) 13:41, 13 November 2011 (UTC)


 * Can it be that the explanation of Boris Tsirelson above explains the Martingale property for $$ \mathbb{E}[X^{\tau_n}_t | F_s ] $$ for all $$ t $$ greater AND equal to 1, and the explanation in the article for $$ t>1 $$ wrong or misleading? — Preceding unsigned comment added by Manalysis (talk • contribs) 09:52, 22 May 2015 (UTC)


 * In principle, everything can be, but what is really the problem? Which argument in the chain of arguments given in the article (mostly, in "Technical detail no. 1") looks wrong or misleading, and why? Boris Tsirelson (talk) 19:26, 22 May 2015 (UTC)