Talk:Martingale central limit theorem

There should be made an index of all the central limit theorems which, among others, include this one. --Steffen Grønneberg 20:41, 21 May 2006 (UTC)

something unclear

There is something not clear :

\tau_v is define as a min. But \sigma_t is a random variable for all t, so the definition is ambiguous. Also the conclusion of the theorem seems a little too good, because there are at the end no condition on the conditional variance. On the other hand, I did not find the theorem in the book you mention as reference. —Preceding unsigned comment added by Le huve (talk • contribs) 12:39, 17 September 2007 (UTC)

Let $D_{t}=X_{t+1}-X_{t}\,$. Then $X_{1},D_{1},\dots D_{t-1}\,$ can be calculated from $X_{1},X_{2},\dots X_{t}\,$ and vice versa. Specifically, $X_{t}=X_{1}+D_{1}+\dots +D_{t-1}\,$. The conditions become:

$\operatorname {E} [D_{t}\vert X_{1},D_{1},\dots D_{t-1}]=0\,$

and

$|D_{t}|\leq k$.

We have

$\sigma _{t}^{2}=\operatorname {E} [D_{t}^{2}\vert X_{1},D_{1},\ldots ,D_{t-1}]$.

The conditionality of these formulas merely means that each of the random variables D being summed may have a distribution which depends on the values of the previous random variables. Remember we are trying to relax the premises of the central limit theorem. There is no conditionality in the conclusion; we are not trying to weaken the conclusion. JRSpriggs (talk) 00:28, 22 May 2008 (UTC)

another thing is unclear

Then

${\frac {X_{\tau _{v}}}{\sqrt {v}}}$

converges in distribution to the normal distribution with mean 0 and variance 1.

Does that mean as v → ∞? Michael Hardy (talk) 19:01, 20 May 2008 (UTC)

Yes, as I indicated in my edit to the article. JRSpriggs (talk) 00:31, 22 May 2008 (UTC)

Mean value

We do not state, that the first X is zero, hence the resulting distribution is not necesserily centered in zero, or am I mistaken with a concept of CLT? My thought is that the martinagle stabilizes around it's starting point, which should be reflected in the normal distribution's parameter. — Preceding unsigned comment added by 82.200.70.124 (talk) 10:54, 13 December 2012 (UTC)

There should be more hypothesis

if every $X_{t}$ is zero a.s., $(X_{t})_{t}$ is a martingale, then for every $v$>0, $\tau _{v}$ is equal to infinite, and distribution of ${\frac {X_{\tau _{v}}}{\sqrt {v}}}$ is equal to zero a.s.

It should be the same if $\sum _{i\geq 0}\sigma _{i}^{2}$ is finite. — Preceding unsigned comment added by 195.83.48.116 (talk) 12:35, 16 October 2013 (UTC)