Re: positive, not uniformly integrable, martingales

On Sep 16, 2:20 pm, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
David C. Ullrich <dullr...@xxxxxxxxxxx> writes:



On Mon, 15 Sep 2008 15:28:41 -0700 (PDT), munuomic...@xxxxxxxxx wrote:

Hi,

Suppose you have a positive martingale, M_n, whose mean is one, but
which isn't uniformly integrable, and tends to zero almost surely as n
goes to infinity:

[i]
E(M_n) = 1 but M_n ---> 0 a.s.

It seems that such martingales experience rare "explosions" (M_n very
large) which as it were reconcile these above two statements.  Does
anyone know of theorems (or obvious results :-) ) that would provide
lower bounds on the probability of large M_n,

[ii]
P ( M_n > something, perhaps a function of n ) >= something else...?

It seems clear that you can't get an inequality like this with
_explicit_ values of the two somethings from the hypotheses
you give - you'd require something quantitative about the rate
of convergence to 0. (For example, if (M_n) satisfies [i]
then so does the martingale consisting of M_1 repeated
as many times as you want, then M_2 repeated as many
tiems as you want, etc. Any explicit version of [ii[ will
fail if you through in enough repetition...)

On the other hand, you might hope that explicit bounds ensuring
that M_n -> 0 in probability really quickly might help.  That does
not seem to be the case either.  Of course, you do have this:

1 = E[M] = int_0^infty P(M > t) dt

So if g(t) is any positive function with int_0^infty g(t) dt < 1,
we know that P(M_n > t) > g(t) for some t (possibly depending on n).

Moreover, if we take 1/2 > epsilon > 0 and delta > 0 such that
P(M_n > epsilon) < delta, then
int_0^epsilon P(M_n > t) dt <= epsilon
int_epsilon^(1/(2 delta)) P(M_n > t) dt < 1/2
so that
int_(1/(2 delta))^infty P(M_n > t) dt > 1/2 - epsilon
and thus if g(t) > 0 with int_{1/(2 delta))^infty g(t) dt < 1/2 - epsilon,
we must have P(M_n > t) > g(t) for some t > 1/(2 delta).  But there is
no way to obtain an explicit t for which this will be true.
--
Robert Israel              isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada


Thanks for these responses. Robert, I guess to make your second bit
operational, you need to be able to find some epsilon and associated
(small) delta; but because epsilon is on the "wrong" side of the mean
to make this easy, it's not so clear how to do this in a vaguely
general case.

As an alternative, it occurred to me that another way of attacking the
problem might be to consider the "maximum-to-date" process

M*_n = max_m<=n M_m

We know that E[M*_infty] = infty because otherwise M* would dominate M
and M would be uniformly integrable. Is there maybe something more we
can say about a lower bound probability that M*_n, or M*_infty, is
greater than some (large) number ? Maybe there's some nice inequality
for increasing processes that can be wheeled out...

Thanks again! :-)
MNO
.

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