Re: Probability of exceeding a specific value

On Sep 26, 10:02 pm, matt271829-n...@xxxxxxxxxxx wrote:
On Sep 23, 1:25 am, David Bernier <david...@xxxxxxxxxxxx> wrote:





matt271829-n...@xxxxxxxxxxx wrote:
On Sep 20, 1:37 am, David Bernier <david...@xxxxxxxxxxxx> wrote:
matt271829-n...@xxxxxxxxxxx wrote:
On Sep 19, 9:54 pm, matt271829-n...@xxxxxxxxxxx wrote:
On Sep 19, 2:27 pm, David Bernier <david...@xxxxxxxxxxxx> wrote:
[snip]
An interesting variation is the distribution (pdf) for the first time
a random walker returns to his starting place, on Z.
The expectation was said to be infinite. The formulation
was equivalent to a random walk on Z, but formulated as follows:
"Suppose we toss a fair coin. Let T ( positive) be first
toss number when we have equal numbers of heads and
tails." Clearly, T is even. It was mentioned that E(T) = oo.
I don't know the distribution of T, but the probability
of n heads and n tails in 2n tosses is C(2n, n)* 2^(-2n).
The pdf of T=2n is just A(2n)/(2^(2n)) , where
A(2n) is the number of random walk paths of length
2n steps where the starting point was visited only
once.
For this one it looks like the pdf of the first time to return to the
origin is, for even t,
C(t, t/2) / ((t - 1)*2^t)
and obviously zero for odd t.
It seems that you are certain to eventually return, but, as you say,
the expected time to do so is infinite.
Here's a tricky one: starting from the origin, what is the
distribution of the first time to return to the origin in the
corresponding continuous (Brownian motion) case? (Note: *return* to
the origin.)
The probabilist Paul Levy studied Brownian motion in the 1930's - 40's.
He found two CDF's
involving the arcsine function. The second arcsine law (theorem) concerns
Brownian motion stopped at time 1. The probability of being at 0 at
time 1 is zero.
So there was a last time T when the motion or particle was at 0.

The cdf for the random variable T ( 0<=T<=1) has the intriguing form:

Prob(T< u) = 2/pi *arcsine( sqrt(u) ), 0<=u<=1 .

So Prob (T<1) = 1.
This means that a return visit to zero is sure to happen between t=0 and
t=1.

Brownian motion from t=0 to t=1/2 is shaped like Brownian motion from
t=0 to t=1, with dilation
by sqrt(2) in the vertical direction because the variance at time t is
t. So again,
the probability of a return visit to zero happening between t=0 and
t=1/2 is 1.

One can divide by 2 any finite number of times. So with probability 1,
there is a return to the origin in the time frame [0, epsilon] for any
epsilon>0.

Yes, that was exactly my conclusion, arrived at in a slightly
different way. There seems, therefore, to be a paradox here concerning
the distribution of the first time of return to the origin (hence my
question).

My explanation is that, with probability 1, there is no first first time
of return to the
origin.

Yes, I think you're right. It's kind of like asking for the smallest
number greater than zero.





Rather, if S = {t: 0<t<=1 and X(t) = 0}, where X(t)_{t in
[0,1]} is a Brownian process, then
with probability 1, infimum(S) = 0. X(t) can be thought of as one
Gaussian per value of t, but
where, for example, X(1/3) and X(1/2) are not independent. There is an
underlying
sigma-algebra and measure, the Wiener measure. The existence and properties
of Wiener measure is not a result whose proof I've seen. I remember a
theorem
about the zero-set of X(t) (t in [0,1]): with probability 1, this set
is a perfect set in
[0,1]. It's a closed subset of [0,1], and there are no isolated points.
It would not surprise me if with probability 1 the zero set has empty
interior.

Also, if X(1)>0, then Prob( measure({t: X(t)>0}) > 1/2 ) is quite
high. In coin-tossing,
one can look at the leader when the game is stopped, and then the
fraction of the
tosses where the leader was ahead. So this fraction is > 1/2 with high
probability.

If you're looking at the limit as the number of tosses goes to
infinity, then I would hazard a guess that this probability is one.
(That's assuming we're not factoring in any information about how far
ahead the leader is, of course... all we know is that he's ahead.) I
would hazard another guess that the probability is also one if 1/2 in
the question is replaced by any number less than one. Unfortunately
simulations aren't much good for nailing this. Probably I should try
to work it out...!

For the record, I now think that this guess is wrong. Let n be the
number of tosses, and let player A be the leader after n tosses. (As
we're letting n go to infinity we can ignore the possibility that the
players are tied after n tosses.) We want to find the probability that
player A was ahead for at least n*r tosses during the course of the
game, where r is some fraction between zero and one (r = 1/2 in your
example).

At the moment I can't see a tractable way to compute this probability
for large n. However, let x be the number of *consecutive* tosses,
leading up to and including the n'th, at which player A was ahead.
Assuming n is even, I think that the probability that x/n >= r is,
deep breath,

2/(2^n - C(n, n/2)) *
sum k = 2, 4, 6 ... n
sum m = 1, 2, 3 ... min((n - k)/2, floor(n*(1 - r)/2))
k * C(2*m, m) * C(n - 2*m - 1, (n - k)/2 - m) / ((n + k)/2 - m)

For r = 1/2 it looks from numerical results as if this might go to 1/2
as n goes to infinity.

I think the only way my original guess could be correct is if this
final run of consecutive tosses at which A is leading came to dominate
as n -> oo. That is what I initially thought, but the results from the
above formula suggest that it doesn't.

.

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