Subject: Re: homework 2 solutions

From: <[email-masked]>
Date: Thu, 18 Jan 2007 16:03:31 -0800 (PST)

This may be an incredibly minor thing, but in the solution to 4e, we are
assuming that the probability that the particle doesn't reach either -m
or n is zero. I am perfectly willing to believe this is true as this is
a finite region, but its proof seems slightly non-trivial at least. I
assume we are not going to require the students to prove it. I just
wanted to double check this.

-Dan

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From: "Gary Lorden" <[email-masked]>
Date: Thu, 18 Jan 2007 21:38:27 -0800

Good point, Dan.   I should have said something about that issue in
the problem.   They're not ready to prove it.  The easiest way is to
use the Strong Law of Large Numbers, which implies that the
probability is one that the particle moves to minus infinity (since p
is less than one-half).

G.L.

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From: Todor Tsankov <[email-masked]>
Date: Fri, 19 Jan 2007 11:35:41 -0800

There is an easier way to see this (easier = not requiring anything the 
students don't know). The probability of never having m+n+1 consecutive 
moves to right is smaller than (1 - p^{m+n+1})^k for each k and 
therefore equal to 0. This approach also has the advantage that it works 
for p = 1/2 as well.

Todor

-------- Original Message --------
Subject: homework 2 solutions
From: Gary Lorden <[email-masked]>
To: [email-masked] <[email-masked]>
Date: Thursday, January 18, 2007 9:38:27 PM

> On 1/18/07, [email-masked] <mailto:[email-masked]>* 
> <[email-masked] <mailto:[email-masked]>> wrote:
>

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From: "Gary Lorden" <[email-masked]>
Date: Fri, 19 Jan 2007 12:55:29 -0800

Nice, Todor.  But let's not require them to comment on that issue!

Your argument is a special case of a "famous" one used to show that
a classic "sequential hypothesis test" can't go on forever.  To get your
bound, as in the famous argument, requires looking at successive
blocks of m+n+1 steps and arguing that events occuring (or not) within
those blocks are independent.   Looking at the first k blocks gives your
bound.

If you'd made this argument (a little more generally, but the same idea)
circa 1949 before a mathematician named Charles Stein did it, it would be
known as "Tsankov's Lemma".

G.L.