Subject: Re: sheet 3



-> 
-> I'm having soem trouble with problem 3.1 on sheet 3.  In part i), I can
-> see that we are summing the different ways to get from i to j (through the
-> fnct G(i,j)) I also recognize the given summation  (ie. Sum
-> y(i)P(i,j)=y(j)) as one of the definitions of a stationary dist.  However,
-> I still don't know where the 2^-n power is coming from in the G(i,j)....
-> 

The 2^{-n} weights are needed to stop the sum from blowing up.  Don't
try to interpret G probabilistically.  Just think of it as a function
that tells you when the chain can get from state i to state j.

-> On part iii)I see that the expectation must be finite, because if we went
-> from some jo, to state i, and visited i an infinite number of times
-> without going back to jo, then jo would have to be transient.  Is this an
-> acceptable method of proving that i must be visited a finite number of
-> times?  

An infinite expected number might not force a non-return for any
particular realization. cf. null recurrence

You should be able to get the result directly from part (ii).

DP

Date: Mon, 22 Feb 1999 12:30:46 -0500

Subject: Re: 6.1.1
-> 
-> is it true?  ex: (1 2 3) -> (2 3 1) but not the other way.
-> 

I think you are right.  The stationary distribution is uniform, but my
suggestion won't get you there.

Thanks.

DP

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Better hint: Show that it is enought to prove \sum_s P(s,t) = 1 for
each t.
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