From pollard@euler Tue Jan 26 11:34:58 1999 Return-Path: Date: Tue, 26 Jan 1999 11:32:24 -0500 -> I have class during your office hours, but I had some problems with the -> homework. If you could give me a nudge in the right direction, I'd be -> indebted. -> -> 2.1) iii) We are told to equate coeffients of the powers of s. I'm sorry, -> but I don't understand which equations we are supposed to be using! -> You should have two power series expansions for F(s). If you equate coefficients of corresponding powers in the two series you should get an expression for f_i. -> 2.3) I first thought of doing the problem with a matrix and using -> gaussian elimination to find the respective values of the stationary -> distribution. However, that won't be possible in an infinite state walk -> on the integers. Should we consider three different chains (ie. i<0, i=0, -> i>0)? Intuitively, the chains for i<0 and i>0 are transient, because they -> would always push towards state 0. -> There is an obvious symmetry between i>0 and i<0, but you need to consider them as a single chain. You might try to express each \pi_i as a combination of \pi_1 and \pi_{-1}, by repeated substitution. Life is easier if you can justify setting \pi_1 equal to \pi_{-1}. Don't forget that the probabilities must be nonnegative and they must sum to one. DP From pollard@euler Wed Feb 3 10:52:30 1999 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 ************************** Better hint: Show that it is enought to prove \sum_s P(s,t) = 1 for each t. ************************** From pollard@euler Fri Feb 26 10:21:25 1999 -> I think with positive probability all balls become BLACK, and under -> this condition T=infinity, so P(T Do you mean T is the stopping time when balls become either all white or -> all black? -> Yes. Take T as the time until all white or all black. You could work with T as I defined it, but then {T= infinity} would correspond to {eventually all black}. Thanks for the correction. DP From pollard Sun Mar 21 18:16:46 1999 Return-Path: Date: Sun, 21 Mar 1999 18:16:45 -0500 From: pollard (David Pollard) To: masahiro.watanabe@yale.edu, kenji.obata@yale.edu, duy.tu@yale.edu Subject: Re: Questions about Stat 551 Prob (8.1) Content-Length: 2637 Guys, My suggestion for the case \sum_i q_i = 0 was pure nonsense. The fact that I had negative components in q for my made up example made it too easy--I gave you an arbitrage opportunity even without the option. (I was focussing just on the option.) I have now revised the question, putting a corrected version on the web site. This time you really do get a martingale measure. I also adjusted T_1 to get nonegative stock prices. Thanks for the comments. I knew I was missing something. My blunder should remind me that a consistent set of prices have to come from some scheme of expectations. DP ****************************** Dear Professor Pollard: How are you? I have a few questions about problem (8.1) in Stat 551. (1) Part (iv). I'm afraid there does not exist a MG measure for given vectors alpha and beta. Specifically, the q vector that satisfies the conditions in part (i) is parallel to (1, 2, -1), at least one element of which will be strictly negative. (2) Again part (iv). Besides the above, the given beta gives T1 = (1, 1, 1) + (3, -2, -1) = (4, -1, 0), where in the second state the price will be negative? (We could make such a security. But that will not be called 'a stock,' which is characterized by limited liability. Kind of trivial mathematically, though.) (3) The numbers xi's are payoff or prices, not rate of 'return,' right? (Like alpha i's and beta i's) Masahiro Watanabe masahiro.watanabe@yale.edu > ps. I am a little worried about question 1. I haven't had to prove > one of the standard facts, which makes me think I have forgotten > something. Please let me know if the problem is not working as advertised. > 8.1.2 - i think you need to add that {\alpha, \beta, (1 1 1)} are LI. kenji Dear Prof Pollard, I believe there's a small mistake in part ii of 8.1. We're asked to show that the vector with components xi - xo must be a linear combination of alpha and beta, with xo defined as in the question. This is not necessarily true in the case xo is defined as 0 if the sum of qi's = 0. For instance, say q=(1,-1,0) and suppose alpha and beta lie in the plane perpendicular to q. Thus q satisfies (i), and the sum of the components of q is zero. In this case, xo is by definition 0. However, this means that the vector components xi - xo are simply xi. Take (x1,x2,x3)=(1,0,0). This vector is not perpendicular to q and hence does not lie in the plane defined by alpha and beta. Therefore it cannot be written as a linear combination of alpha and beta. I'm not sure what should be done, then, in the case where the components of q add up to zero. Sincerely, Duy