--------------------------------------------------------------------------------- Emails sent to Section 3 on Tues/Wed, 4/5 December 2007. The correction is given first, and the original (first) email is given below. --------------------------------------------------------------------------------- Hello again, The previously mentioned error in the text actually was misinterpreted (by me, your all-knowing, infallible instructor!) and can still be used to establish that a matrix is not regular. Here's the original from the text: "If a transition matrix P has some zero entries, and P^2 does as well, you may wonder how far you must compute P^k to be certain that the matrix is not regular. The answer is that if zeros occur in the identical places in both P^k and P^(k+1) for any k, they will appear in those places for all higher powers of P, so P is not regular." The CORRECT way to interpret this is that "...if zeros occur in ALL the identical places in both P^k and P^(k+1) for any k, they will appear in those places for all higher powers of P, so P is not regular." For example, the matrix I gave in the last email (see below) P = .5 .5 0 0 0 0 1 0 0 0 0 1 1 0 0 0 which gives P^2= .25 .25 .5 0 0 0 0 1 1 0 0 0 .5 .5 0 0 we see that not all zeros are in the same locations in both matrices (e.g. the top of column 3). Therefore the statement doesn't apply and we simply need to look at higher powers of P to determine if P is regular (all entries become non-zero) or not (we reach some P^k, P^(k+1) where we can apply the statement correctly). Sorry for the confusion, but hopefully this all makes a little more sense now! :) Take care, Paul --------------------------------------------------------------------------------- --------------------------------------------------------------------------------- Hi folks, I just found a big error in the text regarding the ch 10 Markov Chain material. So prepare your brain to do some "re-learning"! ;) The note on p519 at the start of the "Regular Markov Chains" section (10.2) reads: "If a transition matrix P has some zero entries, and P^2 does as well, you may wonder how far you must compute P^k to be certain that the matrix is not regular. The answer is that if zeros occur in the identical places in both P^k and P^(k+1) for any k, they will appear in those places for all higher powers of P, so P is not regular." BUT, this is NOT true!!! :( Consider a markov chain with N states where the transition matrix looks like (here N=4) P = .5 .5 0 0 0 0 1 0 0 0 0 1 1 0 0 0 That is, things stay in state 1 or move to state 2 with probability 1/2 and 1/2. After that, things move up 1 state with probability 1, and for the last state the process returns to the first state with probability 1. For this matrix, P and P^2 have zeros in the last column in the second row. You'd think that this would mean the matrix is irregular (using the "note"), however you have to go out to P^(2N-2) (P^6 in this case) before you see that entry stay non-zero, and before you see that P is regular (that is, for the above 4x4 matrix P, P^6 is the lowest power of P that has all non-zero entries). BUMMER! That means to show something isn't regular you have to check LOTS of powers of P before you can say it is irregular :( . So what does it all mean in terms of the final? Well, first off you should feel obliged to erase that useless little trick from your brain! It doesn't always work!!! For determining if matrices are regular (or not), well, know the definition of regularity, the consequences of a transition matrix being regular, and of course how to multiply matrices together to find P^2, P^3, etc! ;) (DON'T WORRY, we don't like giving you too much "busy work" on exams, so you wouldn't be asked to check too many powers of matrices.) Happy Studying, Paul