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Unformatted text preview: guess satisﬁes the detailed balance condition, we note that
= ⇡ (x + 1)p(x + 1, x)
(x + 1)!(n x 1)!
n ⇡ (x)p(x, x + 1) = 2 n However the following proof in words is simpler. Create X0 by ﬂipping coins,
with heads = “in the left urn.” The transition from X0 to X1 is done by
picking a coin at random and then ﬂipping it over. It should be clear that
all 2n outcomes of the coin tosses at time 1 are equally likely, so X1 has the
Example 1.32. Three machines, one repairman. Suppose that an o ce
has three machines that each break with probability .1 each day, but when there
is at least one broken, then with probability 0.5 the repairman can ﬁx one of
them for use the next day. If we ignore the possibility of two machines breaking
on the same day, then the number of working machines can be modeled as a
birth and death chain with the following transition matrix:
.7 Rows 0 and 3 are easy to see. To expl...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
- Spring '10
- The Land