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I am working my way through Cox and Miller's
"The Theory of Stochastic Process" to enhance my
understanding of Markov processes and ergodicity.
I am having a problem solving the first exercise
of chapter 3. Here is the direct statement of the
problem from C&M...
"From a realization of a two-state Markov chain with
p_{00} = p_0 and p_{11} = p_1, a new chain is constructed
by calling the pair 01 state 0 and the pair 10 state 1,
ignoring 00 and 11. Only non-overlapping pairs are
considered. Show that for the new chain,
p_{00} = p_{11} = (1 + p_0 + p_1)^{-1}."
When I try to solve this, this is what I get:
The 'original chain' probabilities are (In C&M's notation,
p_{ab} = probability of moving from state a to state b):
p_{00} = p_0 p_{01} = 1 - p_0
p_{10} = 1 - p_1 p_{11} = p_1
Based on these probabilities, I can calculate the
probability that a pair of states [01] stays [01]...
q_{[01][01]} = p_{00}*p_{11} = p_0*p_1
...that a pair of states [01] changes to [10]...
q_{[01][10]} = p_{01}*p_{10} = (1-p_0)*(1-p_1)
...that a pair of states [10] changes to [01]...
q_{[10][01]} = p_{10}*p_{01} = (1-p_1)*(1-p_0)
...or that a pair of states [10] stays [10].
q_{[10][10]} = p_{11}*p_{00} = p_1*p_0.
I am interpreting the statement of the problem to
mean that, for instance, pair-state [01] is only allowed
transitions to [01] or [10]. If this is the case, in the
'new chain', the probability r_{00}, the proabability
that pair state [01] stays [01] should be...
r_{00} = p_0*p_1 / (p_0*p_1 + (1-p_0)*(1-p_1)),
i.e. the probability of one possibility divided by the
probability of all allowed possibilities.
This does not yield the answer given by
Cox and Miller. Clearly I am misinterpreting the
question; can anyone point me in the direction of
the proper interpretation?
-- Andrew
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<pre wrap=""> I am working my way through Cox and Miller's<br>"The Theory of
Stochastic Process" to enhance my<br>understanding of Markov processes and
ergodicity.<br>I am having a problem solving the first exercise<br>of chapter 3. Here
is the direct statement of the<br>problem from C&M...<br><br> "From a realization
of a two-state Markov chain with<br>p_{00} = p_0 and p_{11} = p_1, a new chain is
constructed<br>by calling the pair 01 state 0 and the pair 10 state 1,<br>ignoring 00
and 11. Only non-overlapping pairs are<br>considered. Show that for the new
chain,<br> p_{00} = p_{11} = (1 + p_0 + p_1)^{-1}."<br><br> When I try to
solve this, this is what I get:<br>The 'original chain' probabilities are (In
C&M's notation,<br>p_{ab} = probability of moving from state a to state b):<br>
p_{00} = p_0 p_{01} = 1 - p_0<br> p_{10} = 1 - p_1 p_{11} =
p_1<br><br> Based on these probabilities, I can calculate the<br>probability that a
pair of states [01] stays [01].
.<br> q_{[01][01]} = p_{00}*p_{11} = p_0*p_1<br>...that a pair of states [01]
changes to [10]...<br> q_{[01][10]} = p_{01}*p_{10} = (1-p_0)*(1-p_1)<br>...that a
pair of states [10] changes to [01]...<br> q_{[10][01]} = p_{10}*p_{01} =
(1-p_1)*(1-p_0)<br>...or that a pair of states [10] stays [10].<br> q_{[10][10]}
= p_{11}*p_{00} = p_1*p_0.<br><br> I am interpreting the statement of the problem
to<br>mean that, for instance, pair-state [01] is only allowed<br>transitions to [01]
or [10]. If this is the case, in the<br>'new chain', the probability r_{00}, the
proabability<br>that pair state [01] stays [01] should be...<br> r_{00} =
p_0*p_1 / (p_0*p_1 + (1-p_0)*(1-p_1)),<br>i.e. the probability of one possibility
divided by the<br>probability of all allowed possibilities.<br><br> This does not
yield the answer given by<br>Cox and Miller. Clearly I am misinterpreting
the<br>question; can anyone point me in the direction of<br>the proper
interpretation?<br><br> -- An
drew
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