Re: Markov Chains II




Mark wrote:
Luis wrote:
Hello

Sorry about my english.

You are right, "practise" is an exam or a short test and "in media" is
"the expected number of".
I considered six differents states E(i) = " i quizzes on the desk "
i = 0..5 and the matrix P :
[ 0 1 0 0 0 0 ]
[ 1/3 0 2/3 0 0 0 ]
[ 1/3 0 0 2/3 0 0 ]
[ 1/3 0 0 0 2/3 0 ]
[ 1/3 0 0 0 0 2/3 ]
[ 1 0 0 0 0 0 ]

and I calculated v P = v , v = [ p(0), p(1), ..., p(5) ]
with p(0)+p(1)+...+p(5)=1

After that, the solution is 0*p(0) + 1*p(1) + 2*p(2) +...+ 5*p(5)

Thank you very much for everything.

Your calculations are correct!
Your modelling may be questioned. The random variable whose expected
value you are computing is measured at a specific time every day. For
instance if that time is after the arrival of the new quiz but before
the teacher corrects & returns any quizzes, then there are 5 different
states i.e. there is no E(0). In this scenario, I calculated that the
expected number of quizzes as 473/211

That looks like the number I got; I recognize the 211, at least. 8-)

My transition matrix was

[ 1/3 2/3 0 0 0 ]
[ 1/3 0 2/3 0 0 ]
[ 1/3 0 0 2/3 0 ]
[ 1/3 0 0 0 2/3 ]
[ 1 0 0 0 0 ]

where the states are 0, 1, 2, 3, 4, and represent the number of papers
at the beginning of the day.

--- Christopher Heckman

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