# Re: Markov Chains II

*From*: "Proginoskes" <CCHeckman@xxxxxxxxx>*Date*: 26 Oct 2006 15:45:21 -0700

Mark wrote:

Luis wrote:

HelloYour calculations are correct!

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 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

.

**References**:**Markov Chains II***From:*Luis

**Re: Markov Chains II***From:*Proginoskes

**Re: Markov Chains II***From:*Luis

**Re: Markov Chains II***From:*Mark

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