Markov Chains and absorption probabilites


by macca1994
Tags: absorption, chains, markov, markov chains, probabilites
macca1994
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#1
Feb4-13, 10:29 AM
P: 19
Could someone please help me with this question?

A single-celled organism contains N particles, some of which are of type A, the others of
type B . The cell is said to be in state i where 0<=i<=N if it contains exactly i particles
of type A. Daughter cells are formed by cell division, but rst each particle replicates itself;
the daughter cell inherits N particles chosen at random from the 2i particles of type A
and 2N-2i of type B in the parent cell.

Find the absorption probabilities and expected times to absorption for the case N = 3.


I so far have that the absorbing states are i=0, i=3 but have no idea where to go from there
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mfb
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#2
Feb4-13, 01:24 PM
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For N=3, you can calculate the transition matrix manually. Many entries are 0, and some others follow from symmetry, so you just need 2 interesting entries.
macca1994
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#3
Feb4-13, 04:16 PM
P: 19
how do i calculate the entries though, thats where i'm stuck at the moment, i know of course the lines for starting in state 0 and 3, but have no clue about 1 or 2, once i know that the rest of the question becomes fairly trivial, could you push me in the right direction?

mfb
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#4
Feb4-13, 04:47 PM
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Markov Chains and absorption probabilites


i=1 leads to AABBBB in the cell before splitting. If you randomly pick 3 of them, what is the probability of getting 0 (,1,2,3) times A?
macca1994
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#5
Feb4-13, 04:59 PM
P: 19
oh is that standard binomial? so probability of going from state 1 to 0 would be (2/3)^3 which is 8/27 then do the same for the other states? or am i missing something?
macca1994
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#6
Feb4-13, 05:03 PM
P: 19
i really don't understand the probabilities of getting to the other states, do i not need to also consider what the other cell will contain or is that irrelevant?
macca1994
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#7
Feb5-13, 02:28 AM
P: 19
I think i finally get it, so probability of 0 A's is equal to
(2/3)*(3/5)*(1/2) which is the probability of selecting a B each time
Then follow the same method for 1 A taking into account whether you chose the A first, second or third? I hope thats right
mfb
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#8
Feb5-13, 02:18 PM
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P: 10,840
That is correct.
macca1994
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#9
Feb5-13, 02:20 PM
P: 19
Thanks for the help
Viper7593
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#10
Feb19-14, 04:16 PM
P: 2
I also have difficulty in this question.I have calculated that the probabilty of getting 0 'a's is 1/5...probability of getting 1 'a' is 3/5 and probability of getting 2 'a' is 1/5. What is the transition matrix and what are the absorption probabilities?
thank you
Ray Vickson
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#11
Feb20-14, 12:36 AM
HW Helper
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P: 4,673
Quote Quote by macca1994 View Post
I think i finally get it, so probability of 0 A's is equal to
(2/3)*(3/5)*(1/2) which is the probability of selecting a B each time
Then follow the same method for 1 A taking into account whether you chose the A first, second or third? I hope thats right
If you start with AABBBB and pick three at random, you are looking at the "hypergeometric distribution", which is the probability for choosing k of type A from a population of 2A and 4B, when you choose three altogether. See, eg., http://en.wikipedia.org/wiki/Hyperge...c_distribution or http://mathworld.wolfram.com/Hyperge...tribution.html .
Viper7593
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#12
Feb21-14, 02:39 AM
P: 2
thank you..i have already found the probabilities but how to find the transition matrix please?
Ray Vickson
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#13
Feb21-14, 03:17 AM
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P: 4,673
Quote Quote by Viper7593 View Post
thank you..i have already found the probabilities but how to find the transition matrix please?
Work it out for yourself. If you start in state i = 1, what are the possible next states you can reach in one step? What are the probabilities of going to those various states in one step? Once you have answered those questions you will have worked out what is the i = 1 row of the transition matrix. You get the other rows in a similar way. There are no shortcuts; you have to sit down and do it all, step-by-step; and you will only learn how by doing it yourself.


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