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Markov Chains and absorption probabilites 
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#1
Feb413, 10:29 AM

P: 19

Could someone please help me with this question?
A singlecelled 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 2N2i 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 


#2
Feb413, 01:24 PM

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P: 12,055

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.



#3
Feb413, 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?



#4
Feb413, 04:47 PM

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P: 12,055

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?



#5
Feb413, 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?



#6
Feb413, 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?



#7
Feb513, 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 


#8
Feb513, 02:18 PM

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P: 12,055

That is correct.



#9
Feb513, 02:20 PM

P: 19

Thanks for the help



#10
Feb1914, 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 


#11
Feb2014, 12:36 AM

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#12
Feb2114, 02:39 AM

P: 2

thank you..i have already found the probabilities but how to find the transition matrix please?



#13
Feb2114, 03:17 AM

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P: 5,195




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