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Find the transition probability matrix. Check my answer.

Transition probability matrix for $X_1 = # heads$, *flip heads* $X_2 = # tails$ * flip tails* $X_3 = # heads$Long term probability in Markov ChainsDeterming a transition probability matrixExplicit transition matrixTransition Probability Matrix of Tossing Three coinsProbability of a fair sequence of tosses ending on two successive tails given the first toss was a head?About the transition matrix of Markov ChainsWhat is the Probability that coin is tossed three timesTransition Matrix problemConstruct transition probability matrix for markov chain

0

$begingroup$

A coin is tossed continuously until 2 heads or 2 tails appear respectively. Let the result of first toss is tail. The game is over when we get 2 heads respectively.

Determine the transition probability matrix.

Check my answer is correct or not.

Let $X_n$ denote the number of tail that appear.
Let the state $S=0,1,2$.

So, the transition probability matrix is
$$P=
beginbmatrix
1&0&0\
0&dfrac12&dfrac12\
0&0&1
endbmatrix.
$$

probability stochastic-processes markov-chains

share|cite|improve this question

asked Mar 14 at 6:30

Ongky Denny WijayaOngky Denny Wijaya

4018

$endgroup$

add a comment | 

0

$begingroup$

A coin is tossed continuously until 2 heads or 2 tails appear respectively. Let the result of first toss is tail. The game is over when we get 2 heads respectively.

Determine the transition probability matrix.

Check my answer is correct or not.

Let $X_n$ denote the number of tail that appear.
Let the state $S=0,1,2$.

So, the transition probability matrix is
$$P=
beginbmatrix
1&0&0\
0&dfrac12&dfrac12\
0&0&1
endbmatrix.
$$

probability stochastic-processes markov-chains

share|cite|improve this question

asked Mar 14 at 6:30

Ongky Denny WijayaOngky Denny Wijaya

4018

$endgroup$

add a comment | 

$begingroup$

A coin is tossed continuously until 2 heads or 2 tails appear respectively. Let the result of first toss is tail. The game is over when we get 2 heads respectively.

Determine the transition probability matrix.

Check my answer is correct or not.

Let $X_n$ denote the number of tail that appear.
Let the state $S=0,1,2$.

So, the transition probability matrix is
$$P=
beginbmatrix
1&0&0\
0&dfrac12&dfrac12\
0&0&1
endbmatrix.
$$

probability stochastic-processes markov-chains

share|cite|improve this question

asked Mar 14 at 6:30

Ongky Denny WijayaOngky Denny Wijaya

4018

$endgroup$

share|cite|improve this question

asked Mar 14 at 6:30

Ongky Denny WijayaOngky Denny Wijaya

4018

share|cite|improve this question

asked Mar 14 at 6:30

Ongky Denny WijayaOngky Denny Wijaya

4018

asked Mar 14 at 6:30

Ongky Denny WijayaOngky Denny Wijaya

4018

asked Mar 14 at 6:30

Ongky Denny WijayaOngky Denny Wijaya

4018

2 Answers
2

active

oldest

votes

0

$begingroup$

Notice that $a_i,j$ element in matrix describes the probability of transition from state $i$ to $j$. Your matrix describes process in which '$0$' is absorbing state (which is obviously not true).

If the matrix shall describe number of consecutive tails (and the two tails is absorbing state) then the matrix is as follows:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$

share|cite|improve this answer

answered Mar 14 at 6:41

vermatorvermator

34110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why $P_10=dfrac12$? I think if we have first toss is tail then it is imposibble the number of tail next toss is $0$.
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 6:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Rows/columns of your matrix are states?
  $endgroup$
  – vermator
  Mar 14 at 7:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. It is state.
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 7:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your matrix shall describe number of consecutive tails (which corresponds to your game), or number of tails in general?
  $endgroup$
  – vermator
  Mar 14 at 7:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  number of consecutive tails
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 7:15
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

I think in this case you need to have 4 states: state 2 tails, state tail, state head and state 2 heads which I will call states 1,2,3 and 4 respectively.

$$P=
beginbmatrix
1&0&0&0\
dfrac12&0&dfrac12&0\
0&dfrac12&0&dfrac12\
0&0&0&1
endbmatrix.
$$

If I understand you correctly you keep playing until you get either two heads or two tails. Then, states 1 and 4 are absorbing states.

Then when you are in state 2 (got a tail) then you can jump after the coin flip to either state 1 (two tail flips) or state 3 (one head).

If you start with a tail then your initial state probability vector $pi$ would be

$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$

share|cite|improve this answer

edited Mar 14 at 12:51

answered Mar 14 at 12:45

AhamAham

414

$endgroup$

add a comment | 

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0

$begingroup$

Notice that $a_i,j$ element in matrix describes the probability of transition from state $i$ to $j$. Your matrix describes process in which '$0$' is absorbing state (which is obviously not true).

If the matrix shall describe number of consecutive tails (and the two tails is absorbing state) then the matrix is as follows:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$

share|cite|improve this answer

answered Mar 14 at 6:41

vermatorvermator

34110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why $P_10=dfrac12$? I think if we have first toss is tail then it is imposibble the number of tail next toss is $0$.
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 6:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Rows/columns of your matrix are states?
  $endgroup$
  – vermator
  Mar 14 at 7:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. It is state.
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 7:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your matrix shall describe number of consecutive tails (which corresponds to your game), or number of tails in general?
  $endgroup$
  – vermator
  Mar 14 at 7:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  number of consecutive tails
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 7:15
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

Notice that $a_i,j$ element in matrix describes the probability of transition from state $i$ to $j$. Your matrix describes process in which '$0$' is absorbing state (which is obviously not true).

If the matrix shall describe number of consecutive tails (and the two tails is absorbing state) then the matrix is as follows:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$

share|cite|improve this answer

answered Mar 14 at 6:41

vermatorvermator

34110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why $P_10=dfrac12$? I think if we have first toss is tail then it is imposibble the number of tail next toss is $0$.
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 6:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Rows/columns of your matrix are states?
  $endgroup$
  – vermator
  Mar 14 at 7:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. It is state.
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 7:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your matrix shall describe number of consecutive tails (which corresponds to your game), or number of tails in general?
  $endgroup$
  – vermator
  Mar 14 at 7:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  number of consecutive tails
  $endgroup$
  – Ongky Denny Wijaya
  Mar 14 at 7:15
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Notice that $a_i,j$ element in matrix describes the probability of transition from state $i$ to $j$. Your matrix describes process in which '$0$' is absorbing state (which is obviously not true).

If the matrix shall describe number of consecutive tails (and the two tails is absorbing state) then the matrix is as follows:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$

share|cite|improve this answer

answered Mar 14 at 6:41

vermatorvermator

34110

$endgroup$

share|cite|improve this answer

answered Mar 14 at 6:41

vermatorvermator

34110

answered Mar 14 at 6:41

vermatorvermator

34110

answered Mar 14 at 6:41

vermatorvermator

34110

1

$begingroup$

I think in this case you need to have 4 states: state 2 tails, state tail, state head and state 2 heads which I will call states 1,2,3 and 4 respectively.

$$P=
beginbmatrix
1&0&0&0\
dfrac12&0&dfrac12&0\
0&dfrac12&0&dfrac12\
0&0&0&1
endbmatrix.
$$

If I understand you correctly you keep playing until you get either two heads or two tails. Then, states 1 and 4 are absorbing states.

Then when you are in state 2 (got a tail) then you can jump after the coin flip to either state 1 (two tail flips) or state 3 (one head).

If you start with a tail then your initial state probability vector $pi$ would be

$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$

share|cite|improve this answer

edited Mar 14 at 12:51

answered Mar 14 at 12:45

AhamAham

414

$endgroup$

add a comment | 

1

$begingroup$

I think in this case you need to have 4 states: state 2 tails, state tail, state head and state 2 heads which I will call states 1,2,3 and 4 respectively.

$$P=
beginbmatrix
1&0&0&0\
dfrac12&0&dfrac12&0\
0&dfrac12&0&dfrac12\
0&0&0&1
endbmatrix.
$$

If I understand you correctly you keep playing until you get either two heads or two tails. Then, states 1 and 4 are absorbing states.

Then when you are in state 2 (got a tail) then you can jump after the coin flip to either state 1 (two tail flips) or state 3 (one head).

If you start with a tail then your initial state probability vector $pi$ would be

$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$

share|cite|improve this answer

edited Mar 14 at 12:51

answered Mar 14 at 12:45

AhamAham

414

$endgroup$

add a comment | 

$begingroup$

I think in this case you need to have 4 states: state 2 tails, state tail, state head and state 2 heads which I will call states 1,2,3 and 4 respectively.

$$P=
beginbmatrix
1&0&0&0\
dfrac12&0&dfrac12&0\
0&dfrac12&0&dfrac12\
0&0&0&1
endbmatrix.
$$

If I understand you correctly you keep playing until you get either two heads or two tails. Then, states 1 and 4 are absorbing states.

Then when you are in state 2 (got a tail) then you can jump after the coin flip to either state 1 (two tail flips) or state 3 (one head).

If you start with a tail then your initial state probability vector $pi$ would be

$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$

share|cite|improve this answer

edited Mar 14 at 12:51

answered Mar 14 at 12:45

AhamAham

414

$endgroup$

share|cite|improve this answer

edited Mar 14 at 12:51

answered Mar 14 at 12:45

AhamAham

414

answered Mar 14 at 12:45

AhamAham

414

answered Mar 14 at 12:45

AhamAham

414