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
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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
$begingroup$
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
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
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
asked Mar 14 at 6:30
Ongky Denny WijayaOngky Denny Wijaya
4018
$endgroup$
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
probability stochastic-processes markov-chains
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
asked Mar 14 at 6:30
Ongky Denny WijayaOngky Denny Wijaya
4018
asked Mar 14 at 6:30
Ongky Denny WijayaOngky Denny Wijaya
4018
4018
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$
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
$begingroup$
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
$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$
answered Mar 14 at 12:45
AhamAham
414
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$
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:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$
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:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$
answered Mar 14 at 6:41
vermatorvermator
34110
$endgroup$
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:
$$P=
beginbmatrix
dfrac12&dfrac12&0\
dfrac12&0&dfrac12\
0&0&1
endbmatrix.
$$
answered Mar 14 at 6:41
vermatorvermator
34110
answered Mar 14 at 6:41
vermatorvermator
34110
answered Mar 14 at 6:41
vermatorvermator
34110
answered Mar 14 at 6:41
vermatorvermator
34110
34110
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
number of consecutive tails
$endgroup$
– Ongky Denny Wijaya
Mar 14 at 7:15
|
show 2 more comments
$begingroup$
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
$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$
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
$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$
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
$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$
answered Mar 14 at 12:45
AhamAham
414
$endgroup$
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
$$pi=
beginbmatrix
0\
1\
0\
0
endbmatrix.
$$
answered Mar 14 at 12:45
AhamAham
414
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
414
add a comment |
add a comment |
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