Transition matrix for a two state Markov Chain The 2019 Stack Overflow Developer Survey Results Are InMarkov chain transition matrixTwo different ways of constructing a continuous time Markov chain from discrete time oneEquivalence Classes of a Markov Chain with Transition MatrixChanges in the transition matrix of a Markov chainExpected payoff of a 2-State Markov ChainReduce the mixing time by switching a regular state to an absorbing state in a Markov chainHow to model a markov chain that randomly switches between two transition matrices?Markov Chain never reaches a statePeriodic markov chain - finding initial conditions causing convergence to steady state?Find probability of markov chain ended in state $0$.
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Transition matrix for a two state Markov Chain
The 2019 Stack Overflow Developer Survey Results Are InMarkov chain transition matrixTwo different ways of constructing a continuous time Markov chain from discrete time oneEquivalence Classes of a Markov Chain with Transition MatrixChanges in the transition matrix of a Markov chainExpected payoff of a 2-State Markov ChainReduce the mixing time by switching a regular state to an absorbing state in a Markov chainHow to model a markov chain that randomly switches between two transition matrices?Markov Chain never reaches a statePeriodic markov chain - finding initial conditions causing convergence to steady state?Find probability of markov chain ended in state $0$.
$begingroup$
What is the most general transition matrix for a two state Markov Chain? (both Markov and homogeneous). And show that any such chain has an equilibrium vector.
Is it should be the following matrix? If it is, how could we show it alway exists an equilibrium vector? beginbmatrix
p & 1-p \
q & 1-q
endbmatrix
stochastic-processes
asked Apr 6 at 20:11
EricEric
577
$endgroup$
add a comment |
$begingroup$
What is the most general transition matrix for a two state Markov Chain? (both Markov and homogeneous). And show that any such chain has an equilibrium vector.
Is it should be the following matrix? If it is, how could we show it alway exists an equilibrium vector? beginbmatrix
p & 1-p \
q & 1-q
endbmatrix
stochastic-processes
asked Apr 6 at 20:11
EricEric
577
$endgroup$
add a comment |
$begingroup$
What is the most general transition matrix for a two state Markov Chain? (both Markov and homogeneous). And show that any such chain has an equilibrium vector.
Is it should be the following matrix? If it is, how could we show it alway exists an equilibrium vector? beginbmatrix
p & 1-p \
q & 1-q
endbmatrix
stochastic-processes
asked Apr 6 at 20:11
EricEric
577
$endgroup$
What is the most general transition matrix for a two state Markov Chain? (both Markov and homogeneous). And show that any such chain has an equilibrium vector.
Is it should be the following matrix? If it is, how could we show it alway exists an equilibrium vector? beginbmatrix
p & 1-p \
q & 1-q
endbmatrix
stochastic-processes
stochastic-processes
asked Apr 6 at 20:11
EricEric
577
asked Apr 6 at 20:11
EricEric
577
asked Apr 6 at 20:11
EricEric
577
asked Apr 6 at 20:11
EricEric
577
asked Apr 6 at 20:11
EricEric
577
577
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're right, that's the two-state transition matrix in full generality. To find an equilibrium vector, we solve:
beginalign
beginbmatrix
x & y
endbmatrix
cdot
beginbmatrix
p & 1-p \ q & 1-q
endbmatrix
=
beginbmatrix
x & y
endbmatrix,
endalign
which yields the system
beginequation
left{
beginaligned
px+qy&=x
\(1-p)x+(1-q)y&=y
endaligned
right.
.
endequation
The second equation is redundant,
beginalign
y
&=(1-p)x+(1-q)y
\&=x+y-(px+qy)
\&=x+y-(x)
\&=y,
endalign
so we only have to solve the first one. Now, pick any non-negative value of $x$, then
beginequation
y=frac(1-p)xq
endequation
is a non-negative number, and the pair $(x,y)$ solve the first equation (check). Finally, if $(x,y)$ solve the first equation, then $(sx,sy)$ for any $s$ also solve the first equation, because
beginalign
sx &=psx+qsy
\&=s(px+qy)
\&=s(x)
endalign
since $px+qy=x$. By letting $s=frac1x+y$ we ensure $sx+sy=1$, which means beginbmatrix
sx & sy
endbmatrix is the equilibrium vector we want.
Hope this helps!
answered Apr 6 at 20:41
R_BR_B
839111
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're right, that's the two-state transition matrix in full generality. To find an equilibrium vector, we solve:
beginalign
beginbmatrix
x & y
endbmatrix
cdot
beginbmatrix
p & 1-p \ q & 1-q
endbmatrix
=
beginbmatrix
x & y
endbmatrix,
endalign
which yields the system
beginequation
left{
beginaligned
px+qy&=x
\(1-p)x+(1-q)y&=y
endaligned
right.
.
endequation
The second equation is redundant,
beginalign
y
&=(1-p)x+(1-q)y
\&=x+y-(px+qy)
\&=x+y-(x)
\&=y,
endalign
so we only have to solve the first one. Now, pick any non-negative value of $x$, then
beginequation
y=frac(1-p)xq
endequation
is a non-negative number, and the pair $(x,y)$ solve the first equation (check). Finally, if $(x,y)$ solve the first equation, then $(sx,sy)$ for any $s$ also solve the first equation, because
beginalign
sx &=psx+qsy
\&=s(px+qy)
\&=s(x)
endalign
since $px+qy=x$. By letting $s=frac1x+y$ we ensure $sx+sy=1$, which means beginbmatrix
sx & sy
endbmatrix is the equilibrium vector we want.
Hope this helps!
answered Apr 6 at 20:41
R_BR_B
839111
$endgroup$
add a comment |
$begingroup$
You're right, that's the two-state transition matrix in full generality. To find an equilibrium vector, we solve:
beginalign
beginbmatrix
x & y
endbmatrix
cdot
beginbmatrix
p & 1-p \ q & 1-q
endbmatrix
=
beginbmatrix
x & y
endbmatrix,
endalign
which yields the system
beginequation
left{
beginaligned
px+qy&=x
\(1-p)x+(1-q)y&=y
endaligned
right.
.
endequation
The second equation is redundant,
beginalign
y
&=(1-p)x+(1-q)y
\&=x+y-(px+qy)
\&=x+y-(x)
\&=y,
endalign
so we only have to solve the first one. Now, pick any non-negative value of $x$, then
beginequation
y=frac(1-p)xq
endequation
is a non-negative number, and the pair $(x,y)$ solve the first equation (check). Finally, if $(x,y)$ solve the first equation, then $(sx,sy)$ for any $s$ also solve the first equation, because
beginalign
sx &=psx+qsy
\&=s(px+qy)
\&=s(x)
endalign
since $px+qy=x$. By letting $s=frac1x+y$ we ensure $sx+sy=1$, which means beginbmatrix
sx & sy
endbmatrix is the equilibrium vector we want.
Hope this helps!
answered Apr 6 at 20:41
R_BR_B
839111
$endgroup$
add a comment |
$begingroup$
You're right, that's the two-state transition matrix in full generality. To find an equilibrium vector, we solve:
beginalign
beginbmatrix
x & y
endbmatrix
cdot
beginbmatrix
p & 1-p \ q & 1-q
endbmatrix
=
beginbmatrix
x & y
endbmatrix,
endalign
which yields the system
beginequation
left{
beginaligned
px+qy&=x
\(1-p)x+(1-q)y&=y
endaligned
right.
.
endequation
The second equation is redundant,
beginalign
y
&=(1-p)x+(1-q)y
\&=x+y-(px+qy)
\&=x+y-(x)
\&=y,
endalign
so we only have to solve the first one. Now, pick any non-negative value of $x$, then
beginequation
y=frac(1-p)xq
endequation
is a non-negative number, and the pair $(x,y)$ solve the first equation (check). Finally, if $(x,y)$ solve the first equation, then $(sx,sy)$ for any $s$ also solve the first equation, because
beginalign
sx &=psx+qsy
\&=s(px+qy)
\&=s(x)
endalign
since $px+qy=x$. By letting $s=frac1x+y$ we ensure $sx+sy=1$, which means beginbmatrix
sx & sy
endbmatrix is the equilibrium vector we want.
Hope this helps!
answered Apr 6 at 20:41
R_BR_B
839111
$endgroup$
You're right, that's the two-state transition matrix in full generality. To find an equilibrium vector, we solve:
beginalign
beginbmatrix
x & y
endbmatrix
cdot
beginbmatrix
p & 1-p \ q & 1-q
endbmatrix
=
beginbmatrix
x & y
endbmatrix,
endalign
which yields the system
beginequation
left{
beginaligned
px+qy&=x
\(1-p)x+(1-q)y&=y
endaligned
right.
.
endequation
The second equation is redundant,
beginalign
y
&=(1-p)x+(1-q)y
\&=x+y-(px+qy)
\&=x+y-(x)
\&=y,
endalign
so we only have to solve the first one. Now, pick any non-negative value of $x$, then
beginequation
y=frac(1-p)xq
endequation
is a non-negative number, and the pair $(x,y)$ solve the first equation (check). Finally, if $(x,y)$ solve the first equation, then $(sx,sy)$ for any $s$ also solve the first equation, because
beginalign
sx &=psx+qsy
\&=s(px+qy)
\&=s(x)
endalign
since $px+qy=x$. By letting $s=frac1x+y$ we ensure $sx+sy=1$, which means beginbmatrix
sx & sy
endbmatrix is the equilibrium vector we want.
Hope this helps!
answered Apr 6 at 20:41
R_BR_B
839111
edited Apr 7 at 12:37
answered Apr 6 at 20:41
R_BR_B
839111
answered Apr 6 at 20:41
R_BR_B
839111
answered Apr 6 at 20:41
R_BR_B
839111
839111
add a comment |
add a comment |
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