Does a continuous-time stochastic process satisfying this evolution equation have the Markov property?
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It is well-known that, for a continuous-time Markov process, the transition probabilities $mathbf{P} = mathbb{P}left[X(t) = j |X(0) = iright]$ satisfy the following evolution equation:
$$frac{dmathbf{P}}{dt} = mathbf{A}{mathbf{P}}$$
where $mathbf{A}$ is a time-independent matrix usually called the generator matrix.
If I were to identify a stochastic process whose transition probabilities evolve according to this equation, would this immediately imply the process possesses the Markov property?
probability stochastic-processes markov-process
asked Jan 4 at 22:41
aghostinthefiguresaghostinthefigures
1,2891217
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add a comment |
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It is well-known that, for a continuous-time Markov process, the transition probabilities $mathbf{P} = mathbb{P}left[X(t) = j |X(0) = iright]$ satisfy the following evolution equation:
$$frac{dmathbf{P}}{dt} = mathbf{A}{mathbf{P}}$$
where $mathbf{A}$ is a time-independent matrix usually called the generator matrix.
If I were to identify a stochastic process whose transition probabilities evolve according to this equation, would this immediately imply the process possesses the Markov property?
probability stochastic-processes markov-process
asked Jan 4 at 22:41
aghostinthefiguresaghostinthefigures
1,2891217
$endgroup$
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No, because this says nothing about higher dimensional marginals of the process, for example, about whether $$P(X(2)in Bmid X(1),X(0))=P(X(2)in Bmid X(1))$$
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– Did
Jan 8 at 22:23
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Additional to @Did's comment, if you consider a stochastic differential equation for $X_t$: ${rm d}X_t=b(X_t),{rm d}t+sigma(X_t),{rm d}W_t$, where $b$ and $sigma$ are, say, smooth functions, and $W_t$ denotes the Wiener process, then $X_t$ has the Markov property. However, this does not hold true if ${rm d}X_t=b(X_t,t),{rm d}t+sigma(X_t,t),{rm d}W_t$, where $b$ and $sigma$ are time-dependent.
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– hypernova
Jan 9 at 0:36
add a comment |
$begingroup$
It is well-known that, for a continuous-time Markov process, the transition probabilities $mathbf{P} = mathbb{P}left[X(t) = j |X(0) = iright]$ satisfy the following evolution equation:
$$frac{dmathbf{P}}{dt} = mathbf{A}{mathbf{P}}$$
where $mathbf{A}$ is a time-independent matrix usually called the generator matrix.
If I were to identify a stochastic process whose transition probabilities evolve according to this equation, would this immediately imply the process possesses the Markov property?
probability stochastic-processes markov-process
asked Jan 4 at 22:41
aghostinthefiguresaghostinthefigures
1,2891217
$endgroup$
It is well-known that, for a continuous-time Markov process, the transition probabilities $mathbf{P} = mathbb{P}left[X(t) = j |X(0) = iright]$ satisfy the following evolution equation:
$$frac{dmathbf{P}}{dt} = mathbf{A}{mathbf{P}}$$
where $mathbf{A}$ is a time-independent matrix usually called the generator matrix.
If I were to identify a stochastic process whose transition probabilities evolve according to this equation, would this immediately imply the process possesses the Markov property?
probability stochastic-processes markov-process
probability stochastic-processes markov-process
asked Jan 4 at 22:41
aghostinthefiguresaghostinthefigures
1,2891217
asked Jan 4 at 22:41
aghostinthefiguresaghostinthefigures
1,2891217
edited Jan 8 at 22:20
aghostinthefigures
asked Jan 4 at 22:41
aghostinthefiguresaghostinthefigures
1,2891217
asked Jan 4 at 22:41
aghostinthefiguresaghostinthefigures
1,2891217
asked Jan 4 at 22:41
aghostinthefiguresaghostinthefigures
1,2891217
1,2891217
$begingroup$
No, because this says nothing about higher dimensional marginals of the process, for example, about whether $$P(X(2)in Bmid X(1),X(0))=P(X(2)in Bmid X(1))$$
$endgroup$
– Did
Jan 8 at 22:23
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Additional to @Did's comment, if you consider a stochastic differential equation for $X_t$: ${rm d}X_t=b(X_t),{rm d}t+sigma(X_t),{rm d}W_t$, where $b$ and $sigma$ are, say, smooth functions, and $W_t$ denotes the Wiener process, then $X_t$ has the Markov property. However, this does not hold true if ${rm d}X_t=b(X_t,t),{rm d}t+sigma(X_t,t),{rm d}W_t$, where $b$ and $sigma$ are time-dependent.
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– hypernova
Jan 9 at 0:36
add a comment |
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No, because this says nothing about higher dimensional marginals of the process, for example, about whether $$P(X(2)in Bmid X(1),X(0))=P(X(2)in Bmid X(1))$$
$endgroup$
– Did
Jan 8 at 22:23
$begingroup$
Additional to @Did's comment, if you consider a stochastic differential equation for $X_t$: ${rm d}X_t=b(X_t),{rm d}t+sigma(X_t),{rm d}W_t$, where $b$ and $sigma$ are, say, smooth functions, and $W_t$ denotes the Wiener process, then $X_t$ has the Markov property. However, this does not hold true if ${rm d}X_t=b(X_t,t),{rm d}t+sigma(X_t,t),{rm d}W_t$, where $b$ and $sigma$ are time-dependent.
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– hypernova
Jan 9 at 0:36
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No, because this says nothing about higher dimensional marginals of the process, for example, about whether $$P(X(2)in Bmid X(1),X(0))=P(X(2)in Bmid X(1))$$
$endgroup$
– Did
Jan 8 at 22:23
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No, because this says nothing about higher dimensional marginals of the process, for example, about whether $$P(X(2)in Bmid X(1),X(0))=P(X(2)in Bmid X(1))$$
$endgroup$
– Did
Jan 8 at 22:23
$begingroup$
Additional to @Did's comment, if you consider a stochastic differential equation for $X_t$: ${rm d}X_t=b(X_t),{rm d}t+sigma(X_t),{rm d}W_t$, where $b$ and $sigma$ are, say, smooth functions, and $W_t$ denotes the Wiener process, then $X_t$ has the Markov property. However, this does not hold true if ${rm d}X_t=b(X_t,t),{rm d}t+sigma(X_t,t),{rm d}W_t$, where $b$ and $sigma$ are time-dependent.
$endgroup$
– hypernova
Jan 9 at 0:36
$begingroup$
Additional to @Did's comment, if you consider a stochastic differential equation for $X_t$: ${rm d}X_t=b(X_t),{rm d}t+sigma(X_t),{rm d}W_t$, where $b$ and $sigma$ are, say, smooth functions, and $W_t$ denotes the Wiener process, then $X_t$ has the Markov property. However, this does not hold true if ${rm d}X_t=b(X_t,t),{rm d}t+sigma(X_t,t),{rm d}W_t$, where $b$ and $sigma$ are time-dependent.
$endgroup$
– hypernova
Jan 9 at 0:36
add a comment |
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No, because this says nothing about higher dimensional marginals of the process, for example, about whether $$P(X(2)in Bmid X(1),X(0))=P(X(2)in Bmid X(1))$$
$endgroup$
– Did
Jan 8 at 22:23
$begingroup$
Additional to @Did's comment, if you consider a stochastic differential equation for $X_t$: ${rm d}X_t=b(X_t),{rm d}t+sigma(X_t),{rm d}W_t$, where $b$ and $sigma$ are, say, smooth functions, and $W_t$ denotes the Wiener process, then $X_t$ has the Markov property. However, this does not hold true if ${rm d}X_t=b(X_t,t),{rm d}t+sigma(X_t,t),{rm d}W_t$, where $b$ and $sigma$ are time-dependent.
$endgroup$
– hypernova
Jan 9 at 0:36