Alternative formulation of a markov process
I'm wondering how the markov property can be specified as follows, if anyone can provide more details (this looks awfully like the definition for a martingale):
$$E[f(X_t)|mathcal{F}_s]=E[f(X_t)|sigma(X_s)]$$
for all $t geq s geq 0$ and $f:S to mathbb{R}$ bounded and measurable.
I'm used to the markov property definition in terms conditional probabilities and time-homogeneity.
Also, isn't $sigma(X_s)$ the sigma algebra generated by $X_s$ which is just the filtration $mathcal{F}_s$?
martingales markov-process filtrations
asked Dec 1 at 11:50
user175199
1
add a comment |
I'm wondering how the markov property can be specified as follows, if anyone can provide more details (this looks awfully like the definition for a martingale):
$$E[f(X_t)|mathcal{F}_s]=E[f(X_t)|sigma(X_s)]$$
for all $t geq s geq 0$ and $f:S to mathbb{R}$ bounded and measurable.
I'm used to the markov property definition in terms conditional probabilities and time-homogeneity.
Also, isn't $sigma(X_s)$ the sigma algebra generated by $X_s$ which is just the filtration $mathcal{F}_s$?
martingales markov-process filtrations
asked Dec 1 at 11:50
user175199
1
It's pretty hard to answer your question without knowing exactly the definition which you are used to...
– saz
Dec 1 at 12:40
add a comment |
I'm wondering how the markov property can be specified as follows, if anyone can provide more details (this looks awfully like the definition for a martingale):
$$E[f(X_t)|mathcal{F}_s]=E[f(X_t)|sigma(X_s)]$$
for all $t geq s geq 0$ and $f:S to mathbb{R}$ bounded and measurable.
I'm used to the markov property definition in terms conditional probabilities and time-homogeneity.
Also, isn't $sigma(X_s)$ the sigma algebra generated by $X_s$ which is just the filtration $mathcal{F}_s$?
martingales markov-process filtrations
asked Dec 1 at 11:50
user175199
1
I'm wondering how the markov property can be specified as follows, if anyone can provide more details (this looks awfully like the definition for a martingale):
$$E[f(X_t)|mathcal{F}_s]=E[f(X_t)|sigma(X_s)]$$
for all $t geq s geq 0$ and $f:S to mathbb{R}$ bounded and measurable.
I'm used to the markov property definition in terms conditional probabilities and time-homogeneity.
Also, isn't $sigma(X_s)$ the sigma algebra generated by $X_s$ which is just the filtration $mathcal{F}_s$?
martingales markov-process filtrations
martingales markov-process filtrations
asked Dec 1 at 11:50
user175199
1
asked Dec 1 at 11:50
user175199
1
asked Dec 1 at 11:50
user175199
1
asked Dec 1 at 11:50
user175199
1
asked Dec 1 at 11:50
user175199
1
1
It's pretty hard to answer your question without knowing exactly the definition which you are used to...
– saz
Dec 1 at 12:40
add a comment |
It's pretty hard to answer your question without knowing exactly the definition which you are used to...
– saz
Dec 1 at 12:40
It's pretty hard to answer your question without knowing exactly the definition which you are used to...
– saz
Dec 1 at 12:40
It's pretty hard to answer your question without knowing exactly the definition which you are used to...
– saz
Dec 1 at 12:40
add a comment |
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It's pretty hard to answer your question without knowing exactly the definition which you are used to...
– saz
Dec 1 at 12:40