Equivalence of independence and a conditional expectation equality
Let $E_1$, $E_2$ be Polish, let $D_{E_2}([0,infty])$ be the space of cadlag functions with values in $E_2$ and set $mathcal{F}_t^Y$ to be the $sigma$-algebra generated by $Y_s$, $s leq t$, similarly for $X$ and finally $mathcal{F}_t^{X,Y} := mathcal{F}_t^Y vee mathcal{F}_t^X$. Let $X$ be an $E_1$-valued stochastic process and $Y$ an $E_2$-valued stochastic process.
I am looking for a proof of the following statement:
Assume $Y$ to have independent increments and paths in $D_{E_2}([0,infty])$. Then the following are equivalent.
(i) $(Y_{t+s}-Y_t)_{s geq 0}$ is independent of $mathcal{F}_t^{X,Y}$ for every $t geq 0$.
(ii) $mathbb{E}[h(Y)|mathcal{F}_t^{X,Y}] = mathbb{E}[h(Y)|mathcal{F}_t^{Y}]$ for every $t geq 0$ and every measurable and bounded $h:D_{E_i}([0,infty]) to mathbb{R}$.
Intuively, at least $(i) implies (ii)$ "feels" right and easy to believe to me. However, I could not come up with a rigorous proof. I would appreciate your help very much!
measure-theory stochastic-processes conditional-expectation independence
asked Nov 29 at 16:01
Marco
485
add a comment |
Let $E_1$, $E_2$ be Polish, let $D_{E_2}([0,infty])$ be the space of cadlag functions with values in $E_2$ and set $mathcal{F}_t^Y$ to be the $sigma$-algebra generated by $Y_s$, $s leq t$, similarly for $X$ and finally $mathcal{F}_t^{X,Y} := mathcal{F}_t^Y vee mathcal{F}_t^X$. Let $X$ be an $E_1$-valued stochastic process and $Y$ an $E_2$-valued stochastic process.
I am looking for a proof of the following statement:
Assume $Y$ to have independent increments and paths in $D_{E_2}([0,infty])$. Then the following are equivalent.
(i) $(Y_{t+s}-Y_t)_{s geq 0}$ is independent of $mathcal{F}_t^{X,Y}$ for every $t geq 0$.
(ii) $mathbb{E}[h(Y)|mathcal{F}_t^{X,Y}] = mathbb{E}[h(Y)|mathcal{F}_t^{Y}]$ for every $t geq 0$ and every measurable and bounded $h:D_{E_i}([0,infty]) to mathbb{R}$.
Intuively, at least $(i) implies (ii)$ "feels" right and easy to believe to me. However, I could not come up with a rigorous proof. I would appreciate your help very much!
measure-theory stochastic-processes conditional-expectation independence
asked Nov 29 at 16:01
Marco
485
add a comment |
Let $E_1$, $E_2$ be Polish, let $D_{E_2}([0,infty])$ be the space of cadlag functions with values in $E_2$ and set $mathcal{F}_t^Y$ to be the $sigma$-algebra generated by $Y_s$, $s leq t$, similarly for $X$ and finally $mathcal{F}_t^{X,Y} := mathcal{F}_t^Y vee mathcal{F}_t^X$. Let $X$ be an $E_1$-valued stochastic process and $Y$ an $E_2$-valued stochastic process.
I am looking for a proof of the following statement:
Assume $Y$ to have independent increments and paths in $D_{E_2}([0,infty])$. Then the following are equivalent.
(i) $(Y_{t+s}-Y_t)_{s geq 0}$ is independent of $mathcal{F}_t^{X,Y}$ for every $t geq 0$.
(ii) $mathbb{E}[h(Y)|mathcal{F}_t^{X,Y}] = mathbb{E}[h(Y)|mathcal{F}_t^{Y}]$ for every $t geq 0$ and every measurable and bounded $h:D_{E_i}([0,infty]) to mathbb{R}$.
Intuively, at least $(i) implies (ii)$ "feels" right and easy to believe to me. However, I could not come up with a rigorous proof. I would appreciate your help very much!
measure-theory stochastic-processes conditional-expectation independence
asked Nov 29 at 16:01
Marco
485
Let $E_1$, $E_2$ be Polish, let $D_{E_2}([0,infty])$ be the space of cadlag functions with values in $E_2$ and set $mathcal{F}_t^Y$ to be the $sigma$-algebra generated by $Y_s$, $s leq t$, similarly for $X$ and finally $mathcal{F}_t^{X,Y} := mathcal{F}_t^Y vee mathcal{F}_t^X$. Let $X$ be an $E_1$-valued stochastic process and $Y$ an $E_2$-valued stochastic process.
I am looking for a proof of the following statement:
Assume $Y$ to have independent increments and paths in $D_{E_2}([0,infty])$. Then the following are equivalent.
(i) $(Y_{t+s}-Y_t)_{s geq 0}$ is independent of $mathcal{F}_t^{X,Y}$ for every $t geq 0$.
(ii) $mathbb{E}[h(Y)|mathcal{F}_t^{X,Y}] = mathbb{E}[h(Y)|mathcal{F}_t^{Y}]$ for every $t geq 0$ and every measurable and bounded $h:D_{E_i}([0,infty]) to mathbb{R}$.
Intuively, at least $(i) implies (ii)$ "feels" right and easy to believe to me. However, I could not come up with a rigorous proof. I would appreciate your help very much!
measure-theory stochastic-processes conditional-expectation independence
measure-theory stochastic-processes conditional-expectation independence
asked Nov 29 at 16:01
Marco
485
asked Nov 29 at 16:01
Marco
485
asked Nov 29 at 16:01
Marco
485
asked Nov 29 at 16:01
Marco
485
asked Nov 29 at 16:01
Marco
485
485
add a comment |
add a comment |
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