Mixing
Is Conditional Expectation of stochastic process a martingale?
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I have the following:
Let $Y$ be an integrable and $F_T$ -measurable random variable. Define ${X_t}_{t in [0,T]}$ by $X_t =E(Y|F_t), forall t in T$. The ${F_t}_{t in [0,T]}$ should be the filtration generated by $Y$ (or $X$?). Show that ${X_t}_{t in [0,T]}$ is a martingale.
Can please someone help me? I can show the property of $E(X_t|F_s)=X_s$ for $s leq t$ and that $X_t$ in in $L^1$ but I have troubles in showing that $X_t$ is adapted to the filtration since $Y$ is only $F_T$-measurable.
conditional-expectation martingales filtrations
asked Jan 5 at 10:54
MatteoMatteo
154
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add a comment |
$begingroup$
I have the following:
Let $Y$ be an integrable and $F_T$ -measurable random variable. Define ${X_t}_{t in [0,T]}$ by $X_t =E(Y|F_t), forall t in T$. The ${F_t}_{t in [0,T]}$ should be the filtration generated by $Y$ (or $X$?). Show that ${X_t}_{t in [0,T]}$ is a martingale.
Can please someone help me? I can show the property of $E(X_t|F_s)=X_s$ for $s leq t$ and that $X_t$ in in $L^1$ but I have troubles in showing that $X_t$ is adapted to the filtration since $Y$ is only $F_T$-measurable.
conditional-expectation martingales filtrations
asked Jan 5 at 10:54
MatteoMatteo
154
$endgroup$
2
$begingroup$
By the very definition of conditional expectation, $mathbb{E}(Y mid mathcal{F})$ is $mathcal{F}$-measurable for any $sigma$-algebra $mathcal{F}$...that's all you need to show that $X_t$ is $mathcal{F}_t$-measurable.
$endgroup$
– saz
Jan 5 at 11:38
add a comment |
$begingroup$
I have the following:
Let $Y$ be an integrable and $F_T$ -measurable random variable. Define ${X_t}_{t in [0,T]}$ by $X_t =E(Y|F_t), forall t in T$. The ${F_t}_{t in [0,T]}$ should be the filtration generated by $Y$ (or $X$?). Show that ${X_t}_{t in [0,T]}$ is a martingale.
Can please someone help me? I can show the property of $E(X_t|F_s)=X_s$ for $s leq t$ and that $X_t$ in in $L^1$ but I have troubles in showing that $X_t$ is adapted to the filtration since $Y$ is only $F_T$-measurable.
conditional-expectation martingales filtrations
asked Jan 5 at 10:54
MatteoMatteo
154
$endgroup$
I have the following:
Let $Y$ be an integrable and $F_T$ -measurable random variable. Define ${X_t}_{t in [0,T]}$ by $X_t =E(Y|F_t), forall t in T$. The ${F_t}_{t in [0,T]}$ should be the filtration generated by $Y$ (or $X$?). Show that ${X_t}_{t in [0,T]}$ is a martingale.
Can please someone help me? I can show the property of $E(X_t|F_s)=X_s$ for $s leq t$ and that $X_t$ in in $L^1$ but I have troubles in showing that $X_t$ is adapted to the filtration since $Y$ is only $F_T$-measurable.
conditional-expectation martingales filtrations
conditional-expectation martingales filtrations
asked Jan 5 at 10:54
MatteoMatteo
154
asked Jan 5 at 10:54
MatteoMatteo
154
asked Jan 5 at 10:54
MatteoMatteo
154
asked Jan 5 at 10:54
MatteoMatteo
154
asked Jan 5 at 10:54
MatteoMatteo
154
154
2
$begingroup$
By the very definition of conditional expectation, $mathbb{E}(Y mid mathcal{F})$ is $mathcal{F}$-measurable for any $sigma$-algebra $mathcal{F}$...that's all you need to show that $X_t$ is $mathcal{F}_t$-measurable.
$endgroup$
– saz
Jan 5 at 11:38
add a comment |
2
$begingroup$
By the very definition of conditional expectation, $mathbb{E}(Y mid mathcal{F})$ is $mathcal{F}$-measurable for any $sigma$-algebra $mathcal{F}$...that's all you need to show that $X_t$ is $mathcal{F}_t$-measurable.
$endgroup$
– saz
Jan 5 at 11:38
2
2
$begingroup$
By the very definition of conditional expectation, $mathbb{E}(Y mid mathcal{F})$ is $mathcal{F}$-measurable for any $sigma$-algebra $mathcal{F}$...that's all you need to show that $X_t$ is $mathcal{F}_t$-measurable.
$endgroup$
– saz
Jan 5 at 11:38
$begingroup$
By the very definition of conditional expectation, $mathbb{E}(Y mid mathcal{F})$ is $mathcal{F}$-measurable for any $sigma$-algebra $mathcal{F}$...that's all you need to show that $X_t$ is $mathcal{F}_t$-measurable.
$endgroup$
– saz
Jan 5 at 11:38
add a comment |
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$begingroup$
By the very definition of conditional expectation, $mathbb{E}(Y mid mathcal{F})$ is $mathcal{F}$-measurable for any $sigma$-algebra $mathcal{F}$...that's all you need to show that $X_t$ is $mathcal{F}_t$-measurable.
$endgroup$
– saz
Jan 5 at 11:38