Mixing
$mathbb{E}[(Y_{n+1} - {Y_n})Z] =0.$ implies $Y_n = mathbb{E}[Y_{n+1}|H_n]$?
$begingroup$
Suppose $Y_n = mathbb{E}[Y|X_1 dots X_n]$ is a Doob's Martingale sequence . let $H_n$ be $sigma$-algebra generate by $X_1 dots X_n$,then if for any random variable $Z in H_n$ we have :$mathbb{E}[(Y_{n+1} - {Y_n})Z] =0.$
then we have $Y_n = mathbb{E}[Y_{n+1}|H_n]$
how to see that ?
conditional-expectation martingales
asked Jan 21 at 3:02
ShaoyuPeiShaoyuPei
1778
$endgroup$
add a comment |
$begingroup$
Suppose $Y_n = mathbb{E}[Y|X_1 dots X_n]$ is a Doob's Martingale sequence . let $H_n$ be $sigma$-algebra generate by $X_1 dots X_n$,then if for any random variable $Z in H_n$ we have :$mathbb{E}[(Y_{n+1} - {Y_n})Z] =0.$
then we have $Y_n = mathbb{E}[Y_{n+1}|H_n]$
how to see that ?
conditional-expectation martingales
asked Jan 21 at 3:02
ShaoyuPeiShaoyuPei
1778
$endgroup$
add a comment |
$begingroup$
Suppose $Y_n = mathbb{E}[Y|X_1 dots X_n]$ is a Doob's Martingale sequence . let $H_n$ be $sigma$-algebra generate by $X_1 dots X_n$,then if for any random variable $Z in H_n$ we have :$mathbb{E}[(Y_{n+1} - {Y_n})Z] =0.$
then we have $Y_n = mathbb{E}[Y_{n+1}|H_n]$
how to see that ?
conditional-expectation martingales
asked Jan 21 at 3:02
ShaoyuPeiShaoyuPei
1778
$endgroup$
Suppose $Y_n = mathbb{E}[Y|X_1 dots X_n]$ is a Doob's Martingale sequence . let $H_n$ be $sigma$-algebra generate by $X_1 dots X_n$,then if for any random variable $Z in H_n$ we have :$mathbb{E}[(Y_{n+1} - {Y_n})Z] =0.$
then we have $Y_n = mathbb{E}[Y_{n+1}|H_n]$
how to see that ?
conditional-expectation martingales
conditional-expectation martingales
asked Jan 21 at 3:02
ShaoyuPeiShaoyuPei
1778
asked Jan 21 at 3:02
ShaoyuPeiShaoyuPei
1778
asked Jan 21 at 3:02
ShaoyuPeiShaoyuPei
1778
asked Jan 21 at 3:02
ShaoyuPeiShaoyuPei
1778
asked Jan 21 at 3:02
ShaoyuPeiShaoyuPei
1778
1778
add a comment |
add a comment |
2 Answers
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$begingroup$
By the hypothesis,
$$0 = E[(Y_{n+1} - Y_n) Z] = E[E[Y_{n+1} mid H_n]Z] - E[Y_n Z].$$
Now recall the measure-theoretic definition of the conditional expectation $E[Y_{n+1} mid H_n]$.
answered Jan 21 at 3:23
angryavianangryavian
42k23381
$endgroup$
add a comment |
$begingroup$
If $Z = mathbf{1}_mathrm{X}$ for $mathrm{X} in sigma(X_1, ldots, X_n)$ then $mathbf{E}((Y_{n+1}-Y_n)Z) = 0$ signifies $intlimits_mathrm{X} Y_{n+1} dmathbf{P} = intlimits_mathrm{X} Y_n dmathbf{P},$ and since $Y_n$ is $mathscr{H}_n$-measurable, the result follows upon remembering what the conditional expectation means.
answered Jan 21 at 3:25
Will M.Will M.
2,835315
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
By the hypothesis,
$$0 = E[(Y_{n+1} - Y_n) Z] = E[E[Y_{n+1} mid H_n]Z] - E[Y_n Z].$$
Now recall the measure-theoretic definition of the conditional expectation $E[Y_{n+1} mid H_n]$.
answered Jan 21 at 3:23
angryavianangryavian
42k23381
$endgroup$
add a comment |
$begingroup$
By the hypothesis,
$$0 = E[(Y_{n+1} - Y_n) Z] = E[E[Y_{n+1} mid H_n]Z] - E[Y_n Z].$$
Now recall the measure-theoretic definition of the conditional expectation $E[Y_{n+1} mid H_n]$.
answered Jan 21 at 3:23
angryavianangryavian
42k23381
$endgroup$
add a comment |
$begingroup$
By the hypothesis,
$$0 = E[(Y_{n+1} - Y_n) Z] = E[E[Y_{n+1} mid H_n]Z] - E[Y_n Z].$$
Now recall the measure-theoretic definition of the conditional expectation $E[Y_{n+1} mid H_n]$.
answered Jan 21 at 3:23
angryavianangryavian
42k23381
$endgroup$
By the hypothesis,
$$0 = E[(Y_{n+1} - Y_n) Z] = E[E[Y_{n+1} mid H_n]Z] - E[Y_n Z].$$
Now recall the measure-theoretic definition of the conditional expectation $E[Y_{n+1} mid H_n]$.
answered Jan 21 at 3:23
angryavianangryavian
42k23381
answered Jan 21 at 3:23
angryavianangryavian
42k23381
answered Jan 21 at 3:23
angryavianangryavian
42k23381
answered Jan 21 at 3:23
angryavianangryavian
42k23381
42k23381
add a comment |
add a comment |
$begingroup$
If $Z = mathbf{1}_mathrm{X}$ for $mathrm{X} in sigma(X_1, ldots, X_n)$ then $mathbf{E}((Y_{n+1}-Y_n)Z) = 0$ signifies $intlimits_mathrm{X} Y_{n+1} dmathbf{P} = intlimits_mathrm{X} Y_n dmathbf{P},$ and since $Y_n$ is $mathscr{H}_n$-measurable, the result follows upon remembering what the conditional expectation means.
answered Jan 21 at 3:25
Will M.Will M.
2,835315
$endgroup$
add a comment |
$begingroup$
If $Z = mathbf{1}_mathrm{X}$ for $mathrm{X} in sigma(X_1, ldots, X_n)$ then $mathbf{E}((Y_{n+1}-Y_n)Z) = 0$ signifies $intlimits_mathrm{X} Y_{n+1} dmathbf{P} = intlimits_mathrm{X} Y_n dmathbf{P},$ and since $Y_n$ is $mathscr{H}_n$-measurable, the result follows upon remembering what the conditional expectation means.
answered Jan 21 at 3:25
Will M.Will M.
2,835315
$endgroup$
add a comment |
$begingroup$
If $Z = mathbf{1}_mathrm{X}$ for $mathrm{X} in sigma(X_1, ldots, X_n)$ then $mathbf{E}((Y_{n+1}-Y_n)Z) = 0$ signifies $intlimits_mathrm{X} Y_{n+1} dmathbf{P} = intlimits_mathrm{X} Y_n dmathbf{P},$ and since $Y_n$ is $mathscr{H}_n$-measurable, the result follows upon remembering what the conditional expectation means.
answered Jan 21 at 3:25
Will M.Will M.
2,835315
$endgroup$
If $Z = mathbf{1}_mathrm{X}$ for $mathrm{X} in sigma(X_1, ldots, X_n)$ then $mathbf{E}((Y_{n+1}-Y_n)Z) = 0$ signifies $intlimits_mathrm{X} Y_{n+1} dmathbf{P} = intlimits_mathrm{X} Y_n dmathbf{P},$ and since $Y_n$ is $mathscr{H}_n$-measurable, the result follows upon remembering what the conditional expectation means.
answered Jan 21 at 3:25
Will M.Will M.
2,835315
answered Jan 21 at 3:25
Will M.Will M.
2,835315
answered Jan 21 at 3:25
Will M.Will M.
2,835315
answered Jan 21 at 3:25
Will M.Will M.
2,835315
2,835315
add a comment |
add a comment |
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