Question on Gÿongy' lemma proofDeterministic interpretation of stochastic differential equationquestion on Leif Andersen's “Interest Rate Modeling, vol 2 Term Structure Models”Ito, Stochastic Exponential and GirsanovDupire's formula proofIs the 'constant weight in the risky asset' portfolio-strategy self-financing?Ito representation unique up to indistinguishability? Proof?Calculating the stochastic integral of $exp(-rt)S_t$Conditional Expectation with Indicator Functions for Poisson Process First Jump Time (Option Pricing PDE)On quadratic covariationDerivation and expectation interchange
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Question on Gÿongy' lemma proof
Deterministic interpretation of stochastic differential equationquestion on Leif Andersen's “Interest Rate Modeling, vol 2 Term Structure Models”Ito, Stochastic Exponential and GirsanovDupire's formula proofIs the 'constant weight in the risky asset' portfolio-strategy self-financing?Ito representation unique up to indistinguishability? Proof?Calculating the stochastic integral of $exp(-rt)S_t$Conditional Expectation with Indicator Functions for Poisson Process First Jump Time (Option Pricing PDE)On quadratic covariationDerivation and expectation interchange
$begingroup$
I have some questions regarding a proof of Gÿongy's lemma given in 1
I would like to understand the following passage:
$$
int_s=t_0^s=tmathbbEleft[delta(X_s-K)langle dX_srangle^2 right]=
int_s=t_0^s=tmathbbEleft[delta(X_s-K) right] mathbbEleft[langle dX_srangle^2|X_s=K right]
$$
Thanks
stochastic-calculus
asked Apr 22 at 9:17
AguelmameAguelmame
511
$endgroup$
add a comment |
$begingroup$
I have some questions regarding a proof of Gÿongy's lemma given in 1
I would like to understand the following passage:
$$
int_s=t_0^s=tmathbbEleft[delta(X_s-K)langle dX_srangle^2 right]=
int_s=t_0^s=tmathbbEleft[delta(X_s-K) right] mathbbEleft[langle dX_srangle^2|X_s=K right]
$$
Thanks
stochastic-calculus
asked Apr 22 at 9:17
AguelmameAguelmame
511
$endgroup$
add a comment |
$begingroup$
I have some questions regarding a proof of Gÿongy's lemma given in 1
I would like to understand the following passage:
$$
int_s=t_0^s=tmathbbEleft[delta(X_s-K)langle dX_srangle^2 right]=
int_s=t_0^s=tmathbbEleft[delta(X_s-K) right] mathbbEleft[langle dX_srangle^2|X_s=K right]
$$
Thanks
stochastic-calculus
asked Apr 22 at 9:17
AguelmameAguelmame
511
$endgroup$
I have some questions regarding a proof of Gÿongy's lemma given in 1
I would like to understand the following passage:
$$
int_s=t_0^s=tmathbbEleft[delta(X_s-K)langle dX_srangle^2 right]=
int_s=t_0^s=tmathbbEleft[delta(X_s-K) right] mathbbEleft[langle dX_srangle^2|X_s=K right]
$$
Thanks
stochastic-calculus
stochastic-calculus
asked Apr 22 at 9:17
AguelmameAguelmame
511
asked Apr 22 at 9:17
AguelmameAguelmame
511
asked Apr 22 at 9:17
AguelmameAguelmame
511
asked Apr 22 at 9:17
AguelmameAguelmame
511
asked Apr 22 at 9:17
AguelmameAguelmame
511
511
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $X_s neq K $ then the delta function gives zero, and the product is zero. So the term only contributes when $X_s=K$.
Re-comment, the key to understanding this is the conditional expectation:
$ E left[ dX_s^2 mid X_s=K right] =fracEleft[ dX_s^2 delta(X_s-K)right]Eleft [delta(X_s-K)right] $
Where it might be helpful if you interpret the delta as indicator of $X_s=K$
answered Apr 22 at 11:20
Magic is in the chainMagic is in the chain
1,27915
$endgroup$
$begingroup$
What I don't understand if why the expectation of a product equals the product of expectations. Is this true because the quadratic variation $langle dX_srangle^2$ is independant from $X_s$ ?
$endgroup$
– Aguelmame
Apr 22 at 17:04
$begingroup$
Added further explanation in the answer.
$endgroup$
– Magic is in the chain
Apr 22 at 17:29
1
$begingroup$
Ok, that makes sense. Thanks
$endgroup$
– Aguelmame
Apr 22 at 18:09
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
If $X_s neq K $ then the delta function gives zero, and the product is zero. So the term only contributes when $X_s=K$.
Re-comment, the key to understanding this is the conditional expectation:
$ E left[ dX_s^2 mid X_s=K right] =fracEleft[ dX_s^2 delta(X_s-K)right]Eleft [delta(X_s-K)right] $
Where it might be helpful if you interpret the delta as indicator of $X_s=K$
answered Apr 22 at 11:20
Magic is in the chainMagic is in the chain
1,27915
$endgroup$
$begingroup$
What I don't understand if why the expectation of a product equals the product of expectations. Is this true because the quadratic variation $langle dX_srangle^2$ is independant from $X_s$ ?
$endgroup$
– Aguelmame
Apr 22 at 17:04
$begingroup$
Added further explanation in the answer.
$endgroup$
– Magic is in the chain
Apr 22 at 17:29
1
$begingroup$
Ok, that makes sense. Thanks
$endgroup$
– Aguelmame
Apr 22 at 18:09
add a comment |
$begingroup$
If $X_s neq K $ then the delta function gives zero, and the product is zero. So the term only contributes when $X_s=K$.
Re-comment, the key to understanding this is the conditional expectation:
$ E left[ dX_s^2 mid X_s=K right] =fracEleft[ dX_s^2 delta(X_s-K)right]Eleft [delta(X_s-K)right] $
Where it might be helpful if you interpret the delta as indicator of $X_s=K$
answered Apr 22 at 11:20
Magic is in the chainMagic is in the chain
1,27915
$endgroup$
$begingroup$
What I don't understand if why the expectation of a product equals the product of expectations. Is this true because the quadratic variation $langle dX_srangle^2$ is independant from $X_s$ ?
$endgroup$
– Aguelmame
Apr 22 at 17:04
$begingroup$
Added further explanation in the answer.
$endgroup$
– Magic is in the chain
Apr 22 at 17:29
1
$begingroup$
Ok, that makes sense. Thanks
$endgroup$
– Aguelmame
Apr 22 at 18:09
add a comment |
$begingroup$
If $X_s neq K $ then the delta function gives zero, and the product is zero. So the term only contributes when $X_s=K$.
Re-comment, the key to understanding this is the conditional expectation:
$ E left[ dX_s^2 mid X_s=K right] =fracEleft[ dX_s^2 delta(X_s-K)right]Eleft [delta(X_s-K)right] $
Where it might be helpful if you interpret the delta as indicator of $X_s=K$
answered Apr 22 at 11:20
Magic is in the chainMagic is in the chain
1,27915
$endgroup$
If $X_s neq K $ then the delta function gives zero, and the product is zero. So the term only contributes when $X_s=K$.
Re-comment, the key to understanding this is the conditional expectation:
$ E left[ dX_s^2 mid X_s=K right] =fracEleft[ dX_s^2 delta(X_s-K)right]Eleft [delta(X_s-K)right] $
Where it might be helpful if you interpret the delta as indicator of $X_s=K$
answered Apr 22 at 11:20
Magic is in the chainMagic is in the chain
1,27915
edited Apr 22 at 17:28
answered Apr 22 at 11:20
Magic is in the chainMagic is in the chain
1,27915
answered Apr 22 at 11:20
Magic is in the chainMagic is in the chain
1,27915
answered Apr 22 at 11:20
Magic is in the chainMagic is in the chain
1,27915
1,27915
$begingroup$
What I don't understand if why the expectation of a product equals the product of expectations. Is this true because the quadratic variation $langle dX_srangle^2$ is independant from $X_s$ ?
$endgroup$
– Aguelmame
Apr 22 at 17:04
$begingroup$
Added further explanation in the answer.
$endgroup$
– Magic is in the chain
Apr 22 at 17:29
1
$begingroup$
Ok, that makes sense. Thanks
$endgroup$
– Aguelmame
Apr 22 at 18:09
add a comment |
$begingroup$
What I don't understand if why the expectation of a product equals the product of expectations. Is this true because the quadratic variation $langle dX_srangle^2$ is independant from $X_s$ ?
$endgroup$
– Aguelmame
Apr 22 at 17:04
$begingroup$
Added further explanation in the answer.
$endgroup$
– Magic is in the chain
Apr 22 at 17:29
1
$begingroup$
Ok, that makes sense. Thanks
$endgroup$
– Aguelmame
Apr 22 at 18:09
$begingroup$
What I don't understand if why the expectation of a product equals the product of expectations. Is this true because the quadratic variation $langle dX_srangle^2$ is independant from $X_s$ ?
$endgroup$
– Aguelmame
Apr 22 at 17:04
$begingroup$
What I don't understand if why the expectation of a product equals the product of expectations. Is this true because the quadratic variation $langle dX_srangle^2$ is independant from $X_s$ ?
$endgroup$
– Aguelmame
Apr 22 at 17:04
$begingroup$
Added further explanation in the answer.
$endgroup$
– Magic is in the chain
Apr 22 at 17:29
$begingroup$
Added further explanation in the answer.
$endgroup$
– Magic is in the chain
Apr 22 at 17:29
1
1
$begingroup$
Ok, that makes sense. Thanks
$endgroup$
– Aguelmame
Apr 22 at 18:09
$begingroup$
Ok, that makes sense. Thanks
$endgroup$
– Aguelmame
Apr 22 at 18:09
add a comment |
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