HJM in infinite dimensionsMusiela parameterizationHJM simulation problemHJM framework problem - showing that...
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HJM in infinite dimensions
Musiela parameterizationHJM simulation problemHJM framework problem - showing that HJM drift condition implies that $b(z)=b+βz$ and $(ρ)^2=α$Understanding the HJM drift condition's dimensionsBaxter & Rennie HJM: differentiating Ito integralHJM or Short rates model?HJM model Baxter Rennie: differentiating the discounted asset price using Ito
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I recently started reading Filipovic's Consistency problems for HJM interest rate models and came across the Musiela reparametrization
$$r_t(x)=f(t,x+t)$$ so the forward curve can be thought of as a map $xto r_t(x)$. The book goes on to tell that this maybe thought of as an infinite dimensional state variable.
Does anyone have a good explanation for this? Is it because we pick $r_t$ from a space of functions?
interest-rates stochastic-calculus heath-jarrow-morton
asked 14 hours ago
HeisenbergHeisenberg
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I recently started reading Filipovic's Consistency problems for HJM interest rate models and came across the Musiela reparametrization
$$r_t(x)=f(t,x+t)$$ so the forward curve can be thought of as a map $xto r_t(x)$. The book goes on to tell that this maybe thought of as an infinite dimensional state variable.
Does anyone have a good explanation for this? Is it because we pick $r_t$ from a space of functions?
interest-rates stochastic-calculus heath-jarrow-morton
asked 14 hours ago
HeisenbergHeisenberg
1135 bronze badges
New contributor
Heisenberg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I recently started reading Filipovic's Consistency problems for HJM interest rate models and came across the Musiela reparametrization
$$r_t(x)=f(t,x+t)$$ so the forward curve can be thought of as a map $xto r_t(x)$. The book goes on to tell that this maybe thought of as an infinite dimensional state variable.
Does anyone have a good explanation for this? Is it because we pick $r_t$ from a space of functions?
interest-rates stochastic-calculus heath-jarrow-morton
asked 14 hours ago
HeisenbergHeisenberg
1135 bronze badges
New contributor
Heisenberg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I recently started reading Filipovic's Consistency problems for HJM interest rate models and came across the Musiela reparametrization
$$r_t(x)=f(t,x+t)$$ so the forward curve can be thought of as a map $xto r_t(x)$. The book goes on to tell that this maybe thought of as an infinite dimensional state variable.
Does anyone have a good explanation for this? Is it because we pick $r_t$ from a space of functions?
interest-rates stochastic-calculus heath-jarrow-morton
interest-rates stochastic-calculus heath-jarrow-morton
asked 14 hours ago
HeisenbergHeisenberg
1135 bronze badges
New contributor
Heisenberg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 hours ago
HeisenbergHeisenberg
1135 bronze badges
New contributor
Heisenberg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
Heisenberg
asked 14 hours ago
HeisenbergHeisenberg
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New contributor
Heisenberg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 14 hours ago
HeisenbergHeisenberg
1135 bronze badges
asked 14 hours ago
HeisenbergHeisenberg
1135 bronze badges
1135 bronze badges
New contributor
Heisenberg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Heisenberg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
Keeping it simple, you know HJM SDE gives the dynamics of an instantaneous forward referencing a fixed maturity T, $fleft(t, Tright)$, but there is a continuum of such maturities - the whole forward curve as a function of T.
You can have each forward driven by a different brownian for example, so in general the HJM approach will be infinite dimensional. to visualise infinite dimensions, it may be helpful to recall one dimensional SDE, two dimensional SDE, and so on. But there are special cases for which it an be viewed as finite dimensional.
Please also see the discussion here: Musiela parameterization
answered 14 hours ago
Magic is in the chainMagic is in the chain
2,6991 gold badge4 silver badges7 bronze badges
$endgroup$
$begingroup$
Say we want to price a set of bonds with maturities $T$ varying on a certain interval so we have a continuum of bonds. In this case I see why we end up with a continuum of forwards. But what if we just take a finite set of bonds?
$endgroup$
– Heisenberg
12 hours ago
1
$begingroup$
But then even a single maturity zero coupon depends on the entire range of instantaneous forwards, remember the relationship: $B=e^{-int_{0}^{T}{f(0,u)du}}$
$endgroup$
– Magic is in the chain
12 hours ago
1
$begingroup$
Oh yes!! Thank you so much it makes sense now!
$endgroup$
– Heisenberg
12 hours ago
add a comment |
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1 Answer
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$begingroup$
Keeping it simple, you know HJM SDE gives the dynamics of an instantaneous forward referencing a fixed maturity T, $fleft(t, Tright)$, but there is a continuum of such maturities - the whole forward curve as a function of T.
You can have each forward driven by a different brownian for example, so in general the HJM approach will be infinite dimensional. to visualise infinite dimensions, it may be helpful to recall one dimensional SDE, two dimensional SDE, and so on. But there are special cases for which it an be viewed as finite dimensional.
Please also see the discussion here: Musiela parameterization
answered 14 hours ago
Magic is in the chainMagic is in the chain
2,6991 gold badge4 silver badges7 bronze badges
$endgroup$
$begingroup$
Say we want to price a set of bonds with maturities $T$ varying on a certain interval so we have a continuum of bonds. In this case I see why we end up with a continuum of forwards. But what if we just take a finite set of bonds?
$endgroup$
– Heisenberg
12 hours ago
1
$begingroup$
But then even a single maturity zero coupon depends on the entire range of instantaneous forwards, remember the relationship: $B=e^{-int_{0}^{T}{f(0,u)du}}$
$endgroup$
– Magic is in the chain
12 hours ago
1
$begingroup$
Oh yes!! Thank you so much it makes sense now!
$endgroup$
– Heisenberg
12 hours ago
add a comment |
$begingroup$
Keeping it simple, you know HJM SDE gives the dynamics of an instantaneous forward referencing a fixed maturity T, $fleft(t, Tright)$, but there is a continuum of such maturities - the whole forward curve as a function of T.
You can have each forward driven by a different brownian for example, so in general the HJM approach will be infinite dimensional. to visualise infinite dimensions, it may be helpful to recall one dimensional SDE, two dimensional SDE, and so on. But there are special cases for which it an be viewed as finite dimensional.
Please also see the discussion here: Musiela parameterization
answered 14 hours ago
Magic is in the chainMagic is in the chain
2,6991 gold badge4 silver badges7 bronze badges
$endgroup$
$begingroup$
Say we want to price a set of bonds with maturities $T$ varying on a certain interval so we have a continuum of bonds. In this case I see why we end up with a continuum of forwards. But what if we just take a finite set of bonds?
$endgroup$
– Heisenberg
12 hours ago
1
$begingroup$
But then even a single maturity zero coupon depends on the entire range of instantaneous forwards, remember the relationship: $B=e^{-int_{0}^{T}{f(0,u)du}}$
$endgroup$
– Magic is in the chain
12 hours ago
1
$begingroup$
Oh yes!! Thank you so much it makes sense now!
$endgroup$
– Heisenberg
12 hours ago
add a comment |
$begingroup$
Keeping it simple, you know HJM SDE gives the dynamics of an instantaneous forward referencing a fixed maturity T, $fleft(t, Tright)$, but there is a continuum of such maturities - the whole forward curve as a function of T.
You can have each forward driven by a different brownian for example, so in general the HJM approach will be infinite dimensional. to visualise infinite dimensions, it may be helpful to recall one dimensional SDE, two dimensional SDE, and so on. But there are special cases for which it an be viewed as finite dimensional.
Please also see the discussion here: Musiela parameterization
answered 14 hours ago
Magic is in the chainMagic is in the chain
2,6991 gold badge4 silver badges7 bronze badges
$endgroup$
Keeping it simple, you know HJM SDE gives the dynamics of an instantaneous forward referencing a fixed maturity T, $fleft(t, Tright)$, but there is a continuum of such maturities - the whole forward curve as a function of T.
You can have each forward driven by a different brownian for example, so in general the HJM approach will be infinite dimensional. to visualise infinite dimensions, it may be helpful to recall one dimensional SDE, two dimensional SDE, and so on. But there are special cases for which it an be viewed as finite dimensional.
Please also see the discussion here: Musiela parameterization
answered 14 hours ago
Magic is in the chainMagic is in the chain
2,6991 gold badge4 silver badges7 bronze badges
answered 14 hours ago
Magic is in the chainMagic is in the chain
2,6991 gold badge4 silver badges7 bronze badges
answered 14 hours ago
Magic is in the chainMagic is in the chain
2,6991 gold badge4 silver badges7 bronze badges
answered 14 hours ago
Magic is in the chainMagic is in the chain
2,6991 gold badge4 silver badges7 bronze badges
2,6991 gold badge4 silver badges7 bronze badges
$begingroup$
Say we want to price a set of bonds with maturities $T$ varying on a certain interval so we have a continuum of bonds. In this case I see why we end up with a continuum of forwards. But what if we just take a finite set of bonds?
$endgroup$
– Heisenberg
12 hours ago
1
$begingroup$
But then even a single maturity zero coupon depends on the entire range of instantaneous forwards, remember the relationship: $B=e^{-int_{0}^{T}{f(0,u)du}}$
$endgroup$
– Magic is in the chain
12 hours ago
1
$begingroup$
Oh yes!! Thank you so much it makes sense now!
$endgroup$
– Heisenberg
12 hours ago
add a comment |
$begingroup$
Say we want to price a set of bonds with maturities $T$ varying on a certain interval so we have a continuum of bonds. In this case I see why we end up with a continuum of forwards. But what if we just take a finite set of bonds?
$endgroup$
– Heisenberg
12 hours ago
1
$begingroup$
But then even a single maturity zero coupon depends on the entire range of instantaneous forwards, remember the relationship: $B=e^{-int_{0}^{T}{f(0,u)du}}$
$endgroup$
– Magic is in the chain
12 hours ago
1
$begingroup$
Oh yes!! Thank you so much it makes sense now!
$endgroup$
– Heisenberg
12 hours ago
$begingroup$
Say we want to price a set of bonds with maturities $T$ varying on a certain interval so we have a continuum of bonds. In this case I see why we end up with a continuum of forwards. But what if we just take a finite set of bonds?
$endgroup$
– Heisenberg
12 hours ago
$begingroup$
Say we want to price a set of bonds with maturities $T$ varying on a certain interval so we have a continuum of bonds. In this case I see why we end up with a continuum of forwards. But what if we just take a finite set of bonds?
$endgroup$
– Heisenberg
12 hours ago
1
1
$begingroup$
But then even a single maturity zero coupon depends on the entire range of instantaneous forwards, remember the relationship: $B=e^{-int_{0}^{T}{f(0,u)du}}$
$endgroup$
– Magic is in the chain
12 hours ago
$begingroup$
But then even a single maturity zero coupon depends on the entire range of instantaneous forwards, remember the relationship: $B=e^{-int_{0}^{T}{f(0,u)du}}$
$endgroup$
– Magic is in the chain
12 hours ago
1
1
$begingroup$
Oh yes!! Thank you so much it makes sense now!
$endgroup$
– Heisenberg
12 hours ago
$begingroup$
Oh yes!! Thank you so much it makes sense now!
$endgroup$
– Heisenberg
12 hours ago
add a comment |
Heisenberg is a new contributor. Be nice, and check out our Code of Conduct.
Heisenberg is a new contributor. Be nice, and check out our Code of Conduct.
Heisenberg is a new contributor. Be nice, and check out our Code of Conduct.
Heisenberg is a new contributor. Be nice, and check out our Code of Conduct.
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