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Ito's formula, and the relationship between dt and dB(t)
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reading Bernt Oksendal Stochastic Differential Equations.
I have just seen Ito's Formula, after this the author then says where $dX(t)^2$ is calculated using
$$dtcdot dt=dt cdot dB(t) =dB(t)cdot dt =0$$
and
$$dB(t) cdot dB(t) = dt$$
On page 45 at the top. Where do these relations come from, what is the intuition behind it.
stochastic-processes stochastic-calculus stochastic-integrals
asked 23 hours ago
Monty
15212
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favorite
reading Bernt Oksendal Stochastic Differential Equations.
I have just seen Ito's Formula, after this the author then says where $dX(t)^2$ is calculated using
$$dtcdot dt=dt cdot dB(t) =dB(t)cdot dt =0$$
and
$$dB(t) cdot dB(t) = dt$$
On page 45 at the top. Where do these relations come from, what is the intuition behind it.
stochastic-processes stochastic-calculus stochastic-integrals
asked 23 hours ago
Monty
15212
It comes from the fact that the variance of the (suitably normalised) stochastic process $B_t$ known as "Brownian motion" is equal to $t$.
– uniquesolution
22 hours ago
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
reading Bernt Oksendal Stochastic Differential Equations.
I have just seen Ito's Formula, after this the author then says where $dX(t)^2$ is calculated using
$$dtcdot dt=dt cdot dB(t) =dB(t)cdot dt =0$$
and
$$dB(t) cdot dB(t) = dt$$
On page 45 at the top. Where do these relations come from, what is the intuition behind it.
stochastic-processes stochastic-calculus stochastic-integrals
asked 23 hours ago
Monty
15212
reading Bernt Oksendal Stochastic Differential Equations.
I have just seen Ito's Formula, after this the author then says where $dX(t)^2$ is calculated using
$$dtcdot dt=dt cdot dB(t) =dB(t)cdot dt =0$$
and
$$dB(t) cdot dB(t) = dt$$
On page 45 at the top. Where do these relations come from, what is the intuition behind it.
stochastic-processes stochastic-calculus stochastic-integrals
asked 23 hours ago
Monty
15212
asked 23 hours ago
Monty
15212
asked 23 hours ago
Monty
15212
asked 23 hours ago
Monty
15212
15212
It comes from the fact that the variance of the (suitably normalised) stochastic process $B_t$ known as "Brownian motion" is equal to $t$.
– uniquesolution
22 hours ago
add a comment |Â
It comes from the fact that the variance of the (suitably normalised) stochastic process $B_t$ known as "Brownian motion" is equal to $t$.
– uniquesolution
22 hours ago
It comes from the fact that the variance of the (suitably normalised) stochastic process $B_t$ known as "Brownian motion" is equal to $t$.
– uniquesolution
22 hours ago
It comes from the fact that the variance of the (suitably normalised) stochastic process $B_t$ known as "Brownian motion" is equal to $t$.
– uniquesolution
22 hours ago
add a comment |Â
2 Answers
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For suitable $mu$ and $sigma$, remember that
$$
dX_t = mu(t, X_t) dt + sigma(t, X_t) dB_t, qquad X_0 = x,
$$
is shorthand for
$$
X_t = x + int_0^t mu(s, X_s) ds + int_0^t sigma(s, X_s) dB_s.
$$
Itô's formula for a suitable $f$ tells us that
$$
f(t, X_t)
= f(t, x) + int_0^t f_t(s, X_s) dt + int_0^t f_x(s, X_s) dX_t
+ frac 1 2 int_0^t f_xx(x, X_s) d langle X rangle_t,
$$
where $langle X rangle = (langle X rangle_t)_t geq 0$ is the quadratic variation process of $X$ and is equal to
$$
langle X rangle_t
= left langle int_0^cdot sigma(s, X_s) d W_s right rangle_t
= int_0^t sigma^2(s, X_s) d langle W rangle_s
= int_0^t sigma^2(s, X_s) ds.
$$
The multiplication rules in Øksendal are stated in the way you wrote, so that the version of Itô's formula in that book yields the correct result. In essence, this is because the quadratic variation for a Brownain motion is $langle B rangle_t = t$. It is a good exercise to verify this for yourself. I hope this answers your question.
answered 42 mins ago
K. Brix
4661211
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up vote
0
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$dtcdot dt=0$ is shorthand notation for the fact that the quadratic variation of the identity function with itself is $0$.
$dtcdot dB(t)=0$ is shorthand notation for the fact that the cross variation of Brownian motion with the identity function is $0$.
$dB(t)cdot dB(t)=dt$ is shorthand notation for the fact that the quadratic variation of Brownian motion with itself over a time interval of length $t$ is equal to $t$.
answered 33 mins ago
gs.co
586
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For suitable $mu$ and $sigma$, remember that
$$
dX_t = mu(t, X_t) dt + sigma(t, X_t) dB_t, qquad X_0 = x,
$$
is shorthand for
$$
X_t = x + int_0^t mu(s, X_s) ds + int_0^t sigma(s, X_s) dB_s.
$$
Itô's formula for a suitable $f$ tells us that
$$
f(t, X_t)
= f(t, x) + int_0^t f_t(s, X_s) dt + int_0^t f_x(s, X_s) dX_t
+ frac 1 2 int_0^t f_xx(x, X_s) d langle X rangle_t,
$$
where $langle X rangle = (langle X rangle_t)_t geq 0$ is the quadratic variation process of $X$ and is equal to
$$
langle X rangle_t
= left langle int_0^cdot sigma(s, X_s) d W_s right rangle_t
= int_0^t sigma^2(s, X_s) d langle W rangle_s
= int_0^t sigma^2(s, X_s) ds.
$$
The multiplication rules in Øksendal are stated in the way you wrote, so that the version of Itô's formula in that book yields the correct result. In essence, this is because the quadratic variation for a Brownain motion is $langle B rangle_t = t$. It is a good exercise to verify this for yourself. I hope this answers your question.
answered 42 mins ago
K. Brix
4661211
add a comment |Â
up vote
0
down vote
For suitable $mu$ and $sigma$, remember that
$$
dX_t = mu(t, X_t) dt + sigma(t, X_t) dB_t, qquad X_0 = x,
$$
is shorthand for
$$
X_t = x + int_0^t mu(s, X_s) ds + int_0^t sigma(s, X_s) dB_s.
$$
Itô's formula for a suitable $f$ tells us that
$$
f(t, X_t)
= f(t, x) + int_0^t f_t(s, X_s) dt + int_0^t f_x(s, X_s) dX_t
+ frac 1 2 int_0^t f_xx(x, X_s) d langle X rangle_t,
$$
where $langle X rangle = (langle X rangle_t)_t geq 0$ is the quadratic variation process of $X$ and is equal to
$$
langle X rangle_t
= left langle int_0^cdot sigma(s, X_s) d W_s right rangle_t
= int_0^t sigma^2(s, X_s) d langle W rangle_s
= int_0^t sigma^2(s, X_s) ds.
$$
The multiplication rules in Øksendal are stated in the way you wrote, so that the version of Itô's formula in that book yields the correct result. In essence, this is because the quadratic variation for a Brownain motion is $langle B rangle_t = t$. It is a good exercise to verify this for yourself. I hope this answers your question.
answered 42 mins ago
K. Brix
4661211
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For suitable $mu$ and $sigma$, remember that
$$
dX_t = mu(t, X_t) dt + sigma(t, X_t) dB_t, qquad X_0 = x,
$$
is shorthand for
$$
X_t = x + int_0^t mu(s, X_s) ds + int_0^t sigma(s, X_s) dB_s.
$$
Itô's formula for a suitable $f$ tells us that
$$
f(t, X_t)
= f(t, x) + int_0^t f_t(s, X_s) dt + int_0^t f_x(s, X_s) dX_t
+ frac 1 2 int_0^t f_xx(x, X_s) d langle X rangle_t,
$$
where $langle X rangle = (langle X rangle_t)_t geq 0$ is the quadratic variation process of $X$ and is equal to
$$
langle X rangle_t
= left langle int_0^cdot sigma(s, X_s) d W_s right rangle_t
= int_0^t sigma^2(s, X_s) d langle W rangle_s
= int_0^t sigma^2(s, X_s) ds.
$$
The multiplication rules in Øksendal are stated in the way you wrote, so that the version of Itô's formula in that book yields the correct result. In essence, this is because the quadratic variation for a Brownain motion is $langle B rangle_t = t$. It is a good exercise to verify this for yourself. I hope this answers your question.
answered 42 mins ago
K. Brix
4661211
For suitable $mu$ and $sigma$, remember that
$$
dX_t = mu(t, X_t) dt + sigma(t, X_t) dB_t, qquad X_0 = x,
$$
is shorthand for
$$
X_t = x + int_0^t mu(s, X_s) ds + int_0^t sigma(s, X_s) dB_s.
$$
Itô's formula for a suitable $f$ tells us that
$$
f(t, X_t)
= f(t, x) + int_0^t f_t(s, X_s) dt + int_0^t f_x(s, X_s) dX_t
+ frac 1 2 int_0^t f_xx(x, X_s) d langle X rangle_t,
$$
where $langle X rangle = (langle X rangle_t)_t geq 0$ is the quadratic variation process of $X$ and is equal to
$$
langle X rangle_t
= left langle int_0^cdot sigma(s, X_s) d W_s right rangle_t
= int_0^t sigma^2(s, X_s) d langle W rangle_s
= int_0^t sigma^2(s, X_s) ds.
$$
The multiplication rules in Øksendal are stated in the way you wrote, so that the version of Itô's formula in that book yields the correct result. In essence, this is because the quadratic variation for a Brownain motion is $langle B rangle_t = t$. It is a good exercise to verify this for yourself. I hope this answers your question.
answered 42 mins ago
K. Brix
4661211
edited 37 mins ago
answered 42 mins ago
K. Brix
4661211
answered 42 mins ago
K. Brix
4661211
answered 42 mins ago
K. Brix
4661211
4661211
add a comment |Â
add a comment |Â
up vote
0
down vote
$dtcdot dt=0$ is shorthand notation for the fact that the quadratic variation of the identity function with itself is $0$.
$dtcdot dB(t)=0$ is shorthand notation for the fact that the cross variation of Brownian motion with the identity function is $0$.
$dB(t)cdot dB(t)=dt$ is shorthand notation for the fact that the quadratic variation of Brownian motion with itself over a time interval of length $t$ is equal to $t$.
answered 33 mins ago
gs.co
586
add a comment |Â
up vote
0
down vote
$dtcdot dt=0$ is shorthand notation for the fact that the quadratic variation of the identity function with itself is $0$.
$dtcdot dB(t)=0$ is shorthand notation for the fact that the cross variation of Brownian motion with the identity function is $0$.
$dB(t)cdot dB(t)=dt$ is shorthand notation for the fact that the quadratic variation of Brownian motion with itself over a time interval of length $t$ is equal to $t$.
answered 33 mins ago
gs.co
586
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$dtcdot dt=0$ is shorthand notation for the fact that the quadratic variation of the identity function with itself is $0$.
$dtcdot dB(t)=0$ is shorthand notation for the fact that the cross variation of Brownian motion with the identity function is $0$.
$dB(t)cdot dB(t)=dt$ is shorthand notation for the fact that the quadratic variation of Brownian motion with itself over a time interval of length $t$ is equal to $t$.
answered 33 mins ago
gs.co
586
$dtcdot dt=0$ is shorthand notation for the fact that the quadratic variation of the identity function with itself is $0$.
$dtcdot dB(t)=0$ is shorthand notation for the fact that the cross variation of Brownian motion with the identity function is $0$.
$dB(t)cdot dB(t)=dt$ is shorthand notation for the fact that the quadratic variation of Brownian motion with itself over a time interval of length $t$ is equal to $t$.
answered 33 mins ago
gs.co
586
answered 33 mins ago
gs.co
586
answered 33 mins ago
gs.co
586
answered 33 mins ago
gs.co
586
586
add a comment |Â
add a comment |Â
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It comes from the fact that the variance of the (suitably normalised) stochastic process $B_t$ known as "Brownian motion" is equal to $t$.
– uniquesolution
22 hours ago