Show $[int _0 ^t B_{2,s} dB_{1,s}]^2 - int_0^t (B_{2,s})^2 ds$ is a martingale
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I want to show the following:
$$[int _0 ^t B_{2,s} dB_{1,s}]^2 - int_0^t (B_{2,s})^2 ds$$ is a martingale where $ (B_{2,s},B_{1,s}) $ is a two dim'l Brownian motion.
My attempt:
By Ito formula, $$[int _0 ^t B_{2,s} dB_{1,s}]^2 =2 int_0^t (int_0^u B_{2,s} dB_{1,s} )B_{2,u} dB_{1,u} + int_0^t (B_{2,u})^2 du$$
so it suffices to show
$ int_0^t (int_0^u B_{2,s} dB_{1,s} )B_{2,u} dB_{1,u}$ is a martingale. I tried to show the integral of the expectation of the square of the integrand is finite, but it gets me nowhere.
Any help is appreciated.
stochastic-calculus brownian-motion martingales local-martingales
asked Nov 20 at 1:55
izimath
337110
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up vote
1
down vote
favorite
I want to show the following:
$$[int _0 ^t B_{2,s} dB_{1,s}]^2 - int_0^t (B_{2,s})^2 ds$$ is a martingale where $ (B_{2,s},B_{1,s}) $ is a two dim'l Brownian motion.
My attempt:
By Ito formula, $$[int _0 ^t B_{2,s} dB_{1,s}]^2 =2 int_0^t (int_0^u B_{2,s} dB_{1,s} )B_{2,u} dB_{1,u} + int_0^t (B_{2,u})^2 du$$
so it suffices to show
$ int_0^t (int_0^u B_{2,s} dB_{1,s} )B_{2,u} dB_{1,u}$ is a martingale. I tried to show the integral of the expectation of the square of the integrand is finite, but it gets me nowhere.
Any help is appreciated.
stochastic-calculus brownian-motion martingales local-martingales
asked Nov 20 at 1:55
izimath
337110
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to show the following:
$$[int _0 ^t B_{2,s} dB_{1,s}]^2 - int_0^t (B_{2,s})^2 ds$$ is a martingale where $ (B_{2,s},B_{1,s}) $ is a two dim'l Brownian motion.
My attempt:
By Ito formula, $$[int _0 ^t B_{2,s} dB_{1,s}]^2 =2 int_0^t (int_0^u B_{2,s} dB_{1,s} )B_{2,u} dB_{1,u} + int_0^t (B_{2,u})^2 du$$
so it suffices to show
$ int_0^t (int_0^u B_{2,s} dB_{1,s} )B_{2,u} dB_{1,u}$ is a martingale. I tried to show the integral of the expectation of the square of the integrand is finite, but it gets me nowhere.
Any help is appreciated.
stochastic-calculus brownian-motion martingales local-martingales
asked Nov 20 at 1:55
izimath
337110
I want to show the following:
$$[int _0 ^t B_{2,s} dB_{1,s}]^2 - int_0^t (B_{2,s})^2 ds$$ is a martingale where $ (B_{2,s},B_{1,s}) $ is a two dim'l Brownian motion.
My attempt:
By Ito formula, $$[int _0 ^t B_{2,s} dB_{1,s}]^2 =2 int_0^t (int_0^u B_{2,s} dB_{1,s} )B_{2,u} dB_{1,u} + int_0^t (B_{2,u})^2 du$$
so it suffices to show
$ int_0^t (int_0^u B_{2,s} dB_{1,s} )B_{2,u} dB_{1,u}$ is a martingale. I tried to show the integral of the expectation of the square of the integrand is finite, but it gets me nowhere.
Any help is appreciated.
stochastic-calculus brownian-motion martingales local-martingales
stochastic-calculus brownian-motion martingales local-martingales
asked Nov 20 at 1:55
izimath
337110
asked Nov 20 at 1:55
izimath
337110
asked Nov 20 at 1:55
izimath
337110
asked Nov 20 at 1:55
izimath
337110
asked Nov 20 at 1:55
izimath
337110
337110
add a comment |
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1 Answer
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Hint: Note that $$int_0^t B_{2,s}dB_{1,s}$$ is an $L^2$-martingale since $E[int_0^t |B_{2,s}|^2ds] = int_0^t E|B_{2,s}|^2ds <infty$.
And observe that its quadratic variatio process is $int_0^t B_{2,s}^2ds$. It follows the given process is a continuous martingale.
answered Nov 20 at 5:05
Song
5439
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: Note that $$int_0^t B_{2,s}dB_{1,s}$$ is an $L^2$-martingale since $E[int_0^t |B_{2,s}|^2ds] = int_0^t E|B_{2,s}|^2ds <infty$.
And observe that its quadratic variatio process is $int_0^t B_{2,s}^2ds$. It follows the given process is a continuous martingale.
answered Nov 20 at 5:05
Song
5439
add a comment |
up vote
0
down vote
Hint: Note that $$int_0^t B_{2,s}dB_{1,s}$$ is an $L^2$-martingale since $E[int_0^t |B_{2,s}|^2ds] = int_0^t E|B_{2,s}|^2ds <infty$.
And observe that its quadratic variatio process is $int_0^t B_{2,s}^2ds$. It follows the given process is a continuous martingale.
answered Nov 20 at 5:05
Song
5439
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: Note that $$int_0^t B_{2,s}dB_{1,s}$$ is an $L^2$-martingale since $E[int_0^t |B_{2,s}|^2ds] = int_0^t E|B_{2,s}|^2ds <infty$.
And observe that its quadratic variatio process is $int_0^t B_{2,s}^2ds$. It follows the given process is a continuous martingale.
answered Nov 20 at 5:05
Song
5439
Hint: Note that $$int_0^t B_{2,s}dB_{1,s}$$ is an $L^2$-martingale since $E[int_0^t |B_{2,s}|^2ds] = int_0^t E|B_{2,s}|^2ds <infty$.
And observe that its quadratic variatio process is $int_0^t B_{2,s}^2ds$. It follows the given process is a continuous martingale.
answered Nov 20 at 5:05
Song
5439
answered Nov 20 at 5:05
Song
5439
answered Nov 20 at 5:05
Song
5439
answered Nov 20 at 5:05
Song
5439
5439
add a comment |
add a comment |
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