Filtering Sum of Brownian Motions The 2019 Stack Overflow Developer Survey Results Are Ind-dimensional Brownian motion and martingalesLaws and Moments of two dimensional brownian motionsAbsorbed brownian motion is a Markov processFrom brownian bridge to brownian motion proofBrownian motions under different filtrations, quadratic covariation, covergence of approximating sumIs there a way to construct Brownian motion from a given Brownian motion?How to prove differential of product correlated Brownian Motions?Brownian motion in random landscapeCan we re-write the natural filtration of a Brownian Motion?Noisy Brownian Noise
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Filtering Sum of Brownian Motions
The 2019 Stack Overflow Developer Survey Results Are Ind-dimensional Brownian motion and martingalesLaws and Moments of two dimensional brownian motionsAbsorbed brownian motion is a Markov processFrom brownian bridge to brownian motion proofBrownian motions under different filtrations, quadratic covariation, covergence of approximating sumIs there a way to construct Brownian motion from a given Brownian motion?How to prove differential of product correlated Brownian Motions?Brownian motion in random landscapeCan we re-write the natural filtration of a Brownian Motion?Noisy Brownian Noise
$begingroup$
Let us assume that there exist two independent Brownian Motions $B_t$ and $W_t$, and consider their sum $Y_t=B_t + W_t$. Next, define the filtration generated by the sum, $mathcalF_t^Y=sigma(Y_u)_0
leq u leq t$.
How would one compute the filter $E[ B_t | mathcalF_t^Y]$?
My intuition tells me that the solution to this filtering problem should just be $E[B_t | mathcalF_t^Y] = frac12 Y_t$, although I cannot prove it. As a secondary question, can we generalize to having independent continuous martingales instead of two Brownian Motions?
Any help is appreciated!
stochastic-processes stochastic-calculus brownian-motion
edited 19 hours ago
Henno Brandsma
116k349127
asked Apr 8 at 0:51
Hotdog2000Hotdog2000
678
$endgroup$
add a comment |
$begingroup$
Let us assume that there exist two independent Brownian Motions $B_t$ and $W_t$, and consider their sum $Y_t=B_t + W_t$. Next, define the filtration generated by the sum, $mathcalF_t^Y=sigma(Y_u)_0
leq u leq t$.
How would one compute the filter $E[ B_t | mathcalF_t^Y]$?
My intuition tells me that the solution to this filtering problem should just be $E[B_t | mathcalF_t^Y] = frac12 Y_t$, although I cannot prove it. As a secondary question, can we generalize to having independent continuous martingales instead of two Brownian Motions?
Any help is appreciated!
stochastic-processes stochastic-calculus brownian-motion
edited 19 hours ago
Henno Brandsma
116k349127
asked Apr 8 at 0:51
Hotdog2000Hotdog2000
678
$endgroup$
add a comment |
$begingroup$
Let us assume that there exist two independent Brownian Motions $B_t$ and $W_t$, and consider their sum $Y_t=B_t + W_t$. Next, define the filtration generated by the sum, $mathcalF_t^Y=sigma(Y_u)_0
leq u leq t$.
How would one compute the filter $E[ B_t | mathcalF_t^Y]$?
My intuition tells me that the solution to this filtering problem should just be $E[B_t | mathcalF_t^Y] = frac12 Y_t$, although I cannot prove it. As a secondary question, can we generalize to having independent continuous martingales instead of two Brownian Motions?
Any help is appreciated!
stochastic-processes stochastic-calculus brownian-motion
edited 19 hours ago
Henno Brandsma
116k349127
asked Apr 8 at 0:51
Hotdog2000Hotdog2000
678
$endgroup$
Let us assume that there exist two independent Brownian Motions $B_t$ and $W_t$, and consider their sum $Y_t=B_t + W_t$. Next, define the filtration generated by the sum, $mathcalF_t^Y=sigma(Y_u)_0
leq u leq t$.
How would one compute the filter $E[ B_t | mathcalF_t^Y]$?
My intuition tells me that the solution to this filtering problem should just be $E[B_t | mathcalF_t^Y] = frac12 Y_t$, although I cannot prove it. As a secondary question, can we generalize to having independent continuous martingales instead of two Brownian Motions?
Any help is appreciated!
stochastic-processes stochastic-calculus brownian-motion
stochastic-processes stochastic-calculus brownian-motion
edited 19 hours ago
Henno Brandsma
116k349127
asked Apr 8 at 0:51
Hotdog2000Hotdog2000
678
edited 19 hours ago
Henno Brandsma
116k349127
asked Apr 8 at 0:51
Hotdog2000Hotdog2000
678
edited 19 hours ago
Henno Brandsma
116k349127
edited 19 hours ago
Henno Brandsma
116k349127
edited 19 hours ago
Henno Brandsma
116k349127
116k349127
asked Apr 8 at 0:51
Hotdog2000Hotdog2000
678
asked Apr 8 at 0:51
Hotdog2000Hotdog2000
678
asked Apr 8 at 0:51
Hotdog2000Hotdog2000
678
678
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Looks like I solved this one already, so I will share the solution.
$$
Y_t = E[Y_t | mathcalF_t^Y] = E[B_t | mathcalF_t^Y] + E[W_t | mathcalF_t^Y]
$$
since $B_t$ and $W_t$ are identically distributed, we should have
$$
E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]
$$
and so we get that $frac12 Y_t = E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]$.
answered Apr 8 at 1:04
Hotdog2000Hotdog2000
678
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Looks like I solved this one already, so I will share the solution.
$$
Y_t = E[Y_t | mathcalF_t^Y] = E[B_t | mathcalF_t^Y] + E[W_t | mathcalF_t^Y]
$$
since $B_t$ and $W_t$ are identically distributed, we should have
$$
E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]
$$
and so we get that $frac12 Y_t = E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]$.
answered Apr 8 at 1:04
Hotdog2000Hotdog2000
678
$endgroup$
add a comment |
$begingroup$
Looks like I solved this one already, so I will share the solution.
$$
Y_t = E[Y_t | mathcalF_t^Y] = E[B_t | mathcalF_t^Y] + E[W_t | mathcalF_t^Y]
$$
since $B_t$ and $W_t$ are identically distributed, we should have
$$
E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]
$$
and so we get that $frac12 Y_t = E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]$.
answered Apr 8 at 1:04
Hotdog2000Hotdog2000
678
$endgroup$
add a comment |
$begingroup$
Looks like I solved this one already, so I will share the solution.
$$
Y_t = E[Y_t | mathcalF_t^Y] = E[B_t | mathcalF_t^Y] + E[W_t | mathcalF_t^Y]
$$
since $B_t$ and $W_t$ are identically distributed, we should have
$$
E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]
$$
and so we get that $frac12 Y_t = E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]$.
answered Apr 8 at 1:04
Hotdog2000Hotdog2000
678
$endgroup$
Looks like I solved this one already, so I will share the solution.
$$
Y_t = E[Y_t | mathcalF_t^Y] = E[B_t | mathcalF_t^Y] + E[W_t | mathcalF_t^Y]
$$
since $B_t$ and $W_t$ are identically distributed, we should have
$$
E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]
$$
and so we get that $frac12 Y_t = E[B_t | mathcalF_t^Y] = E[W_t | mathcalF_t^Y]$.
answered Apr 8 at 1:04
Hotdog2000Hotdog2000
678
answered Apr 8 at 1:04
Hotdog2000Hotdog2000
678
answered Apr 8 at 1:04
Hotdog2000Hotdog2000
678
answered Apr 8 at 1:04
Hotdog2000Hotdog2000
678
678
add a comment |
add a comment |
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