Mixing
Compute $Eleft((B_t−1)^2int ^t_0(B_s+1)^2 dB_sright)$, where $(B_t)$ is a standard Brownian motion
$begingroup$
Compute $E((B_t−1)^2int ^t_0(B_s+1)^2 dB_s)$ for $t≥0$ given that $(B_t)_{t≥0}$ is a Standard Brownian Motion.
Presume we will need to compute $E((B_t+B_s)-(B_s-1))^2$ to get some independent terms but really stuck on what to do with the integral part. Thanks for any help with this question.
probability-theory stochastic-calculus brownian-motion
edited Jan 6 at 15:01
Did
247k23223459
asked Jan 6 at 14:36
ZugzwangerzZugzwangerz
203
$endgroup$
add a comment |
$begingroup$
Compute $E((B_t−1)^2int ^t_0(B_s+1)^2 dB_s)$ for $t≥0$ given that $(B_t)_{t≥0}$ is a Standard Brownian Motion.
Presume we will need to compute $E((B_t+B_s)-(B_s-1))^2$ to get some independent terms but really stuck on what to do with the integral part. Thanks for any help with this question.
probability-theory stochastic-calculus brownian-motion
edited Jan 6 at 15:01
Did
247k23223459
asked Jan 6 at 14:36
ZugzwangerzZugzwangerz
203
$endgroup$
$begingroup$
@Did could you clarify why $E(B_{t}X_{t})=E(int_0^tY_{s}ds)$.
$endgroup$
– Zugzwangerz
Jan 10 at 15:35
add a comment |
$begingroup$
Compute $E((B_t−1)^2int ^t_0(B_s+1)^2 dB_s)$ for $t≥0$ given that $(B_t)_{t≥0}$ is a Standard Brownian Motion.
Presume we will need to compute $E((B_t+B_s)-(B_s-1))^2$ to get some independent terms but really stuck on what to do with the integral part. Thanks for any help with this question.
probability-theory stochastic-calculus brownian-motion
edited Jan 6 at 15:01
Did
247k23223459
asked Jan 6 at 14:36
ZugzwangerzZugzwangerz
203
$endgroup$
Compute $E((B_t−1)^2int ^t_0(B_s+1)^2 dB_s)$ for $t≥0$ given that $(B_t)_{t≥0}$ is a Standard Brownian Motion.
Presume we will need to compute $E((B_t+B_s)-(B_s-1))^2$ to get some independent terms but really stuck on what to do with the integral part. Thanks for any help with this question.
probability-theory stochastic-calculus brownian-motion
probability-theory stochastic-calculus brownian-motion
edited Jan 6 at 15:01
Did
247k23223459
asked Jan 6 at 14:36
ZugzwangerzZugzwangerz
203
edited Jan 6 at 15:01
Did
247k23223459
asked Jan 6 at 14:36
ZugzwangerzZugzwangerz
203
edited Jan 6 at 15:01
Did
247k23223459
edited Jan 6 at 15:01
Did
247k23223459
edited Jan 6 at 15:01
Did
247k23223459
247k23223459
asked Jan 6 at 14:36
ZugzwangerzZugzwangerz
203
asked Jan 6 at 14:36
ZugzwangerzZugzwangerz
203
asked Jan 6 at 14:36
ZugzwangerzZugzwangerz
203
203
$begingroup$
@Did could you clarify why $E(B_{t}X_{t})=E(int_0^tY_{s}ds)$.
$endgroup$
– Zugzwangerz
Jan 10 at 15:35
add a comment |
$begingroup$
@Did could you clarify why $E(B_{t}X_{t})=E(int_0^tY_{s}ds)$.
$endgroup$
– Zugzwangerz
Jan 10 at 15:35
$begingroup$
@Did could you clarify why $E(B_{t}X_{t})=E(int_0^tY_{s}ds)$.
$endgroup$
– Zugzwangerz
Jan 10 at 15:35
$begingroup$
@Did could you clarify why $E(B_{t}X_{t})=E(int_0^tY_{s}ds)$.
$endgroup$
– Zugzwangerz
Jan 10 at 15:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the processes $$X_t=int ^t_0Y_s dB_sqquad Y_t=(B_t+1)^2$$ By repeated applications of Itô isometry, one gets:
- $E(X_t)=0$
$B_tX_t=displaystyleint_0^tdB_scdotint ^t_0Y_sdB_s$ hence $$E(B_tX_t)=Eleft(int_0^tY_s dsright)=int_0^tE(Y_s)ds$$
$(B_t^2-t)X_t=displaystyleint_0^t2B_sdB_scdotint ^t_0Y_sdB_s$ hence $$E((B_t^2-t)X_t)=Eleft(int_0^t2B_sY_s dsright)=2int_0^tE(B_sY_s)ds$$
Finally, if one can compute $E(Y_t)$ and $E(B_tY_t)$, the proof is complete. Can you?
answered Jan 6 at 14:57
DidDid
247k23223459
$endgroup$
$begingroup$
Like at every other step, use Itô isometry, based on the fact that $$dlangle B,Brangle_t=dt$$ hence, for every suitable processes $(u_t)$ and $(v_t)$, $$Eleft(int_0^tu_sdB_scdotint_0^tv_sdB_sright)=Eleft(int_0^tu_sv_sdsright)$$
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
((This is answering a comment formerly here but now on main, by the OP.))
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
ah I get it now, thank you!
$endgroup$
– Zugzwangerz
Jan 10 at 15:39
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Consider the processes $$X_t=int ^t_0Y_s dB_sqquad Y_t=(B_t+1)^2$$ By repeated applications of Itô isometry, one gets:
- $E(X_t)=0$
$B_tX_t=displaystyleint_0^tdB_scdotint ^t_0Y_sdB_s$ hence $$E(B_tX_t)=Eleft(int_0^tY_s dsright)=int_0^tE(Y_s)ds$$
$(B_t^2-t)X_t=displaystyleint_0^t2B_sdB_scdotint ^t_0Y_sdB_s$ hence $$E((B_t^2-t)X_t)=Eleft(int_0^t2B_sY_s dsright)=2int_0^tE(B_sY_s)ds$$
Finally, if one can compute $E(Y_t)$ and $E(B_tY_t)$, the proof is complete. Can you?
answered Jan 6 at 14:57
DidDid
247k23223459
$endgroup$
$begingroup$
Like at every other step, use Itô isometry, based on the fact that $$dlangle B,Brangle_t=dt$$ hence, for every suitable processes $(u_t)$ and $(v_t)$, $$Eleft(int_0^tu_sdB_scdotint_0^tv_sdB_sright)=Eleft(int_0^tu_sv_sdsright)$$
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
((This is answering a comment formerly here but now on main, by the OP.))
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
ah I get it now, thank you!
$endgroup$
– Zugzwangerz
Jan 10 at 15:39
add a comment |
$begingroup$
Consider the processes $$X_t=int ^t_0Y_s dB_sqquad Y_t=(B_t+1)^2$$ By repeated applications of Itô isometry, one gets:
- $E(X_t)=0$
$B_tX_t=displaystyleint_0^tdB_scdotint ^t_0Y_sdB_s$ hence $$E(B_tX_t)=Eleft(int_0^tY_s dsright)=int_0^tE(Y_s)ds$$
$(B_t^2-t)X_t=displaystyleint_0^t2B_sdB_scdotint ^t_0Y_sdB_s$ hence $$E((B_t^2-t)X_t)=Eleft(int_0^t2B_sY_s dsright)=2int_0^tE(B_sY_s)ds$$
Finally, if one can compute $E(Y_t)$ and $E(B_tY_t)$, the proof is complete. Can you?
answered Jan 6 at 14:57
DidDid
247k23223459
$endgroup$
$begingroup$
Like at every other step, use Itô isometry, based on the fact that $$dlangle B,Brangle_t=dt$$ hence, for every suitable processes $(u_t)$ and $(v_t)$, $$Eleft(int_0^tu_sdB_scdotint_0^tv_sdB_sright)=Eleft(int_0^tu_sv_sdsright)$$
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
((This is answering a comment formerly here but now on main, by the OP.))
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
ah I get it now, thank you!
$endgroup$
– Zugzwangerz
Jan 10 at 15:39
add a comment |
$begingroup$
Consider the processes $$X_t=int ^t_0Y_s dB_sqquad Y_t=(B_t+1)^2$$ By repeated applications of Itô isometry, one gets:
- $E(X_t)=0$
$B_tX_t=displaystyleint_0^tdB_scdotint ^t_0Y_sdB_s$ hence $$E(B_tX_t)=Eleft(int_0^tY_s dsright)=int_0^tE(Y_s)ds$$
$(B_t^2-t)X_t=displaystyleint_0^t2B_sdB_scdotint ^t_0Y_sdB_s$ hence $$E((B_t^2-t)X_t)=Eleft(int_0^t2B_sY_s dsright)=2int_0^tE(B_sY_s)ds$$
Finally, if one can compute $E(Y_t)$ and $E(B_tY_t)$, the proof is complete. Can you?
answered Jan 6 at 14:57
DidDid
247k23223459
$endgroup$
Consider the processes $$X_t=int ^t_0Y_s dB_sqquad Y_t=(B_t+1)^2$$ By repeated applications of Itô isometry, one gets:
- $E(X_t)=0$
$B_tX_t=displaystyleint_0^tdB_scdotint ^t_0Y_sdB_s$ hence $$E(B_tX_t)=Eleft(int_0^tY_s dsright)=int_0^tE(Y_s)ds$$
$(B_t^2-t)X_t=displaystyleint_0^t2B_sdB_scdotint ^t_0Y_sdB_s$ hence $$E((B_t^2-t)X_t)=Eleft(int_0^t2B_sY_s dsright)=2int_0^tE(B_sY_s)ds$$
Finally, if one can compute $E(Y_t)$ and $E(B_tY_t)$, the proof is complete. Can you?
answered Jan 6 at 14:57
DidDid
247k23223459
answered Jan 6 at 14:57
DidDid
247k23223459
answered Jan 6 at 14:57
DidDid
247k23223459
answered Jan 6 at 14:57
DidDid
247k23223459
247k23223459
$begingroup$
Like at every other step, use Itô isometry, based on the fact that $$dlangle B,Brangle_t=dt$$ hence, for every suitable processes $(u_t)$ and $(v_t)$, $$Eleft(int_0^tu_sdB_scdotint_0^tv_sdB_sright)=Eleft(int_0^tu_sv_sdsright)$$
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
((This is answering a comment formerly here but now on main, by the OP.))
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
ah I get it now, thank you!
$endgroup$
– Zugzwangerz
Jan 10 at 15:39
add a comment |
$begingroup$
Like at every other step, use Itô isometry, based on the fact that $$dlangle B,Brangle_t=dt$$ hence, for every suitable processes $(u_t)$ and $(v_t)$, $$Eleft(int_0^tu_sdB_scdotint_0^tv_sdB_sright)=Eleft(int_0^tu_sv_sdsright)$$
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
((This is answering a comment formerly here but now on main, by the OP.))
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
ah I get it now, thank you!
$endgroup$
– Zugzwangerz
Jan 10 at 15:39
$begingroup$
Like at every other step, use Itô isometry, based on the fact that $$dlangle B,Brangle_t=dt$$ hence, for every suitable processes $(u_t)$ and $(v_t)$, $$Eleft(int_0^tu_sdB_scdotint_0^tv_sdB_sright)=Eleft(int_0^tu_sv_sdsright)$$
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
Like at every other step, use Itô isometry, based on the fact that $$dlangle B,Brangle_t=dt$$ hence, for every suitable processes $(u_t)$ and $(v_t)$, $$Eleft(int_0^tu_sdB_scdotint_0^tv_sdB_sright)=Eleft(int_0^tu_sv_sdsright)$$
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
((This is answering a comment formerly here but now on main, by the OP.))
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
((This is answering a comment formerly here but now on main, by the OP.))
$endgroup$
– Did
Jan 10 at 15:38
$begingroup$
ah I get it now, thank you!
$endgroup$
– Zugzwangerz
Jan 10 at 15:39
$begingroup$
ah I get it now, thank you!
$endgroup$
– Zugzwangerz
Jan 10 at 15:39
add a comment |
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$begingroup$
@Did could you clarify why $E(B_{t}X_{t})=E(int_0^tY_{s}ds)$.
$endgroup$
– Zugzwangerz
Jan 10 at 15:35