Doobs Martingale inequality
$begingroup$
I am confused on how to calculate the expectation of an integral on this question.
Use the Doob martingale inequality to estimate
$mathbb{E}sup_{0leq s leq t} midint_{0}^{s} cos(u)dB(u)mid^2 quad$ (1)
where $B(t)$ is a one-dimensional Brownian motion.
I understand that (1) is $leq 4mathbb{E}midint_{0}^{s} cos(u)dB(u)mid^2$ but not sure where to go from here. Would welcome any help!
stochastic-processes stochastic-calculus brownian-motion
asked Jan 22 at 0:01
MelMel
61
$endgroup$
add a comment |
$begingroup$
I am confused on how to calculate the expectation of an integral on this question.
Use the Doob martingale inequality to estimate
$mathbb{E}sup_{0leq s leq t} midint_{0}^{s} cos(u)dB(u)mid^2 quad$ (1)
where $B(t)$ is a one-dimensional Brownian motion.
I understand that (1) is $leq 4mathbb{E}midint_{0}^{s} cos(u)dB(u)mid^2$ but not sure where to go from here. Would welcome any help!
stochastic-processes stochastic-calculus brownian-motion
asked Jan 22 at 0:01
MelMel
61
$endgroup$
2
$begingroup$
I love the fact that Doob is a last name. It's great :)
$endgroup$
– Zubin Mukerjee
Jan 22 at 0:02
add a comment |
$begingroup$
I am confused on how to calculate the expectation of an integral on this question.
Use the Doob martingale inequality to estimate
$mathbb{E}sup_{0leq s leq t} midint_{0}^{s} cos(u)dB(u)mid^2 quad$ (1)
where $B(t)$ is a one-dimensional Brownian motion.
I understand that (1) is $leq 4mathbb{E}midint_{0}^{s} cos(u)dB(u)mid^2$ but not sure where to go from here. Would welcome any help!
stochastic-processes stochastic-calculus brownian-motion
asked Jan 22 at 0:01
MelMel
61
$endgroup$
I am confused on how to calculate the expectation of an integral on this question.
Use the Doob martingale inequality to estimate
$mathbb{E}sup_{0leq s leq t} midint_{0}^{s} cos(u)dB(u)mid^2 quad$ (1)
where $B(t)$ is a one-dimensional Brownian motion.
I understand that (1) is $leq 4mathbb{E}midint_{0}^{s} cos(u)dB(u)mid^2$ but not sure where to go from here. Would welcome any help!
stochastic-processes stochastic-calculus brownian-motion
stochastic-processes stochastic-calculus brownian-motion
asked Jan 22 at 0:01
MelMel
61
asked Jan 22 at 0:01
MelMel
61
asked Jan 22 at 0:01
MelMel
61
asked Jan 22 at 0:01
MelMel
61
asked Jan 22 at 0:01
MelMel
61
61
2
$begingroup$
I love the fact that Doob is a last name. It's great :)
$endgroup$
– Zubin Mukerjee
Jan 22 at 0:02
add a comment |
2
$begingroup$
I love the fact that Doob is a last name. It's great :)
$endgroup$
– Zubin Mukerjee
Jan 22 at 0:02
2
2
$begingroup$
I love the fact that Doob is a last name. It's great :)
$endgroup$
– Zubin Mukerjee
Jan 22 at 0:02
$begingroup$
I love the fact that Doob is a last name. It's great :)
$endgroup$
– Zubin Mukerjee
Jan 22 at 0:02
add a comment |
1 Answer
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$begingroup$
$E|int_o^{s}cos (u) , dB(u)|^{2}=int_0^{s} cos^{2}(u), du=frac 1 2 int_0^{s} (1+cos(2u)), du =frac 1 2 (s+frac {sin(2s)} 2)$.
answered Jan 22 at 0:41
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
$E|int_o^{s}cos (u) , dB(u)|^{2}=int_0^{s} cos^{2}(u), du=frac 1 2 int_0^{s} (1+cos(2u)), du =frac 1 2 (s+frac {sin(2s)} 2)$.
answered Jan 22 at 0:41
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
$E|int_o^{s}cos (u) , dB(u)|^{2}=int_0^{s} cos^{2}(u), du=frac 1 2 int_0^{s} (1+cos(2u)), du =frac 1 2 (s+frac {sin(2s)} 2)$.
answered Jan 22 at 0:41
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
$E|int_o^{s}cos (u) , dB(u)|^{2}=int_0^{s} cos^{2}(u), du=frac 1 2 int_0^{s} (1+cos(2u)), du =frac 1 2 (s+frac {sin(2s)} 2)$.
answered Jan 22 at 0:41
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$E|int_o^{s}cos (u) , dB(u)|^{2}=int_0^{s} cos^{2}(u), du=frac 1 2 int_0^{s} (1+cos(2u)), du =frac 1 2 (s+frac {sin(2s)} 2)$.
answered Jan 22 at 0:41
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 22 at 0:41
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 22 at 0:41
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 22 at 0:41
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
add a comment |
add a comment |
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$begingroup$
I love the fact that Doob is a last name. It's great :)
$endgroup$
– Zubin Mukerjee
Jan 22 at 0:02