What is $E[W_t ^2 e^{(mu W_t - frac{sigma^2}{2}t)}]$? [closed]
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What is the expected value:
$E[W_t ^2 e^{(mu W_t - frac{sigma^2}{2}t)}]$
where $W_t$ is a standard Brownian Motion and $mu, sigma >0$
One possible hint is: take $d/d mu$ twice.
I don't know how to use it, could somebody help?
normal-distribution stochastic-calculus expected-value
asked Dec 31 '18 at 0:01
Joseph YangJoseph Yang
11
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closed as off-topic by Did, José Carlos Santos, Adrian Keister, Davide Giraudo, A. Pongrácz Jan 2 at 20:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, Adrian Keister, Davide Giraudo, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What is the expected value:
$E[W_t ^2 e^{(mu W_t - frac{sigma^2}{2}t)}]$
where $W_t$ is a standard Brownian Motion and $mu, sigma >0$
One possible hint is: take $d/d mu$ twice.
I don't know how to use it, could somebody help?
normal-distribution stochastic-calculus expected-value
asked Dec 31 '18 at 0:01
Joseph YangJoseph Yang
11
$endgroup$
closed as off-topic by Did, José Carlos Santos, Adrian Keister, Davide Giraudo, A. Pongrácz Jan 2 at 20:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, Adrian Keister, Davide Giraudo, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What is the expected value:
$E[W_t ^2 e^{(mu W_t - frac{sigma^2}{2}t)}]$
where $W_t$ is a standard Brownian Motion and $mu, sigma >0$
One possible hint is: take $d/d mu$ twice.
I don't know how to use it, could somebody help?
normal-distribution stochastic-calculus expected-value
asked Dec 31 '18 at 0:01
Joseph YangJoseph Yang
11
$endgroup$
What is the expected value:
$E[W_t ^2 e^{(mu W_t - frac{sigma^2}{2}t)}]$
where $W_t$ is a standard Brownian Motion and $mu, sigma >0$
One possible hint is: take $d/d mu$ twice.
I don't know how to use it, could somebody help?
normal-distribution stochastic-calculus expected-value
normal-distribution stochastic-calculus expected-value
asked Dec 31 '18 at 0:01
Joseph YangJoseph Yang
11
asked Dec 31 '18 at 0:01
Joseph YangJoseph Yang
11
asked Dec 31 '18 at 0:01
Joseph YangJoseph Yang
11
asked Dec 31 '18 at 0:01
Joseph YangJoseph Yang
11
asked Dec 31 '18 at 0:01
Joseph YangJoseph Yang
11
11
closed as off-topic by Did, José Carlos Santos, Adrian Keister, Davide Giraudo, A. Pongrácz Jan 2 at 20:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, Adrian Keister, Davide Giraudo, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, José Carlos Santos, Adrian Keister, Davide Giraudo, A. Pongrácz Jan 2 at 20:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, Adrian Keister, Davide Giraudo, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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votes
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Hint: let $f(mu)=Ee^{mu W_t-frac {sigma^{2}t} 2}$. Then and $f''(mu)=E W_t^{2}e^{mu W_t-frac {sigma^{2}t} 2}$. Recall that $Ee^{mu W_t}=e^{mu^{2}t/2}$.
answered Dec 31 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
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$begingroup$
Amazing! One minor mistake, 'recall that... expected value should be e to the mu squared times t divided by 2, not t squared, because Wt ~ N(0,t)
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– Joseph Yang
Dec 31 '18 at 0:58
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@JosephYang Very true. Thanks for pointing out.
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– Kavi Rama Murthy
Dec 31 '18 at 5:28
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: let $f(mu)=Ee^{mu W_t-frac {sigma^{2}t} 2}$. Then and $f''(mu)=E W_t^{2}e^{mu W_t-frac {sigma^{2}t} 2}$. Recall that $Ee^{mu W_t}=e^{mu^{2}t/2}$.
answered Dec 31 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
$begingroup$
Amazing! One minor mistake, 'recall that... expected value should be e to the mu squared times t divided by 2, not t squared, because Wt ~ N(0,t)
$endgroup$
– Joseph Yang
Dec 31 '18 at 0:58
$begingroup$
@JosephYang Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 5:28
add a comment |
$begingroup$
Hint: let $f(mu)=Ee^{mu W_t-frac {sigma^{2}t} 2}$. Then and $f''(mu)=E W_t^{2}e^{mu W_t-frac {sigma^{2}t} 2}$. Recall that $Ee^{mu W_t}=e^{mu^{2}t/2}$.
answered Dec 31 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
$begingroup$
Amazing! One minor mistake, 'recall that... expected value should be e to the mu squared times t divided by 2, not t squared, because Wt ~ N(0,t)
$endgroup$
– Joseph Yang
Dec 31 '18 at 0:58
$begingroup$
@JosephYang Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 5:28
add a comment |
$begingroup$
Hint: let $f(mu)=Ee^{mu W_t-frac {sigma^{2}t} 2}$. Then and $f''(mu)=E W_t^{2}e^{mu W_t-frac {sigma^{2}t} 2}$. Recall that $Ee^{mu W_t}=e^{mu^{2}t/2}$.
answered Dec 31 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
Hint: let $f(mu)=Ee^{mu W_t-frac {sigma^{2}t} 2}$. Then and $f''(mu)=E W_t^{2}e^{mu W_t-frac {sigma^{2}t} 2}$. Recall that $Ee^{mu W_t}=e^{mu^{2}t/2}$.
answered Dec 31 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
edited Dec 31 '18 at 5:27
answered Dec 31 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Dec 31 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Dec 31 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
69.5k53170
$begingroup$
Amazing! One minor mistake, 'recall that... expected value should be e to the mu squared times t divided by 2, not t squared, because Wt ~ N(0,t)
$endgroup$
– Joseph Yang
Dec 31 '18 at 0:58
$begingroup$
@JosephYang Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 5:28
add a comment |
$begingroup$
Amazing! One minor mistake, 'recall that... expected value should be e to the mu squared times t divided by 2, not t squared, because Wt ~ N(0,t)
$endgroup$
– Joseph Yang
Dec 31 '18 at 0:58
$begingroup$
@JosephYang Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 5:28
$begingroup$
Amazing! One minor mistake, 'recall that... expected value should be e to the mu squared times t divided by 2, not t squared, because Wt ~ N(0,t)
$endgroup$
– Joseph Yang
Dec 31 '18 at 0:58
$begingroup$
Amazing! One minor mistake, 'recall that... expected value should be e to the mu squared times t divided by 2, not t squared, because Wt ~ N(0,t)
$endgroup$
– Joseph Yang
Dec 31 '18 at 0:58
$begingroup$
@JosephYang Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 5:28
$begingroup$
@JosephYang Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 5:28
add a comment |
