Geometric Brownian Motion - Normal distribution The Next CEO of Stack Overflow“Flattening” a 2D Normal DistributionNormal distributionQuestion of Normal Distribution ??Normal DistributionGeometric Mean of Random VariablesMean of a portion of a normal distribution?Normal distribution and sample distribution standard deviationNormal distribution for heightStandard Normal Distribution Question: Help?intepretation of standard deviation for geometric mean
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Geometric Brownian Motion - Normal distribution
The Next CEO of Stack Overflow“Flattening” a 2D Normal DistributionNormal distributionQuestion of Normal Distribution ??Normal DistributionGeometric Mean of Random VariablesMean of a portion of a normal distribution?Normal distribution and sample distribution standard deviationNormal distribution for heightStandard Normal Distribution Question: Help?intepretation of standard deviation for geometric mean
$begingroup$
I have the following formula:
Geometric Brownian Model Formula
Does anyone know if the portion highlighted in yellow, corresponds to a normal distribution with mean = 0 and standard deviation = 1? Or is it a uniform distribution between 0 and 1?
statistics
asked Mar 19 at 20:51
Patriots299Patriots299
1
$endgroup$
add a comment |
$begingroup$
I have the following formula:
Geometric Brownian Model Formula
Does anyone know if the portion highlighted in yellow, corresponds to a normal distribution with mean = 0 and standard deviation = 1? Or is it a uniform distribution between 0 and 1?
statistics
asked Mar 19 at 20:51
Patriots299Patriots299
1
$endgroup$
add a comment |
$begingroup$
I have the following formula:
Geometric Brownian Model Formula
Does anyone know if the portion highlighted in yellow, corresponds to a normal distribution with mean = 0 and standard deviation = 1? Or is it a uniform distribution between 0 and 1?
statistics
asked Mar 19 at 20:51
Patriots299Patriots299
1
$endgroup$
I have the following formula:
Geometric Brownian Model Formula
Does anyone know if the portion highlighted in yellow, corresponds to a normal distribution with mean = 0 and standard deviation = 1? Or is it a uniform distribution between 0 and 1?
statistics
statistics
asked Mar 19 at 20:51
Patriots299Patriots299
1
asked Mar 19 at 20:51
Patriots299Patriots299
1
asked Mar 19 at 20:51
Patriots299Patriots299
1
asked Mar 19 at 20:51
Patriots299Patriots299
1
asked Mar 19 at 20:51
Patriots299Patriots299
1
1
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It refers to the former, i.e. a normal random variable with mean $0$ and standard deviation $1$.
(If you search up a proof of the solution to the geometric brownian motion stochastic differential equation, you will see a "Wiener process"/brownian motion term appearing, which is where the normal distribution comes from.)
answered Mar 20 at 6:31
Minus One-TwelfthMinus One-Twelfth
2,943413
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
It refers to the former, i.e. a normal random variable with mean $0$ and standard deviation $1$.
(If you search up a proof of the solution to the geometric brownian motion stochastic differential equation, you will see a "Wiener process"/brownian motion term appearing, which is where the normal distribution comes from.)
answered Mar 20 at 6:31
Minus One-TwelfthMinus One-Twelfth
2,943413
$endgroup$
add a comment |
$begingroup$
It refers to the former, i.e. a normal random variable with mean $0$ and standard deviation $1$.
(If you search up a proof of the solution to the geometric brownian motion stochastic differential equation, you will see a "Wiener process"/brownian motion term appearing, which is where the normal distribution comes from.)
answered Mar 20 at 6:31
Minus One-TwelfthMinus One-Twelfth
2,943413
$endgroup$
add a comment |
$begingroup$
It refers to the former, i.e. a normal random variable with mean $0$ and standard deviation $1$.
(If you search up a proof of the solution to the geometric brownian motion stochastic differential equation, you will see a "Wiener process"/brownian motion term appearing, which is where the normal distribution comes from.)
answered Mar 20 at 6:31
Minus One-TwelfthMinus One-Twelfth
2,943413
$endgroup$
It refers to the former, i.e. a normal random variable with mean $0$ and standard deviation $1$.
(If you search up a proof of the solution to the geometric brownian motion stochastic differential equation, you will see a "Wiener process"/brownian motion term appearing, which is where the normal distribution comes from.)
answered Mar 20 at 6:31
Minus One-TwelfthMinus One-Twelfth
2,943413
answered Mar 20 at 6:31
Minus One-TwelfthMinus One-Twelfth
2,943413
answered Mar 20 at 6:31
Minus One-TwelfthMinus One-Twelfth
2,943413
answered Mar 20 at 6:31
Minus One-TwelfthMinus One-Twelfth
2,943413
2,943413
add a comment |
add a comment |
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