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Measure of a Brownian motion = normal distribution?
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Consider some model where the process increments are normally distributed, e.g. Vasicek:
$$dr(t) = left(theta - ar(t)right)dt + sigma dW(t).$$
We usually say that $W(t)$ is a Brownian motion under a measure $mathbb P$. $W(t)$ is a Brownian motion if, among other conditions, $W(t) sim N(0, t)$ given $W(0)=0$. Does it mean that the measure $mathbb P$ is actually a normal distribution, i.e.
$$mathbb Pleft(frac{W(t)}{sqrt t} in [a ,b]right) = Phi(b) - Phi(a)$$ where $Phi(cdot)$ denotes the CDF of a standard normal random variable?
brownian-motion risk-neutral-measure normal-distribution self-study
asked 5 hours ago
tosiktosik
26927
$endgroup$
add a comment |
$begingroup$
Consider some model where the process increments are normally distributed, e.g. Vasicek:
$$dr(t) = left(theta - ar(t)right)dt + sigma dW(t).$$
We usually say that $W(t)$ is a Brownian motion under a measure $mathbb P$. $W(t)$ is a Brownian motion if, among other conditions, $W(t) sim N(0, t)$ given $W(0)=0$. Does it mean that the measure $mathbb P$ is actually a normal distribution, i.e.
$$mathbb Pleft(frac{W(t)}{sqrt t} in [a ,b]right) = Phi(b) - Phi(a)$$ where $Phi(cdot)$ denotes the CDF of a standard normal random variable?
brownian-motion risk-neutral-measure normal-distribution self-study
asked 5 hours ago
tosiktosik
26927
$endgroup$
$begingroup$
The equation you wrote is correct. But it is only one example of the properties of the measure P. The measure P is more general than this and cannot be said "to be a normal distribution", but the normal distribution does come up in describing the measure. The measure applies to a stochastic procees, the normal distribution applies to a random variable, so they cannot be identical.
$endgroup$
– noob2
5 hours ago
add a comment |
$begingroup$
Consider some model where the process increments are normally distributed, e.g. Vasicek:
$$dr(t) = left(theta - ar(t)right)dt + sigma dW(t).$$
We usually say that $W(t)$ is a Brownian motion under a measure $mathbb P$. $W(t)$ is a Brownian motion if, among other conditions, $W(t) sim N(0, t)$ given $W(0)=0$. Does it mean that the measure $mathbb P$ is actually a normal distribution, i.e.
$$mathbb Pleft(frac{W(t)}{sqrt t} in [a ,b]right) = Phi(b) - Phi(a)$$ where $Phi(cdot)$ denotes the CDF of a standard normal random variable?
brownian-motion risk-neutral-measure normal-distribution self-study
asked 5 hours ago
tosiktosik
26927
$endgroup$
Consider some model where the process increments are normally distributed, e.g. Vasicek:
$$dr(t) = left(theta - ar(t)right)dt + sigma dW(t).$$
We usually say that $W(t)$ is a Brownian motion under a measure $mathbb P$. $W(t)$ is a Brownian motion if, among other conditions, $W(t) sim N(0, t)$ given $W(0)=0$. Does it mean that the measure $mathbb P$ is actually a normal distribution, i.e.
$$mathbb Pleft(frac{W(t)}{sqrt t} in [a ,b]right) = Phi(b) - Phi(a)$$ where $Phi(cdot)$ denotes the CDF of a standard normal random variable?
brownian-motion risk-neutral-measure normal-distribution self-study
brownian-motion risk-neutral-measure normal-distribution self-study
asked 5 hours ago
tosiktosik
26927
asked 5 hours ago
tosiktosik
26927
edited 5 hours ago
tosik
asked 5 hours ago
tosiktosik
26927
asked 5 hours ago
tosiktosik
26927
asked 5 hours ago
tosiktosik
26927
26927
$begingroup$
The equation you wrote is correct. But it is only one example of the properties of the measure P. The measure P is more general than this and cannot be said "to be a normal distribution", but the normal distribution does come up in describing the measure. The measure applies to a stochastic procees, the normal distribution applies to a random variable, so they cannot be identical.
$endgroup$
– noob2
5 hours ago
add a comment |
$begingroup$
The equation you wrote is correct. But it is only one example of the properties of the measure P. The measure P is more general than this and cannot be said "to be a normal distribution", but the normal distribution does come up in describing the measure. The measure applies to a stochastic procees, the normal distribution applies to a random variable, so they cannot be identical.
$endgroup$
– noob2
5 hours ago
$begingroup$
The equation you wrote is correct. But it is only one example of the properties of the measure P. The measure P is more general than this and cannot be said "to be a normal distribution", but the normal distribution does come up in describing the measure. The measure applies to a stochastic procees, the normal distribution applies to a random variable, so they cannot be identical.
$endgroup$
– noob2
5 hours ago
$begingroup$
The equation you wrote is correct. But it is only one example of the properties of the measure P. The measure P is more general than this and cannot be said "to be a normal distribution", but the normal distribution does come up in describing the measure. The measure applies to a stochastic procees, the normal distribution applies to a random variable, so they cannot be identical.
$endgroup$
– noob2
5 hours ago
add a comment |
1 Answer
1
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oldest
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$begingroup$
I am not allowed post a comment, so it goes here.
- It is correct that
$$
mathbf{P}(t^{-1/2}W(t) in[a,b])=Φ(b)−Φ(a), forall tin(0,infty)
$$
due to the stationary increments property of the Wiener process and the fact that you normalized the random variable by dividing by its standard deviation.
$mathbf{P}$ is a probability measure on an abstract space, not a random variable. Hence, you probably mean that $W(t)$ is normally distributed under $mathbf{P}$, NOT $mathbf{P}$ is normally distributed. People tend to mention the probability measure, for if you change it the process will no longer be Gaussian.
answered 4 hours ago
phantagarowphantagarow
514
New contributor
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to Quant SE. No need for comments :) This is (along with the rest of your answers) acceptable as an answer to the question.
$endgroup$
– Sanjay
4 hours ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
I am not allowed post a comment, so it goes here.
- It is correct that
$$
mathbf{P}(t^{-1/2}W(t) in[a,b])=Φ(b)−Φ(a), forall tin(0,infty)
$$
due to the stationary increments property of the Wiener process and the fact that you normalized the random variable by dividing by its standard deviation.
$mathbf{P}$ is a probability measure on an abstract space, not a random variable. Hence, you probably mean that $W(t)$ is normally distributed under $mathbf{P}$, NOT $mathbf{P}$ is normally distributed. People tend to mention the probability measure, for if you change it the process will no longer be Gaussian.
answered 4 hours ago
phantagarowphantagarow
514
New contributor
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to Quant SE. No need for comments :) This is (along with the rest of your answers) acceptable as an answer to the question.
$endgroup$
– Sanjay
4 hours ago
add a comment |
$begingroup$
I am not allowed post a comment, so it goes here.
- It is correct that
$$
mathbf{P}(t^{-1/2}W(t) in[a,b])=Φ(b)−Φ(a), forall tin(0,infty)
$$
due to the stationary increments property of the Wiener process and the fact that you normalized the random variable by dividing by its standard deviation.
$mathbf{P}$ is a probability measure on an abstract space, not a random variable. Hence, you probably mean that $W(t)$ is normally distributed under $mathbf{P}$, NOT $mathbf{P}$ is normally distributed. People tend to mention the probability measure, for if you change it the process will no longer be Gaussian.
answered 4 hours ago
phantagarowphantagarow
514
New contributor
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to Quant SE. No need for comments :) This is (along with the rest of your answers) acceptable as an answer to the question.
$endgroup$
– Sanjay
4 hours ago
add a comment |
$begingroup$
I am not allowed post a comment, so it goes here.
- It is correct that
$$
mathbf{P}(t^{-1/2}W(t) in[a,b])=Φ(b)−Φ(a), forall tin(0,infty)
$$
due to the stationary increments property of the Wiener process and the fact that you normalized the random variable by dividing by its standard deviation.
$mathbf{P}$ is a probability measure on an abstract space, not a random variable. Hence, you probably mean that $W(t)$ is normally distributed under $mathbf{P}$, NOT $mathbf{P}$ is normally distributed. People tend to mention the probability measure, for if you change it the process will no longer be Gaussian.
answered 4 hours ago
phantagarowphantagarow
514
New contributor
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am not allowed post a comment, so it goes here.
- It is correct that
$$
mathbf{P}(t^{-1/2}W(t) in[a,b])=Φ(b)−Φ(a), forall tin(0,infty)
$$
due to the stationary increments property of the Wiener process and the fact that you normalized the random variable by dividing by its standard deviation.
$mathbf{P}$ is a probability measure on an abstract space, not a random variable. Hence, you probably mean that $W(t)$ is normally distributed under $mathbf{P}$, NOT $mathbf{P}$ is normally distributed. People tend to mention the probability measure, for if you change it the process will no longer be Gaussian.
answered 4 hours ago
phantagarowphantagarow
514
New contributor
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
answered 4 hours ago
phantagarowphantagarow
514
New contributor
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
phantagarowphantagarow
514
answered 4 hours ago
phantagarowphantagarow
514
514
New contributor
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Welcome to Quant SE. No need for comments :) This is (along with the rest of your answers) acceptable as an answer to the question.
$endgroup$
– Sanjay
4 hours ago
add a comment |
2
$begingroup$
Welcome to Quant SE. No need for comments :) This is (along with the rest of your answers) acceptable as an answer to the question.
$endgroup$
– Sanjay
4 hours ago
2
2
$begingroup$
Welcome to Quant SE. No need for comments :) This is (along with the rest of your answers) acceptable as an answer to the question.
$endgroup$
– Sanjay
4 hours ago
$begingroup$
Welcome to Quant SE. No need for comments :) This is (along with the rest of your answers) acceptable as an answer to the question.
$endgroup$
– Sanjay
4 hours ago
add a comment |
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$begingroup$
The equation you wrote is correct. But it is only one example of the properties of the measure P. The measure P is more general than this and cannot be said "to be a normal distribution", but the normal distribution does come up in describing the measure. The measure applies to a stochastic procees, the normal distribution applies to a random variable, so they cannot be identical.
$endgroup$
– noob2
5 hours ago