Measure of a Brownian motion = normal distribution?












2












$begingroup$


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







share|improve this question











$endgroup$












  • $begingroup$
    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


















2












$begingroup$


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







share|improve this question











$endgroup$












  • $begingroup$
    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
















2












2








2





$begingroup$


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







share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 5 hours ago







tosik

















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




















  • $begingroup$
    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












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I am not allowed post a comment, so it goes here.




  1. 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.


  2. $mathbf{P}$ 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.





share










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phantagarow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 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











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$begingroup$

I am not allowed post a comment, so it goes here.




  1. 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.


  2. $mathbf{P}$ 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.





share










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
















4












$begingroup$

I am not allowed post a comment, so it goes here.




  1. 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.


  2. $mathbf{P}$ 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.





share










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














4












4








4





$begingroup$

I am not allowed post a comment, so it goes here.




  1. 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.


  2. $mathbf{P}$ 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.





share










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.




  1. 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.


  2. $mathbf{P}$ 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.






share










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answered 4 hours ago









phantagarowphantagarow

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  • 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




    $begingroup$
    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


















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