Measure of a Brownian motion = normal distribution?
1
$begingroup$
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 Feb 28 at 19:02
tosiktosik
26927
$endgroup$
add a comment |
1
$begingroup$
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 Feb 28 at 19:02
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
Feb 28 at 19:33
add a comment |
1
1
1
$begingroup$
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 Feb 28 at 19:02
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 Feb 28 at 19:02
tosiktosik
26927
asked Feb 28 at 19:02
tosiktosik
26927
edited Feb 28 at 19:31
tosik
asked Feb 28 at 19:02
tosiktosik
26927
asked Feb 28 at 19:02
tosiktosik
26927
asked Feb 28 at 19:02
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
Feb 28 at 19:33
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
Feb 28 at 19:33
$begingroup$
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
Feb 28 at 19:33
$begingroup$
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
Feb 28 at 19:33
add a comment |
1 Answer
1
active
oldest
votes
6
$begingroup$
- 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.
edited Mar 1 at 5:35
Emma
313112
answered Feb 28 at 20:18
phantagarowphantagarow
715
$endgroup$
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
Feb 28 at 20:31
$begingroup$
Could you elaborate more on “...measure on an abstract space”. Normal distribution is also a measure (of a rv). Also I could change to martingale measure $mathbb Q$ and my process still would be Gaussian, i.e. $dW^{mathbb Q}= dW^{mathbb P} +lambda dt$ by Girsanov.
$endgroup$
– tosik
Mar 1 at 7:07
1
$begingroup$
@tosik. A probability space is a triple $(Omega, mathcal{F}, mathbf{P})$ where $Omega$ is an abstract space. A normal distribution is a very specific type (with a specific density) on $mathbb{R}^n$ for some $n geq 1$. A random variable or vector $X: Omega rightarrow mathbb{R}^n$ is said to be normal if it induces a normal distribution on the range space.
$endgroup$
– phantagarow
Mar 1 at 14:01
1
$begingroup$
@tosik. You might want to be precise with Girsanov's Theorem so as not conflate related but different concepts. If $W_{t}$ is a Brownian Motion under $mathbf{P}$, the newly defined process $Z_{t} = W_{t} - int_{0}^{t} theta_{s} ds$ is no longer a BM under $mathbf{P}$. Girsanov's Thm tells you how to construct a new measure so as that $Z_{t}$ becomes a BM.
$endgroup$
– phantagarow
Mar 1 at 14:40
$begingroup$
I agree that $Z_t$ is no longer BM under $mathbb P$ but I think you agree that it is Gaussian both under $mathbb P$ and $mathbb Q$. I indeed probably confuse measures on $Omega$ with distributions on $mathbf R$. Your answer (and comments) do shed some light but I still don't have a clear picture. So, $Omega$ is an abstract space where for each $omega in Omega$ there is an associated function $f$ that we think of as a stochastic process? Can you elaborate more on the link between $mathbb P$ and normal distribution for the case of BM?
$endgroup$
– tosik
Mar 1 at 15:01
|
show 2 more comments
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6
$begingroup$
- 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.
edited Mar 1 at 5:35
Emma
313112
answered Feb 28 at 20:18
phantagarowphantagarow
715
$endgroup$
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
Feb 28 at 20:31
$begingroup$
Could you elaborate more on “...measure on an abstract space”. Normal distribution is also a measure (of a rv). Also I could change to martingale measure $mathbb Q$ and my process still would be Gaussian, i.e. $dW^{mathbb Q}= dW^{mathbb P} +lambda dt$ by Girsanov.
$endgroup$
– tosik
Mar 1 at 7:07
1
$begingroup$
@tosik. A probability space is a triple $(Omega, mathcal{F}, mathbf{P})$ where $Omega$ is an abstract space. A normal distribution is a very specific type (with a specific density) on $mathbb{R}^n$ for some $n geq 1$. A random variable or vector $X: Omega rightarrow mathbb{R}^n$ is said to be normal if it induces a normal distribution on the range space.
$endgroup$
– phantagarow
Mar 1 at 14:01
1
$begingroup$
@tosik. You might want to be precise with Girsanov's Theorem so as not conflate related but different concepts. If $W_{t}$ is a Brownian Motion under $mathbf{P}$, the newly defined process $Z_{t} = W_{t} - int_{0}^{t} theta_{s} ds$ is no longer a BM under $mathbf{P}$. Girsanov's Thm tells you how to construct a new measure so as that $Z_{t}$ becomes a BM.
$endgroup$
– phantagarow
Mar 1 at 14:40
$begingroup$
I agree that $Z_t$ is no longer BM under $mathbb P$ but I think you agree that it is Gaussian both under $mathbb P$ and $mathbb Q$. I indeed probably confuse measures on $Omega$ with distributions on $mathbf R$. Your answer (and comments) do shed some light but I still don't have a clear picture. So, $Omega$ is an abstract space where for each $omega in Omega$ there is an associated function $f$ that we think of as a stochastic process? Can you elaborate more on the link between $mathbb P$ and normal distribution for the case of BM?
$endgroup$
– tosik
Mar 1 at 15:01
|
show 2 more comments
6
$begingroup$
- 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.
edited Mar 1 at 5:35
Emma
313112
answered Feb 28 at 20:18
phantagarowphantagarow
715
$endgroup$
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
Feb 28 at 20:31
$begingroup$
Could you elaborate more on “...measure on an abstract space”. Normal distribution is also a measure (of a rv). Also I could change to martingale measure $mathbb Q$ and my process still would be Gaussian, i.e. $dW^{mathbb Q}= dW^{mathbb P} +lambda dt$ by Girsanov.
$endgroup$
– tosik
Mar 1 at 7:07
1
$begingroup$
@tosik. A probability space is a triple $(Omega, mathcal{F}, mathbf{P})$ where $Omega$ is an abstract space. A normal distribution is a very specific type (with a specific density) on $mathbb{R}^n$ for some $n geq 1$. A random variable or vector $X: Omega rightarrow mathbb{R}^n$ is said to be normal if it induces a normal distribution on the range space.
$endgroup$
– phantagarow
Mar 1 at 14:01
1
$begingroup$
@tosik. You might want to be precise with Girsanov's Theorem so as not conflate related but different concepts. If $W_{t}$ is a Brownian Motion under $mathbf{P}$, the newly defined process $Z_{t} = W_{t} - int_{0}^{t} theta_{s} ds$ is no longer a BM under $mathbf{P}$. Girsanov's Thm tells you how to construct a new measure so as that $Z_{t}$ becomes a BM.
$endgroup$
– phantagarow
Mar 1 at 14:40
$begingroup$
I agree that $Z_t$ is no longer BM under $mathbb P$ but I think you agree that it is Gaussian both under $mathbb P$ and $mathbb Q$. I indeed probably confuse measures on $Omega$ with distributions on $mathbf R$. Your answer (and comments) do shed some light but I still don't have a clear picture. So, $Omega$ is an abstract space where for each $omega in Omega$ there is an associated function $f$ that we think of as a stochastic process? Can you elaborate more on the link between $mathbb P$ and normal distribution for the case of BM?
$endgroup$
– tosik
Mar 1 at 15:01
|
show 2 more comments
6
6
6
$begingroup$
- 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.
edited Mar 1 at 5:35
Emma
313112
answered Feb 28 at 20:18
phantagarowphantagarow
715
$endgroup$
- 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.
edited Mar 1 at 5:35
Emma
313112
answered Feb 28 at 20:18
phantagarowphantagarow
715
edited Mar 1 at 5:35
Emma
313112
edited Mar 1 at 5:35
Emma
313112
edited Mar 1 at 5:35
Emma
313112
313112
answered Feb 28 at 20:18
phantagarowphantagarow
715
answered Feb 28 at 20:18
phantagarowphantagarow
715
answered Feb 28 at 20:18
phantagarowphantagarow
715
715
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
Feb 28 at 20:31
$begingroup$
Could you elaborate more on “...measure on an abstract space”. Normal distribution is also a measure (of a rv). Also I could change to martingale measure $mathbb Q$ and my process still would be Gaussian, i.e. $dW^{mathbb Q}= dW^{mathbb P} +lambda dt$ by Girsanov.
$endgroup$
– tosik
Mar 1 at 7:07
1
$begingroup$
@tosik. A probability space is a triple $(Omega, mathcal{F}, mathbf{P})$ where $Omega$ is an abstract space. A normal distribution is a very specific type (with a specific density) on $mathbb{R}^n$ for some $n geq 1$. A random variable or vector $X: Omega rightarrow mathbb{R}^n$ is said to be normal if it induces a normal distribution on the range space.
$endgroup$
– phantagarow
Mar 1 at 14:01
1
$begingroup$
@tosik. You might want to be precise with Girsanov's Theorem so as not conflate related but different concepts. If $W_{t}$ is a Brownian Motion under $mathbf{P}$, the newly defined process $Z_{t} = W_{t} - int_{0}^{t} theta_{s} ds$ is no longer a BM under $mathbf{P}$. Girsanov's Thm tells you how to construct a new measure so as that $Z_{t}$ becomes a BM.
$endgroup$
– phantagarow
Mar 1 at 14:40
$begingroup$
I agree that $Z_t$ is no longer BM under $mathbb P$ but I think you agree that it is Gaussian both under $mathbb P$ and $mathbb Q$. I indeed probably confuse measures on $Omega$ with distributions on $mathbf R$. Your answer (and comments) do shed some light but I still don't have a clear picture. So, $Omega$ is an abstract space where for each $omega in Omega$ there is an associated function $f$ that we think of as a stochastic process? Can you elaborate more on the link between $mathbb P$ and normal distribution for the case of BM?
$endgroup$
– tosik
Mar 1 at 15:01
|
show 2 more comments
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
Feb 28 at 20:31
$begingroup$
Could you elaborate more on “...measure on an abstract space”. Normal distribution is also a measure (of a rv). Also I could change to martingale measure $mathbb Q$ and my process still would be Gaussian, i.e. $dW^{mathbb Q}= dW^{mathbb P} +lambda dt$ by Girsanov.
$endgroup$
– tosik
Mar 1 at 7:07
1
$begingroup$
@tosik. A probability space is a triple $(Omega, mathcal{F}, mathbf{P})$ where $Omega$ is an abstract space. A normal distribution is a very specific type (with a specific density) on $mathbb{R}^n$ for some $n geq 1$. A random variable or vector $X: Omega rightarrow mathbb{R}^n$ is said to be normal if it induces a normal distribution on the range space.
$endgroup$
– phantagarow
Mar 1 at 14:01
1
$begingroup$
@tosik. You might want to be precise with Girsanov's Theorem so as not conflate related but different concepts. If $W_{t}$ is a Brownian Motion under $mathbf{P}$, the newly defined process $Z_{t} = W_{t} - int_{0}^{t} theta_{s} ds$ is no longer a BM under $mathbf{P}$. Girsanov's Thm tells you how to construct a new measure so as that $Z_{t}$ becomes a BM.
$endgroup$
– phantagarow
Mar 1 at 14:40
$begingroup$
I agree that $Z_t$ is no longer BM under $mathbb P$ but I think you agree that it is Gaussian both under $mathbb P$ and $mathbb Q$. I indeed probably confuse measures on $Omega$ with distributions on $mathbf R$. Your answer (and comments) do shed some light but I still don't have a clear picture. So, $Omega$ is an abstract space where for each $omega in Omega$ there is an associated function $f$ that we think of as a stochastic process? Can you elaborate more on the link between $mathbb P$ and normal distribution for the case of BM?
$endgroup$
– tosik
Mar 1 at 15:01
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
Feb 28 at 20:31
$begingroup$
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
Feb 28 at 20:31
$begingroup$
Could you elaborate more on “...measure on an abstract space”. Normal distribution is also a measure (of a rv). Also I could change to martingale measure $mathbb Q$ and my process still would be Gaussian, i.e. $dW^{mathbb Q}= dW^{mathbb P} +lambda dt$ by Girsanov.
$endgroup$
– tosik
Mar 1 at 7:07
$begingroup$
Could you elaborate more on “...measure on an abstract space”. Normal distribution is also a measure (of a rv). Also I could change to martingale measure $mathbb Q$ and my process still would be Gaussian, i.e. $dW^{mathbb Q}= dW^{mathbb P} +lambda dt$ by Girsanov.
$endgroup$
– tosik
Mar 1 at 7:07
1
1
$begingroup$
@tosik. A probability space is a triple $(Omega, mathcal{F}, mathbf{P})$ where $Omega$ is an abstract space. A normal distribution is a very specific type (with a specific density) on $mathbb{R}^n$ for some $n geq 1$. A random variable or vector $X: Omega rightarrow mathbb{R}^n$ is said to be normal if it induces a normal distribution on the range space.
$endgroup$
– phantagarow
Mar 1 at 14:01
$begingroup$
@tosik. A probability space is a triple $(Omega, mathcal{F}, mathbf{P})$ where $Omega$ is an abstract space. A normal distribution is a very specific type (with a specific density) on $mathbb{R}^n$ for some $n geq 1$. A random variable or vector $X: Omega rightarrow mathbb{R}^n$ is said to be normal if it induces a normal distribution on the range space.
$endgroup$
– phantagarow
Mar 1 at 14:01
1
1
$begingroup$
@tosik. You might want to be precise with Girsanov's Theorem so as not conflate related but different concepts. If $W_{t}$ is a Brownian Motion under $mathbf{P}$, the newly defined process $Z_{t} = W_{t} - int_{0}^{t} theta_{s} ds$ is no longer a BM under $mathbf{P}$. Girsanov's Thm tells you how to construct a new measure so as that $Z_{t}$ becomes a BM.
$endgroup$
– phantagarow
Mar 1 at 14:40
$begingroup$
@tosik. You might want to be precise with Girsanov's Theorem so as not conflate related but different concepts. If $W_{t}$ is a Brownian Motion under $mathbf{P}$, the newly defined process $Z_{t} = W_{t} - int_{0}^{t} theta_{s} ds$ is no longer a BM under $mathbf{P}$. Girsanov's Thm tells you how to construct a new measure so as that $Z_{t}$ becomes a BM.
$endgroup$
– phantagarow
Mar 1 at 14:40
$begingroup$
I agree that $Z_t$ is no longer BM under $mathbb P$ but I think you agree that it is Gaussian both under $mathbb P$ and $mathbb Q$. I indeed probably confuse measures on $Omega$ with distributions on $mathbf R$. Your answer (and comments) do shed some light but I still don't have a clear picture. So, $Omega$ is an abstract space where for each $omega in Omega$ there is an associated function $f$ that we think of as a stochastic process? Can you elaborate more on the link between $mathbb P$ and normal distribution for the case of BM?
$endgroup$
– tosik
Mar 1 at 15:01
$begingroup$
I agree that $Z_t$ is no longer BM under $mathbb P$ but I think you agree that it is Gaussian both under $mathbb P$ and $mathbb Q$. I indeed probably confuse measures on $Omega$ with distributions on $mathbf R$. Your answer (and comments) do shed some light but I still don't have a clear picture. So, $Omega$ is an abstract space where for each $omega in Omega$ there is an associated function $f$ that we think of as a stochastic process? Can you elaborate more on the link between $mathbb P$ and normal distribution for the case of BM?
$endgroup$
– tosik
Mar 1 at 15:01
|
show 2 more comments
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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.
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– noob2
Feb 28 at 19:33