Find constants $a$ and $b$ such that $X(t)$ is a Brownian Motion
$begingroup$
Let $B(t)$ be a Brownian Motion. Find all constants $A$ and $b$ such that $X(t)=int_0^t(a+bfrac{u}{t})dB(u)$ is also a Brownian Motion.
First we know that if $f in L^2[a,b]$ then $int_a^bfdB(u)$ is a Gaussian random variable with mean $0$ and variance $||f||_{L^2[a,b]}^2$ and also if $X(t)$ is going to be a Brownian motion then $X(t)$ must be Gaussian with mean $0$ and variance $t$. So since $(a+bfrac{u}{t}) in L^2[0,t]$ then this shows that $X(t)$ is Gaussian with mean $0$ and variance $t(a^2+ab+frac{b^2}{3})$ which means that $(a^2+ab+frac{b^2}{3})$ must equal $1$ for $X(t)$ to be a Brownian motion but this isnt enough im guessing?
Next we need to show that $X(0)=0$ almost surely. For this we may use integration by parts and the identity $int_0^tB(u)du=int_0^t(t-u)dB(u)$ to see that $$ X(t)=(a+b)B(t)-frac{b}{t}int_0^t(t-u)dB(u) $$
and as $tto0^+$ we easily see the first term goes to $0$ and the second term also goes to $0$ after applying L'hospital's rule and the stochastic form of Leibniz's rule.
I also need to show that $X(t)$ has independent increments and $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t-s$. I am unsure how to show these two things. Any help is appreciated.
brownian-motion stochastic-integrals stochastic-analysis
asked Dec 17 '18 at 19:18
alpastalpast
468314
$endgroup$
|
show 2 more comments
$begingroup$
Let $B(t)$ be a Brownian Motion. Find all constants $A$ and $b$ such that $X(t)=int_0^t(a+bfrac{u}{t})dB(u)$ is also a Brownian Motion.
First we know that if $f in L^2[a,b]$ then $int_a^bfdB(u)$ is a Gaussian random variable with mean $0$ and variance $||f||_{L^2[a,b]}^2$ and also if $X(t)$ is going to be a Brownian motion then $X(t)$ must be Gaussian with mean $0$ and variance $t$. So since $(a+bfrac{u}{t}) in L^2[0,t]$ then this shows that $X(t)$ is Gaussian with mean $0$ and variance $t(a^2+ab+frac{b^2}{3})$ which means that $(a^2+ab+frac{b^2}{3})$ must equal $1$ for $X(t)$ to be a Brownian motion but this isnt enough im guessing?
Next we need to show that $X(0)=0$ almost surely. For this we may use integration by parts and the identity $int_0^tB(u)du=int_0^t(t-u)dB(u)$ to see that $$ X(t)=(a+b)B(t)-frac{b}{t}int_0^t(t-u)dB(u) $$
and as $tto0^+$ we easily see the first term goes to $0$ and the second term also goes to $0$ after applying L'hospital's rule and the stochastic form of Leibniz's rule.
I also need to show that $X(t)$ has independent increments and $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t-s$. I am unsure how to show these two things. Any help is appreciated.
brownian-motion stochastic-integrals stochastic-analysis
asked Dec 17 '18 at 19:18
alpastalpast
468314
$endgroup$
$begingroup$
It may help to know that one can drop the explicit assumption of independent increments and replace it with the requirement that $E[X(t) X(s)]=min { t,s }$. This should result in another nontrivial equation constraining $a,b$.
$endgroup$
– Ian
Dec 17 '18 at 19:21
$begingroup$
@Ian I forgot, my book mentions that ill try to work with that. But my book says that independent increments can be deduced from the property that $X(t)-X(s)$ is Gaussian with mean 0 and variance $t$ and $E(X(t)X(s)=min{t,s}$. So I still need to show $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t$ and I am stuck on that part
$endgroup$
– alpast
Dec 17 '18 at 19:25
$begingroup$
You don't need to show the increments are Gaussian, that follows for free from $X(t)$ being Gaussian at each time. Similarly the variance of the increments will follow from knowing the covariance function. All you really need to do on top of what you already did is compute $E[X(t) X(s)]$ as a function of $a,b,t,s$.
$endgroup$
– Ian
Dec 17 '18 at 19:30
$begingroup$
Levi theorem should work
$endgroup$
– Makina
Dec 17 '18 at 19:32
$begingroup$
@Ian How do I find $E(X(t)X(s))$ because in my text when this expectation is done it is assumed the increments are independent. and by covariance function do you mean $Cov(X,Y) = E(XY)-E(X)E(Y)$.
$endgroup$
– alpast
Dec 17 '18 at 20:25
|
show 2 more comments
$begingroup$
Let $B(t)$ be a Brownian Motion. Find all constants $A$ and $b$ such that $X(t)=int_0^t(a+bfrac{u}{t})dB(u)$ is also a Brownian Motion.
First we know that if $f in L^2[a,b]$ then $int_a^bfdB(u)$ is a Gaussian random variable with mean $0$ and variance $||f||_{L^2[a,b]}^2$ and also if $X(t)$ is going to be a Brownian motion then $X(t)$ must be Gaussian with mean $0$ and variance $t$. So since $(a+bfrac{u}{t}) in L^2[0,t]$ then this shows that $X(t)$ is Gaussian with mean $0$ and variance $t(a^2+ab+frac{b^2}{3})$ which means that $(a^2+ab+frac{b^2}{3})$ must equal $1$ for $X(t)$ to be a Brownian motion but this isnt enough im guessing?
Next we need to show that $X(0)=0$ almost surely. For this we may use integration by parts and the identity $int_0^tB(u)du=int_0^t(t-u)dB(u)$ to see that $$ X(t)=(a+b)B(t)-frac{b}{t}int_0^t(t-u)dB(u) $$
and as $tto0^+$ we easily see the first term goes to $0$ and the second term also goes to $0$ after applying L'hospital's rule and the stochastic form of Leibniz's rule.
I also need to show that $X(t)$ has independent increments and $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t-s$. I am unsure how to show these two things. Any help is appreciated.
brownian-motion stochastic-integrals stochastic-analysis
asked Dec 17 '18 at 19:18
alpastalpast
468314
$endgroup$
Let $B(t)$ be a Brownian Motion. Find all constants $A$ and $b$ such that $X(t)=int_0^t(a+bfrac{u}{t})dB(u)$ is also a Brownian Motion.
First we know that if $f in L^2[a,b]$ then $int_a^bfdB(u)$ is a Gaussian random variable with mean $0$ and variance $||f||_{L^2[a,b]}^2$ and also if $X(t)$ is going to be a Brownian motion then $X(t)$ must be Gaussian with mean $0$ and variance $t$. So since $(a+bfrac{u}{t}) in L^2[0,t]$ then this shows that $X(t)$ is Gaussian with mean $0$ and variance $t(a^2+ab+frac{b^2}{3})$ which means that $(a^2+ab+frac{b^2}{3})$ must equal $1$ for $X(t)$ to be a Brownian motion but this isnt enough im guessing?
Next we need to show that $X(0)=0$ almost surely. For this we may use integration by parts and the identity $int_0^tB(u)du=int_0^t(t-u)dB(u)$ to see that $$ X(t)=(a+b)B(t)-frac{b}{t}int_0^t(t-u)dB(u) $$
and as $tto0^+$ we easily see the first term goes to $0$ and the second term also goes to $0$ after applying L'hospital's rule and the stochastic form of Leibniz's rule.
I also need to show that $X(t)$ has independent increments and $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t-s$. I am unsure how to show these two things. Any help is appreciated.
brownian-motion stochastic-integrals stochastic-analysis
brownian-motion stochastic-integrals stochastic-analysis
asked Dec 17 '18 at 19:18
alpastalpast
468314
asked Dec 17 '18 at 19:18
alpastalpast
468314
asked Dec 17 '18 at 19:18
alpastalpast
468314
asked Dec 17 '18 at 19:18
alpastalpast
468314
asked Dec 17 '18 at 19:18
alpastalpast
468314
468314
$begingroup$
It may help to know that one can drop the explicit assumption of independent increments and replace it with the requirement that $E[X(t) X(s)]=min { t,s }$. This should result in another nontrivial equation constraining $a,b$.
$endgroup$
– Ian
Dec 17 '18 at 19:21
$begingroup$
@Ian I forgot, my book mentions that ill try to work with that. But my book says that independent increments can be deduced from the property that $X(t)-X(s)$ is Gaussian with mean 0 and variance $t$ and $E(X(t)X(s)=min{t,s}$. So I still need to show $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t$ and I am stuck on that part
$endgroup$
– alpast
Dec 17 '18 at 19:25
$begingroup$
You don't need to show the increments are Gaussian, that follows for free from $X(t)$ being Gaussian at each time. Similarly the variance of the increments will follow from knowing the covariance function. All you really need to do on top of what you already did is compute $E[X(t) X(s)]$ as a function of $a,b,t,s$.
$endgroup$
– Ian
Dec 17 '18 at 19:30
$begingroup$
Levi theorem should work
$endgroup$
– Makina
Dec 17 '18 at 19:32
$begingroup$
@Ian How do I find $E(X(t)X(s))$ because in my text when this expectation is done it is assumed the increments are independent. and by covariance function do you mean $Cov(X,Y) = E(XY)-E(X)E(Y)$.
$endgroup$
– alpast
Dec 17 '18 at 20:25
|
show 2 more comments
$begingroup$
It may help to know that one can drop the explicit assumption of independent increments and replace it with the requirement that $E[X(t) X(s)]=min { t,s }$. This should result in another nontrivial equation constraining $a,b$.
$endgroup$
– Ian
Dec 17 '18 at 19:21
$begingroup$
@Ian I forgot, my book mentions that ill try to work with that. But my book says that independent increments can be deduced from the property that $X(t)-X(s)$ is Gaussian with mean 0 and variance $t$ and $E(X(t)X(s)=min{t,s}$. So I still need to show $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t$ and I am stuck on that part
$endgroup$
– alpast
Dec 17 '18 at 19:25
$begingroup$
You don't need to show the increments are Gaussian, that follows for free from $X(t)$ being Gaussian at each time. Similarly the variance of the increments will follow from knowing the covariance function. All you really need to do on top of what you already did is compute $E[X(t) X(s)]$ as a function of $a,b,t,s$.
$endgroup$
– Ian
Dec 17 '18 at 19:30
$begingroup$
Levi theorem should work
$endgroup$
– Makina
Dec 17 '18 at 19:32
$begingroup$
@Ian How do I find $E(X(t)X(s))$ because in my text when this expectation is done it is assumed the increments are independent. and by covariance function do you mean $Cov(X,Y) = E(XY)-E(X)E(Y)$.
$endgroup$
– alpast
Dec 17 '18 at 20:25
$begingroup$
It may help to know that one can drop the explicit assumption of independent increments and replace it with the requirement that $E[X(t) X(s)]=min { t,s }$. This should result in another nontrivial equation constraining $a,b$.
$endgroup$
– Ian
Dec 17 '18 at 19:21
$begingroup$
It may help to know that one can drop the explicit assumption of independent increments and replace it with the requirement that $E[X(t) X(s)]=min { t,s }$. This should result in another nontrivial equation constraining $a,b$.
$endgroup$
– Ian
Dec 17 '18 at 19:21
$begingroup$
@Ian I forgot, my book mentions that ill try to work with that. But my book says that independent increments can be deduced from the property that $X(t)-X(s)$ is Gaussian with mean 0 and variance $t$ and $E(X(t)X(s)=min{t,s}$. So I still need to show $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t$ and I am stuck on that part
$endgroup$
– alpast
Dec 17 '18 at 19:25
$begingroup$
@Ian I forgot, my book mentions that ill try to work with that. But my book says that independent increments can be deduced from the property that $X(t)-X(s)$ is Gaussian with mean 0 and variance $t$ and $E(X(t)X(s)=min{t,s}$. So I still need to show $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t$ and I am stuck on that part
$endgroup$
– alpast
Dec 17 '18 at 19:25
$begingroup$
You don't need to show the increments are Gaussian, that follows for free from $X(t)$ being Gaussian at each time. Similarly the variance of the increments will follow from knowing the covariance function. All you really need to do on top of what you already did is compute $E[X(t) X(s)]$ as a function of $a,b,t,s$.
$endgroup$
– Ian
Dec 17 '18 at 19:30
$begingroup$
You don't need to show the increments are Gaussian, that follows for free from $X(t)$ being Gaussian at each time. Similarly the variance of the increments will follow from knowing the covariance function. All you really need to do on top of what you already did is compute $E[X(t) X(s)]$ as a function of $a,b,t,s$.
$endgroup$
– Ian
Dec 17 '18 at 19:30
$begingroup$
Levi theorem should work
$endgroup$
– Makina
Dec 17 '18 at 19:32
$begingroup$
Levi theorem should work
$endgroup$
– Makina
Dec 17 '18 at 19:32
$begingroup$
@Ian How do I find $E(X(t)X(s))$ because in my text when this expectation is done it is assumed the increments are independent. and by covariance function do you mean $Cov(X,Y) = E(XY)-E(X)E(Y)$.
$endgroup$
– alpast
Dec 17 '18 at 20:25
$begingroup$
@Ian How do I find $E(X(t)X(s))$ because in my text when this expectation is done it is assumed the increments are independent. and by covariance function do you mean $Cov(X,Y) = E(XY)-E(X)E(Y)$.
$endgroup$
– alpast
Dec 17 '18 at 20:25
|
show 2 more comments
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$begingroup$
It may help to know that one can drop the explicit assumption of independent increments and replace it with the requirement that $E[X(t) X(s)]=min { t,s }$. This should result in another nontrivial equation constraining $a,b$.
$endgroup$
– Ian
Dec 17 '18 at 19:21
$begingroup$
@Ian I forgot, my book mentions that ill try to work with that. But my book says that independent increments can be deduced from the property that $X(t)-X(s)$ is Gaussian with mean 0 and variance $t$ and $E(X(t)X(s)=min{t,s}$. So I still need to show $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t$ and I am stuck on that part
$endgroup$
– alpast
Dec 17 '18 at 19:25
$begingroup$
You don't need to show the increments are Gaussian, that follows for free from $X(t)$ being Gaussian at each time. Similarly the variance of the increments will follow from knowing the covariance function. All you really need to do on top of what you already did is compute $E[X(t) X(s)]$ as a function of $a,b,t,s$.
$endgroup$
– Ian
Dec 17 '18 at 19:30
$begingroup$
Levi theorem should work
$endgroup$
– Makina
Dec 17 '18 at 19:32
$begingroup$
@Ian How do I find $E(X(t)X(s))$ because in my text when this expectation is done it is assumed the increments are independent. and by covariance function do you mean $Cov(X,Y) = E(XY)-E(X)E(Y)$.
$endgroup$
– alpast
Dec 17 '18 at 20:25