Let $B(t)$ be a Brownian Motion. Show that the following processes are Brownian Motions on $[0,T].$
$begingroup$
1) $X(t)=-B(t).$
Since $-X(t)$ is a Gaussian Process with zero mean, $X(t)$ is also Gaussian Process with zero mean. We simply want to show that its covariance function is $min(t,s).$ This is trivial since $$gamma(s,t) = text{cov}(-B(t)times -B(s)) = text{cov}(B(t)B(s))=min(t,s).$$ Thus $X(t) = -B(t)$ is also a Brownian motion.
2) $X(t) = B(T-t)-B(T).$
I am not sure how to argue why $X(t)$ has zero mean. However for the covariance function we have:
$$gamma(s,t) = text{cov}((B(T-s)-B(s))times (B(T-t)-B(t)))$$
$$= E[(B(T-s)-B(s))times (B(T-t)-B(t))] $$
$$= max(t,s)-min(T-s,t)-min(T-t,s)+min(s,t)$$
$$= t + s - 2min(T,s+t).$$
I don't know how to show that this is the same as $min(s,t).$
3) $X(t) = cB(t/c^2)$ where $Tleq infty.$
For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = c^2E(B(s/c^2)B(t/c^2)) = c^2 min(s/c^2,t/c^2)=min(s,t).$$
However, I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$
4) $X(t) = tB(1/t)$ and $X(0)=0.$
For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = tsE(B(1/s)B(1/t)) = ts min(1/t,1/s)=min(s,t).$$
However, for this part too I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$
probability
asked Dec 18 '18 at 21:12
Hello_WorldHello_World
4,13121831
$endgroup$
add a comment |
$begingroup$
1) $X(t)=-B(t).$
Since $-X(t)$ is a Gaussian Process with zero mean, $X(t)$ is also Gaussian Process with zero mean. We simply want to show that its covariance function is $min(t,s).$ This is trivial since $$gamma(s,t) = text{cov}(-B(t)times -B(s)) = text{cov}(B(t)B(s))=min(t,s).$$ Thus $X(t) = -B(t)$ is also a Brownian motion.
2) $X(t) = B(T-t)-B(T).$
I am not sure how to argue why $X(t)$ has zero mean. However for the covariance function we have:
$$gamma(s,t) = text{cov}((B(T-s)-B(s))times (B(T-t)-B(t)))$$
$$= E[(B(T-s)-B(s))times (B(T-t)-B(t))] $$
$$= max(t,s)-min(T-s,t)-min(T-t,s)+min(s,t)$$
$$= t + s - 2min(T,s+t).$$
I don't know how to show that this is the same as $min(s,t).$
3) $X(t) = cB(t/c^2)$ where $Tleq infty.$
For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = c^2E(B(s/c^2)B(t/c^2)) = c^2 min(s/c^2,t/c^2)=min(s,t).$$
However, I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$
4) $X(t) = tB(1/t)$ and $X(0)=0.$
For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = tsE(B(1/s)B(1/t)) = ts min(1/t,1/s)=min(s,t).$$
However, for this part too I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$
probability
asked Dec 18 '18 at 21:12
Hello_WorldHello_World
4,13121831
$endgroup$
add a comment |
$begingroup$
1) $X(t)=-B(t).$
Since $-X(t)$ is a Gaussian Process with zero mean, $X(t)$ is also Gaussian Process with zero mean. We simply want to show that its covariance function is $min(t,s).$ This is trivial since $$gamma(s,t) = text{cov}(-B(t)times -B(s)) = text{cov}(B(t)B(s))=min(t,s).$$ Thus $X(t) = -B(t)$ is also a Brownian motion.
2) $X(t) = B(T-t)-B(T).$
I am not sure how to argue why $X(t)$ has zero mean. However for the covariance function we have:
$$gamma(s,t) = text{cov}((B(T-s)-B(s))times (B(T-t)-B(t)))$$
$$= E[(B(T-s)-B(s))times (B(T-t)-B(t))] $$
$$= max(t,s)-min(T-s,t)-min(T-t,s)+min(s,t)$$
$$= t + s - 2min(T,s+t).$$
I don't know how to show that this is the same as $min(s,t).$
3) $X(t) = cB(t/c^2)$ where $Tleq infty.$
For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = c^2E(B(s/c^2)B(t/c^2)) = c^2 min(s/c^2,t/c^2)=min(s,t).$$
However, I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$
4) $X(t) = tB(1/t)$ and $X(0)=0.$
For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = tsE(B(1/s)B(1/t)) = ts min(1/t,1/s)=min(s,t).$$
However, for this part too I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$
probability
asked Dec 18 '18 at 21:12
Hello_WorldHello_World
4,13121831
$endgroup$
1) $X(t)=-B(t).$
Since $-X(t)$ is a Gaussian Process with zero mean, $X(t)$ is also Gaussian Process with zero mean. We simply want to show that its covariance function is $min(t,s).$ This is trivial since $$gamma(s,t) = text{cov}(-B(t)times -B(s)) = text{cov}(B(t)B(s))=min(t,s).$$ Thus $X(t) = -B(t)$ is also a Brownian motion.
2) $X(t) = B(T-t)-B(T).$
I am not sure how to argue why $X(t)$ has zero mean. However for the covariance function we have:
$$gamma(s,t) = text{cov}((B(T-s)-B(s))times (B(T-t)-B(t)))$$
$$= E[(B(T-s)-B(s))times (B(T-t)-B(t))] $$
$$= max(t,s)-min(T-s,t)-min(T-t,s)+min(s,t)$$
$$= t + s - 2min(T,s+t).$$
I don't know how to show that this is the same as $min(s,t).$
3) $X(t) = cB(t/c^2)$ where $Tleq infty.$
For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = c^2E(B(s/c^2)B(t/c^2)) = c^2 min(s/c^2,t/c^2)=min(s,t).$$
However, I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$
4) $X(t) = tB(1/t)$ and $X(0)=0.$
For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = tsE(B(1/s)B(1/t)) = ts min(1/t,1/s)=min(s,t).$$
However, for this part too I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$
probability
probability
asked Dec 18 '18 at 21:12
Hello_WorldHello_World
4,13121831
asked Dec 18 '18 at 21:12
Hello_WorldHello_World
4,13121831
asked Dec 18 '18 at 21:12
Hello_WorldHello_World
4,13121831
asked Dec 18 '18 at 21:12
Hello_WorldHello_World
4,13121831
asked Dec 18 '18 at 21:12
Hello_WorldHello_World
4,13121831
4,13121831
add a comment |
add a comment |
1 Answer
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$begingroup$
In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
$$
Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
$$
You should also mention that $X(0) = 0$, which is trivial of course.
In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
$$
(X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
$$
The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.
For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.
In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.
answered Dec 19 '18 at 14:17
Tki DenebTki Deneb
32710
$endgroup$
add a comment |
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$begingroup$
In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
$$
Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
$$
You should also mention that $X(0) = 0$, which is trivial of course.
In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
$$
(X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
$$
The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.
For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.
In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.
answered Dec 19 '18 at 14:17
Tki DenebTki Deneb
32710
$endgroup$
add a comment |
$begingroup$
In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
$$
Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
$$
You should also mention that $X(0) = 0$, which is trivial of course.
In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
$$
(X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
$$
The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.
For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.
In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.
answered Dec 19 '18 at 14:17
Tki DenebTki Deneb
32710
$endgroup$
add a comment |
$begingroup$
In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
$$
Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
$$
You should also mention that $X(0) = 0$, which is trivial of course.
In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
$$
(X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
$$
The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.
For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.
In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.
answered Dec 19 '18 at 14:17
Tki DenebTki Deneb
32710
$endgroup$
In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
$$
Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
$$
You should also mention that $X(0) = 0$, which is trivial of course.
In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
$$
(X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
$$
The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.
For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.
In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.
answered Dec 19 '18 at 14:17
Tki DenebTki Deneb
32710
edited Dec 19 '18 at 14:26
answered Dec 19 '18 at 14:17
Tki DenebTki Deneb
32710
answered Dec 19 '18 at 14:17
Tki DenebTki Deneb
32710
answered Dec 19 '18 at 14:17
Tki DenebTki Deneb
32710
32710
add a comment |
add a comment |
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