Mixing
Are $int_0^t text{sign}(W_u) , dW_u$ and $W_t$ independent for a Brownian motion $(W_t)_{t geq 0}$?
$begingroup$
Let $(W_t)_{t geq 0}$ be a Brownian motion. Consider
$$
X_t = int^t_0 text{sign}(W_u) , dW_u
$$
where $$text{sign}(x) := begin{cases} 1, & x geq 0, \ -1, & x>0. end{cases}$$
Prove that $X_t$ and $W_t$ are independent.
How do I prove that? Naturally I need to compute $E[X_tW_t]=0$. I don't know if it helps me but I have computed $X^2_t-t$ which is a martingale so by Levy characterization Theorem $X_t$ is a brownian motion itself.
probability-theory stochastic-processes stochastic-calculus brownian-motion
edited Jan 22 at 14:34
saz
81.6k861127
asked Jan 22 at 11:01
k.dkhkk.dkhk
16410
$endgroup$
add a comment |
$begingroup$
Let $(W_t)_{t geq 0}$ be a Brownian motion. Consider
$$
X_t = int^t_0 text{sign}(W_u) , dW_u
$$
where $$text{sign}(x) := begin{cases} 1, & x geq 0, \ -1, & x>0. end{cases}$$
Prove that $X_t$ and $W_t$ are independent.
How do I prove that? Naturally I need to compute $E[X_tW_t]=0$. I don't know if it helps me but I have computed $X^2_t-t$ which is a martingale so by Levy characterization Theorem $X_t$ is a brownian motion itself.
probability-theory stochastic-processes stochastic-calculus brownian-motion
edited Jan 22 at 14:34
saz
81.6k861127
asked Jan 22 at 11:01
k.dkhkk.dkhk
16410
$endgroup$
1
$begingroup$
$E(X_tW_t)=Eleft(int_0^toperatorname{sgn}(W_s)dW_sint_0^tdW_sright)=Eint_0^toperatorname{sgn}(W_s)ds=0$. But I'm not sure if $X_t$ and $W_t$ are independent.
$endgroup$
– AddSup
Jan 22 at 12:14
1
$begingroup$
As a consequence of Tanaka's formula, $int_0^ttext{sign}(W_s),dW_sle |W_t|$ almost surely. This would seem to preclude the independence of these random variables.
$endgroup$
– John Dawkins
Jan 22 at 15:03
add a comment |
$begingroup$
Let $(W_t)_{t geq 0}$ be a Brownian motion. Consider
$$
X_t = int^t_0 text{sign}(W_u) , dW_u
$$
where $$text{sign}(x) := begin{cases} 1, & x geq 0, \ -1, & x>0. end{cases}$$
Prove that $X_t$ and $W_t$ are independent.
How do I prove that? Naturally I need to compute $E[X_tW_t]=0$. I don't know if it helps me but I have computed $X^2_t-t$ which is a martingale so by Levy characterization Theorem $X_t$ is a brownian motion itself.
probability-theory stochastic-processes stochastic-calculus brownian-motion
edited Jan 22 at 14:34
saz
81.6k861127
asked Jan 22 at 11:01
k.dkhkk.dkhk
16410
$endgroup$
Let $(W_t)_{t geq 0}$ be a Brownian motion. Consider
$$
X_t = int^t_0 text{sign}(W_u) , dW_u
$$
where $$text{sign}(x) := begin{cases} 1, & x geq 0, \ -1, & x>0. end{cases}$$
Prove that $X_t$ and $W_t$ are independent.
How do I prove that? Naturally I need to compute $E[X_tW_t]=0$. I don't know if it helps me but I have computed $X^2_t-t$ which is a martingale so by Levy characterization Theorem $X_t$ is a brownian motion itself.
probability-theory stochastic-processes stochastic-calculus brownian-motion
probability-theory stochastic-processes stochastic-calculus brownian-motion
edited Jan 22 at 14:34
saz
81.6k861127
asked Jan 22 at 11:01
k.dkhkk.dkhk
16410
edited Jan 22 at 14:34
saz
81.6k861127
asked Jan 22 at 11:01
k.dkhkk.dkhk
16410
edited Jan 22 at 14:34
saz
81.6k861127
edited Jan 22 at 14:34
saz
81.6k861127
edited Jan 22 at 14:34
saz
81.6k861127
81.6k861127
asked Jan 22 at 11:01
k.dkhkk.dkhk
16410
asked Jan 22 at 11:01
k.dkhkk.dkhk
16410
asked Jan 22 at 11:01
k.dkhkk.dkhk
16410
16410
1
$begingroup$
$E(X_tW_t)=Eleft(int_0^toperatorname{sgn}(W_s)dW_sint_0^tdW_sright)=Eint_0^toperatorname{sgn}(W_s)ds=0$. But I'm not sure if $X_t$ and $W_t$ are independent.
$endgroup$
– AddSup
Jan 22 at 12:14
1
$begingroup$
As a consequence of Tanaka's formula, $int_0^ttext{sign}(W_s),dW_sle |W_t|$ almost surely. This would seem to preclude the independence of these random variables.
$endgroup$
– John Dawkins
Jan 22 at 15:03
add a comment |
1
$begingroup$
$E(X_tW_t)=Eleft(int_0^toperatorname{sgn}(W_s)dW_sint_0^tdW_sright)=Eint_0^toperatorname{sgn}(W_s)ds=0$. But I'm not sure if $X_t$ and $W_t$ are independent.
$endgroup$
– AddSup
Jan 22 at 12:14
1
$begingroup$
As a consequence of Tanaka's formula, $int_0^ttext{sign}(W_s),dW_sle |W_t|$ almost surely. This would seem to preclude the independence of these random variables.
$endgroup$
– John Dawkins
Jan 22 at 15:03
1
1
$begingroup$
$E(X_tW_t)=Eleft(int_0^toperatorname{sgn}(W_s)dW_sint_0^tdW_sright)=Eint_0^toperatorname{sgn}(W_s)ds=0$. But I'm not sure if $X_t$ and $W_t$ are independent.
$endgroup$
– AddSup
Jan 22 at 12:14
$begingroup$
$E(X_tW_t)=Eleft(int_0^toperatorname{sgn}(W_s)dW_sint_0^tdW_sright)=Eint_0^toperatorname{sgn}(W_s)ds=0$. But I'm not sure if $X_t$ and $W_t$ are independent.
$endgroup$
– AddSup
Jan 22 at 12:14
1
1
$begingroup$
As a consequence of Tanaka's formula, $int_0^ttext{sign}(W_s),dW_sle |W_t|$ almost surely. This would seem to preclude the independence of these random variables.
$endgroup$
– John Dawkins
Jan 22 at 15:03
$begingroup$
As a consequence of Tanaka's formula, $int_0^ttext{sign}(W_s),dW_sle |W_t|$ almost surely. This would seem to preclude the independence of these random variables.
$endgroup$
– John Dawkins
Jan 22 at 15:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It might be tempting to reason as follows: Since $(X_t)_{t geq 0}$ and $(W_t)_{t geq 0}$ are Brownian motions, we know that $W_t$ and $X_t$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $mathbb{E}(X_t W_t) = 0$, and consequently $X_t$ and $W_t$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $X_t$ and $W_t$ are independent.
The reasoning is wrong. Why? In order to conclude that $W_t$ and $X_t$ are independent, we need to know that the random vector $(X_t,W_t)$ is Gaussian; it is not enough to know that (the marginals) $X_t$ and $W_t$ are Gaussian.
In fact, $W_t$ and $X_t$ are not independent. By Itô's formula, we have $$W_t^2 = 2 int_0^t W_s ,dW_s + t.$$ As $mathbb{E}(X_t)=0$ we thus find
begin{align*} mathbb{E}(W_t^2 X_t)& =2 mathbb{E} left( X_t int_0^t W_s , dW_s right) \ &=2 mathbb{E} left(left[ int_0^t text{sgn}(W_s) , dW_s right] left[ int_0^t W_s , dW_s right] right) end{align*}
Applying Itô's isometry we obtain that
$$mathbb{E}(W_t^2 X_t) =2 mathbb{E} left( int_0^t W_s , text{sgn}(W_s) , ds right) =2int_0^t mathbb{E}(|W_s|) , ds.$$
The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $W_s sim N(0,s)$ entails $mathbb{E}(|W_s|) = sqrt{(2s)/pi}$). As $mathbb{E}(X_t)=0$ this shows that $$mathbb{E}(W_t^2 X_t) neq mathbb{E}(W_t^2) mathbb{E}(X_t),$$ and therefore $W_t$ and $X_t$ are not independent.
answered Jan 22 at 14:31
sazsaz
81.6k861127
$endgroup$
$begingroup$
Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct?
$endgroup$
– k.dkhk
Jan 23 at 12:34
add a comment |
Your Answer
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$begingroup$
It might be tempting to reason as follows: Since $(X_t)_{t geq 0}$ and $(W_t)_{t geq 0}$ are Brownian motions, we know that $W_t$ and $X_t$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $mathbb{E}(X_t W_t) = 0$, and consequently $X_t$ and $W_t$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $X_t$ and $W_t$ are independent.
The reasoning is wrong. Why? In order to conclude that $W_t$ and $X_t$ are independent, we need to know that the random vector $(X_t,W_t)$ is Gaussian; it is not enough to know that (the marginals) $X_t$ and $W_t$ are Gaussian.
In fact, $W_t$ and $X_t$ are not independent. By Itô's formula, we have $$W_t^2 = 2 int_0^t W_s ,dW_s + t.$$ As $mathbb{E}(X_t)=0$ we thus find
begin{align*} mathbb{E}(W_t^2 X_t)& =2 mathbb{E} left( X_t int_0^t W_s , dW_s right) \ &=2 mathbb{E} left(left[ int_0^t text{sgn}(W_s) , dW_s right] left[ int_0^t W_s , dW_s right] right) end{align*}
Applying Itô's isometry we obtain that
$$mathbb{E}(W_t^2 X_t) =2 mathbb{E} left( int_0^t W_s , text{sgn}(W_s) , ds right) =2int_0^t mathbb{E}(|W_s|) , ds.$$
The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $W_s sim N(0,s)$ entails $mathbb{E}(|W_s|) = sqrt{(2s)/pi}$). As $mathbb{E}(X_t)=0$ this shows that $$mathbb{E}(W_t^2 X_t) neq mathbb{E}(W_t^2) mathbb{E}(X_t),$$ and therefore $W_t$ and $X_t$ are not independent.
answered Jan 22 at 14:31
sazsaz
81.6k861127
$endgroup$
$begingroup$
Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct?
$endgroup$
– k.dkhk
Jan 23 at 12:34
add a comment |
$begingroup$
It might be tempting to reason as follows: Since $(X_t)_{t geq 0}$ and $(W_t)_{t geq 0}$ are Brownian motions, we know that $W_t$ and $X_t$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $mathbb{E}(X_t W_t) = 0$, and consequently $X_t$ and $W_t$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $X_t$ and $W_t$ are independent.
The reasoning is wrong. Why? In order to conclude that $W_t$ and $X_t$ are independent, we need to know that the random vector $(X_t,W_t)$ is Gaussian; it is not enough to know that (the marginals) $X_t$ and $W_t$ are Gaussian.
In fact, $W_t$ and $X_t$ are not independent. By Itô's formula, we have $$W_t^2 = 2 int_0^t W_s ,dW_s + t.$$ As $mathbb{E}(X_t)=0$ we thus find
begin{align*} mathbb{E}(W_t^2 X_t)& =2 mathbb{E} left( X_t int_0^t W_s , dW_s right) \ &=2 mathbb{E} left(left[ int_0^t text{sgn}(W_s) , dW_s right] left[ int_0^t W_s , dW_s right] right) end{align*}
Applying Itô's isometry we obtain that
$$mathbb{E}(W_t^2 X_t) =2 mathbb{E} left( int_0^t W_s , text{sgn}(W_s) , ds right) =2int_0^t mathbb{E}(|W_s|) , ds.$$
The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $W_s sim N(0,s)$ entails $mathbb{E}(|W_s|) = sqrt{(2s)/pi}$). As $mathbb{E}(X_t)=0$ this shows that $$mathbb{E}(W_t^2 X_t) neq mathbb{E}(W_t^2) mathbb{E}(X_t),$$ and therefore $W_t$ and $X_t$ are not independent.
answered Jan 22 at 14:31
sazsaz
81.6k861127
$endgroup$
$begingroup$
Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct?
$endgroup$
– k.dkhk
Jan 23 at 12:34
add a comment |
$begingroup$
It might be tempting to reason as follows: Since $(X_t)_{t geq 0}$ and $(W_t)_{t geq 0}$ are Brownian motions, we know that $W_t$ and $X_t$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $mathbb{E}(X_t W_t) = 0$, and consequently $X_t$ and $W_t$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $X_t$ and $W_t$ are independent.
The reasoning is wrong. Why? In order to conclude that $W_t$ and $X_t$ are independent, we need to know that the random vector $(X_t,W_t)$ is Gaussian; it is not enough to know that (the marginals) $X_t$ and $W_t$ are Gaussian.
In fact, $W_t$ and $X_t$ are not independent. By Itô's formula, we have $$W_t^2 = 2 int_0^t W_s ,dW_s + t.$$ As $mathbb{E}(X_t)=0$ we thus find
begin{align*} mathbb{E}(W_t^2 X_t)& =2 mathbb{E} left( X_t int_0^t W_s , dW_s right) \ &=2 mathbb{E} left(left[ int_0^t text{sgn}(W_s) , dW_s right] left[ int_0^t W_s , dW_s right] right) end{align*}
Applying Itô's isometry we obtain that
$$mathbb{E}(W_t^2 X_t) =2 mathbb{E} left( int_0^t W_s , text{sgn}(W_s) , ds right) =2int_0^t mathbb{E}(|W_s|) , ds.$$
The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $W_s sim N(0,s)$ entails $mathbb{E}(|W_s|) = sqrt{(2s)/pi}$). As $mathbb{E}(X_t)=0$ this shows that $$mathbb{E}(W_t^2 X_t) neq mathbb{E}(W_t^2) mathbb{E}(X_t),$$ and therefore $W_t$ and $X_t$ are not independent.
answered Jan 22 at 14:31
sazsaz
81.6k861127
$endgroup$
It might be tempting to reason as follows: Since $(X_t)_{t geq 0}$ and $(W_t)_{t geq 0}$ are Brownian motions, we know that $W_t$ and $X_t$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $mathbb{E}(X_t W_t) = 0$, and consequently $X_t$ and $W_t$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $X_t$ and $W_t$ are independent.
The reasoning is wrong. Why? In order to conclude that $W_t$ and $X_t$ are independent, we need to know that the random vector $(X_t,W_t)$ is Gaussian; it is not enough to know that (the marginals) $X_t$ and $W_t$ are Gaussian.
In fact, $W_t$ and $X_t$ are not independent. By Itô's formula, we have $$W_t^2 = 2 int_0^t W_s ,dW_s + t.$$ As $mathbb{E}(X_t)=0$ we thus find
begin{align*} mathbb{E}(W_t^2 X_t)& =2 mathbb{E} left( X_t int_0^t W_s , dW_s right) \ &=2 mathbb{E} left(left[ int_0^t text{sgn}(W_s) , dW_s right] left[ int_0^t W_s , dW_s right] right) end{align*}
Applying Itô's isometry we obtain that
$$mathbb{E}(W_t^2 X_t) =2 mathbb{E} left( int_0^t W_s , text{sgn}(W_s) , ds right) =2int_0^t mathbb{E}(|W_s|) , ds.$$
The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $W_s sim N(0,s)$ entails $mathbb{E}(|W_s|) = sqrt{(2s)/pi}$). As $mathbb{E}(X_t)=0$ this shows that $$mathbb{E}(W_t^2 X_t) neq mathbb{E}(W_t^2) mathbb{E}(X_t),$$ and therefore $W_t$ and $X_t$ are not independent.
answered Jan 22 at 14:31
sazsaz
81.6k861127
edited Jan 22 at 19:09
answered Jan 22 at 14:31
sazsaz
81.6k861127
answered Jan 22 at 14:31
sazsaz
81.6k861127
answered Jan 22 at 14:31
sazsaz
81.6k861127
81.6k861127
$begingroup$
Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct?
$endgroup$
– k.dkhk
Jan 23 at 12:34
add a comment |
$begingroup$
Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct?
$endgroup$
– k.dkhk
Jan 23 at 12:34
$begingroup$
Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct?
$endgroup$
– k.dkhk
Jan 23 at 12:34
$begingroup$
Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct?
$endgroup$
– k.dkhk
Jan 23 at 12:34
add a comment |
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$E(X_tW_t)=Eleft(int_0^toperatorname{sgn}(W_s)dW_sint_0^tdW_sright)=Eint_0^toperatorname{sgn}(W_s)ds=0$. But I'm not sure if $X_t$ and $W_t$ are independent.
$endgroup$
– AddSup
Jan 22 at 12:14
1
$begingroup$
As a consequence of Tanaka's formula, $int_0^ttext{sign}(W_s),dW_sle |W_t|$ almost surely. This would seem to preclude the independence of these random variables.
$endgroup$
– John Dawkins
Jan 22 at 15:03