Mixing
Lévy process + scaling property $implies$ Brownian motion
$begingroup$
How can I show that if $xi_t$ is a Lévy process distributed as $xi_{t+s}- xi_s$ for all $t,s in [0,infty)$ and has independence of increments, and also is distributed as $lambdaxi_{lambda^{-2}t}$ for all $lambdain (0,infty)$, then $xi_t$ is a Brownian motion of some diffusivity $kin[0,infty)$?
probability-theory stochastic-processes brownian-motion levy-processes
edited Jan 15 at 20:08
saz
80.9k861127
asked Jan 13 at 14:55
Claudio DelfinoClaudio Delfino
43
$endgroup$
add a comment |
$begingroup$
How can I show that if $xi_t$ is a Lévy process distributed as $xi_{t+s}- xi_s$ for all $t,s in [0,infty)$ and has independence of increments, and also is distributed as $lambdaxi_{lambda^{-2}t}$ for all $lambdain (0,infty)$, then $xi_t$ is a Brownian motion of some diffusivity $kin[0,infty)$?
probability-theory stochastic-processes brownian-motion levy-processes
edited Jan 15 at 20:08
saz
80.9k861127
asked Jan 13 at 14:55
Claudio DelfinoClaudio Delfino
43
$endgroup$
$begingroup$
The only thing which you need to show is that $xi_t$ is Gaussian (with appropriate constants) for any $t>0$. To this end you can study the characteristic function of $xi_t$ using the fact that $xi_t$ is distributed as $lambda xi_{lambda^{-2} t}$.
$endgroup$
– saz
Jan 13 at 18:41
$begingroup$
If we call $phi_{xi_t}$ the characteristic function we have $phi_{xi_t}(s) = mathbb{E} [e^{is xi_t}]=mathbb{E}[e^{islambda xi_{lambda^{-2}t}}]=phi_{xi_{lambda^{-2}t}}(slambda)$, i don't understand how it should help...
$endgroup$
– Claudio Delfino
Jan 14 at 17:50
add a comment |
$begingroup$
How can I show that if $xi_t$ is a Lévy process distributed as $xi_{t+s}- xi_s$ for all $t,s in [0,infty)$ and has independence of increments, and also is distributed as $lambdaxi_{lambda^{-2}t}$ for all $lambdain (0,infty)$, then $xi_t$ is a Brownian motion of some diffusivity $kin[0,infty)$?
probability-theory stochastic-processes brownian-motion levy-processes
edited Jan 15 at 20:08
saz
80.9k861127
asked Jan 13 at 14:55
Claudio DelfinoClaudio Delfino
43
$endgroup$
How can I show that if $xi_t$ is a Lévy process distributed as $xi_{t+s}- xi_s$ for all $t,s in [0,infty)$ and has independence of increments, and also is distributed as $lambdaxi_{lambda^{-2}t}$ for all $lambdain (0,infty)$, then $xi_t$ is a Brownian motion of some diffusivity $kin[0,infty)$?
probability-theory stochastic-processes brownian-motion levy-processes
probability-theory stochastic-processes brownian-motion levy-processes
edited Jan 15 at 20:08
saz
80.9k861127
asked Jan 13 at 14:55
Claudio DelfinoClaudio Delfino
43
edited Jan 15 at 20:08
saz
80.9k861127
asked Jan 13 at 14:55
Claudio DelfinoClaudio Delfino
43
edited Jan 15 at 20:08
saz
80.9k861127
edited Jan 15 at 20:08
saz
80.9k861127
edited Jan 15 at 20:08
saz
80.9k861127
80.9k861127
asked Jan 13 at 14:55
Claudio DelfinoClaudio Delfino
43
asked Jan 13 at 14:55
Claudio DelfinoClaudio Delfino
43
asked Jan 13 at 14:55
Claudio DelfinoClaudio Delfino
43
43
$begingroup$
The only thing which you need to show is that $xi_t$ is Gaussian (with appropriate constants) for any $t>0$. To this end you can study the characteristic function of $xi_t$ using the fact that $xi_t$ is distributed as $lambda xi_{lambda^{-2} t}$.
$endgroup$
– saz
Jan 13 at 18:41
$begingroup$
If we call $phi_{xi_t}$ the characteristic function we have $phi_{xi_t}(s) = mathbb{E} [e^{is xi_t}]=mathbb{E}[e^{islambda xi_{lambda^{-2}t}}]=phi_{xi_{lambda^{-2}t}}(slambda)$, i don't understand how it should help...
$endgroup$
– Claudio Delfino
Jan 14 at 17:50
add a comment |
$begingroup$
The only thing which you need to show is that $xi_t$ is Gaussian (with appropriate constants) for any $t>0$. To this end you can study the characteristic function of $xi_t$ using the fact that $xi_t$ is distributed as $lambda xi_{lambda^{-2} t}$.
$endgroup$
– saz
Jan 13 at 18:41
$begingroup$
If we call $phi_{xi_t}$ the characteristic function we have $phi_{xi_t}(s) = mathbb{E} [e^{is xi_t}]=mathbb{E}[e^{islambda xi_{lambda^{-2}t}}]=phi_{xi_{lambda^{-2}t}}(slambda)$, i don't understand how it should help...
$endgroup$
– Claudio Delfino
Jan 14 at 17:50
$begingroup$
The only thing which you need to show is that $xi_t$ is Gaussian (with appropriate constants) for any $t>0$. To this end you can study the characteristic function of $xi_t$ using the fact that $xi_t$ is distributed as $lambda xi_{lambda^{-2} t}$.
$endgroup$
– saz
Jan 13 at 18:41
$begingroup$
The only thing which you need to show is that $xi_t$ is Gaussian (with appropriate constants) for any $t>0$. To this end you can study the characteristic function of $xi_t$ using the fact that $xi_t$ is distributed as $lambda xi_{lambda^{-2} t}$.
$endgroup$
– saz
Jan 13 at 18:41
$begingroup$
If we call $phi_{xi_t}$ the characteristic function we have $phi_{xi_t}(s) = mathbb{E} [e^{is xi_t}]=mathbb{E}[e^{islambda xi_{lambda^{-2}t}}]=phi_{xi_{lambda^{-2}t}}(slambda)$, i don't understand how it should help...
$endgroup$
– Claudio Delfino
Jan 14 at 17:50
$begingroup$
If we call $phi_{xi_t}$ the characteristic function we have $phi_{xi_t}(s) = mathbb{E} [e^{is xi_t}]=mathbb{E}[e^{islambda xi_{lambda^{-2}t}}]=phi_{xi_{lambda^{-2}t}}(slambda)$, i don't understand how it should help...
$endgroup$
– Claudio Delfino
Jan 14 at 17:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $(xi_t)_{t geq 0}$ has, by assumption, stationary and independent increments, it suffices to show that $xi_t sim N(0,t)$ for any $t>0$. Denote by $psi$ the characteristic exponent of $(xi_t)_{t geq 0}$, i.e. $$mathbb{E}exp(i eta xi_t) = exp(-t psi(eta)), qquad t geq 0, eta in mathbb{R}. tag{1}$$ Fix $t>0$. As
$$xi_t sim lambda xi_{lambda^{-2}t}, qquad lambda>0 tag{2}$$
we have
$$mathbb{E}exp(i eta xi_t) = mathbb{E}exp(i eta lambda xi_{t lambda^{-2}})$$
i.e.
$$exp(-t psi(eta)) = exp(- t lambda^{-2} psi(lambda eta)).$$
Because of the uniqueness of the characteristic exponent $psi$ this implies
$$psi(eta) = lambda^{-2} psi(lambda eta) quad text{for all $lambda>0$, $eta in mathbb{R}$}. tag{3}$$
If we choose $eta:=1$ it follows that the limit $$d := lim_{lambda to 0} frac{psi(lambda)}{lambda^2}$$ exists and $$d = psi(1).$$
As
$$frac{psi(eta)}{eta^2} stackrel{(3)}{=} frac{psi(lambda eta)}{(lambda eta)^2} xrightarrow{ lambda to 0} d = psi(1)$$
we conclude that
$$psi(eta) = psi(1) eta^2, qquad eta in mathbb{R},$$
which means that $(xi_t)_{t geq 0}$ is a Brownian motion with scaling parameter $sigma geq 0$.
Remark: In the multidimensional framework, i.e. if $(xi_t)_{t geq 0}$ is a $d$-dimensional Lévy process, we can apply the above reasoning to each of the coordinates.
answered Jan 15 at 20:05
sazsaz
80.9k861127
$endgroup$
add a comment |
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$begingroup$
Since $(xi_t)_{t geq 0}$ has, by assumption, stationary and independent increments, it suffices to show that $xi_t sim N(0,t)$ for any $t>0$. Denote by $psi$ the characteristic exponent of $(xi_t)_{t geq 0}$, i.e. $$mathbb{E}exp(i eta xi_t) = exp(-t psi(eta)), qquad t geq 0, eta in mathbb{R}. tag{1}$$ Fix $t>0$. As
$$xi_t sim lambda xi_{lambda^{-2}t}, qquad lambda>0 tag{2}$$
we have
$$mathbb{E}exp(i eta xi_t) = mathbb{E}exp(i eta lambda xi_{t lambda^{-2}})$$
i.e.
$$exp(-t psi(eta)) = exp(- t lambda^{-2} psi(lambda eta)).$$
Because of the uniqueness of the characteristic exponent $psi$ this implies
$$psi(eta) = lambda^{-2} psi(lambda eta) quad text{for all $lambda>0$, $eta in mathbb{R}$}. tag{3}$$
If we choose $eta:=1$ it follows that the limit $$d := lim_{lambda to 0} frac{psi(lambda)}{lambda^2}$$ exists and $$d = psi(1).$$
As
$$frac{psi(eta)}{eta^2} stackrel{(3)}{=} frac{psi(lambda eta)}{(lambda eta)^2} xrightarrow{ lambda to 0} d = psi(1)$$
we conclude that
$$psi(eta) = psi(1) eta^2, qquad eta in mathbb{R},$$
which means that $(xi_t)_{t geq 0}$ is a Brownian motion with scaling parameter $sigma geq 0$.
Remark: In the multidimensional framework, i.e. if $(xi_t)_{t geq 0}$ is a $d$-dimensional Lévy process, we can apply the above reasoning to each of the coordinates.
answered Jan 15 at 20:05
sazsaz
80.9k861127
$endgroup$
add a comment |
$begingroup$
Since $(xi_t)_{t geq 0}$ has, by assumption, stationary and independent increments, it suffices to show that $xi_t sim N(0,t)$ for any $t>0$. Denote by $psi$ the characteristic exponent of $(xi_t)_{t geq 0}$, i.e. $$mathbb{E}exp(i eta xi_t) = exp(-t psi(eta)), qquad t geq 0, eta in mathbb{R}. tag{1}$$ Fix $t>0$. As
$$xi_t sim lambda xi_{lambda^{-2}t}, qquad lambda>0 tag{2}$$
we have
$$mathbb{E}exp(i eta xi_t) = mathbb{E}exp(i eta lambda xi_{t lambda^{-2}})$$
i.e.
$$exp(-t psi(eta)) = exp(- t lambda^{-2} psi(lambda eta)).$$
Because of the uniqueness of the characteristic exponent $psi$ this implies
$$psi(eta) = lambda^{-2} psi(lambda eta) quad text{for all $lambda>0$, $eta in mathbb{R}$}. tag{3}$$
If we choose $eta:=1$ it follows that the limit $$d := lim_{lambda to 0} frac{psi(lambda)}{lambda^2}$$ exists and $$d = psi(1).$$
As
$$frac{psi(eta)}{eta^2} stackrel{(3)}{=} frac{psi(lambda eta)}{(lambda eta)^2} xrightarrow{ lambda to 0} d = psi(1)$$
we conclude that
$$psi(eta) = psi(1) eta^2, qquad eta in mathbb{R},$$
which means that $(xi_t)_{t geq 0}$ is a Brownian motion with scaling parameter $sigma geq 0$.
Remark: In the multidimensional framework, i.e. if $(xi_t)_{t geq 0}$ is a $d$-dimensional Lévy process, we can apply the above reasoning to each of the coordinates.
answered Jan 15 at 20:05
sazsaz
80.9k861127
$endgroup$
add a comment |
$begingroup$
Since $(xi_t)_{t geq 0}$ has, by assumption, stationary and independent increments, it suffices to show that $xi_t sim N(0,t)$ for any $t>0$. Denote by $psi$ the characteristic exponent of $(xi_t)_{t geq 0}$, i.e. $$mathbb{E}exp(i eta xi_t) = exp(-t psi(eta)), qquad t geq 0, eta in mathbb{R}. tag{1}$$ Fix $t>0$. As
$$xi_t sim lambda xi_{lambda^{-2}t}, qquad lambda>0 tag{2}$$
we have
$$mathbb{E}exp(i eta xi_t) = mathbb{E}exp(i eta lambda xi_{t lambda^{-2}})$$
i.e.
$$exp(-t psi(eta)) = exp(- t lambda^{-2} psi(lambda eta)).$$
Because of the uniqueness of the characteristic exponent $psi$ this implies
$$psi(eta) = lambda^{-2} psi(lambda eta) quad text{for all $lambda>0$, $eta in mathbb{R}$}. tag{3}$$
If we choose $eta:=1$ it follows that the limit $$d := lim_{lambda to 0} frac{psi(lambda)}{lambda^2}$$ exists and $$d = psi(1).$$
As
$$frac{psi(eta)}{eta^2} stackrel{(3)}{=} frac{psi(lambda eta)}{(lambda eta)^2} xrightarrow{ lambda to 0} d = psi(1)$$
we conclude that
$$psi(eta) = psi(1) eta^2, qquad eta in mathbb{R},$$
which means that $(xi_t)_{t geq 0}$ is a Brownian motion with scaling parameter $sigma geq 0$.
Remark: In the multidimensional framework, i.e. if $(xi_t)_{t geq 0}$ is a $d$-dimensional Lévy process, we can apply the above reasoning to each of the coordinates.
answered Jan 15 at 20:05
sazsaz
80.9k861127
$endgroup$
Since $(xi_t)_{t geq 0}$ has, by assumption, stationary and independent increments, it suffices to show that $xi_t sim N(0,t)$ for any $t>0$. Denote by $psi$ the characteristic exponent of $(xi_t)_{t geq 0}$, i.e. $$mathbb{E}exp(i eta xi_t) = exp(-t psi(eta)), qquad t geq 0, eta in mathbb{R}. tag{1}$$ Fix $t>0$. As
$$xi_t sim lambda xi_{lambda^{-2}t}, qquad lambda>0 tag{2}$$
we have
$$mathbb{E}exp(i eta xi_t) = mathbb{E}exp(i eta lambda xi_{t lambda^{-2}})$$
i.e.
$$exp(-t psi(eta)) = exp(- t lambda^{-2} psi(lambda eta)).$$
Because of the uniqueness of the characteristic exponent $psi$ this implies
$$psi(eta) = lambda^{-2} psi(lambda eta) quad text{for all $lambda>0$, $eta in mathbb{R}$}. tag{3}$$
If we choose $eta:=1$ it follows that the limit $$d := lim_{lambda to 0} frac{psi(lambda)}{lambda^2}$$ exists and $$d = psi(1).$$
As
$$frac{psi(eta)}{eta^2} stackrel{(3)}{=} frac{psi(lambda eta)}{(lambda eta)^2} xrightarrow{ lambda to 0} d = psi(1)$$
we conclude that
$$psi(eta) = psi(1) eta^2, qquad eta in mathbb{R},$$
which means that $(xi_t)_{t geq 0}$ is a Brownian motion with scaling parameter $sigma geq 0$.
Remark: In the multidimensional framework, i.e. if $(xi_t)_{t geq 0}$ is a $d$-dimensional Lévy process, we can apply the above reasoning to each of the coordinates.
answered Jan 15 at 20:05
sazsaz
80.9k861127
answered Jan 15 at 20:05
sazsaz
80.9k861127
answered Jan 15 at 20:05
sazsaz
80.9k861127
answered Jan 15 at 20:05
sazsaz
80.9k861127
80.9k861127
add a comment |
add a comment |
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$begingroup$
The only thing which you need to show is that $xi_t$ is Gaussian (with appropriate constants) for any $t>0$. To this end you can study the characteristic function of $xi_t$ using the fact that $xi_t$ is distributed as $lambda xi_{lambda^{-2} t}$.
$endgroup$
– saz
Jan 13 at 18:41
$begingroup$
If we call $phi_{xi_t}$ the characteristic function we have $phi_{xi_t}(s) = mathbb{E} [e^{is xi_t}]=mathbb{E}[e^{islambda xi_{lambda^{-2}t}}]=phi_{xi_{lambda^{-2}t}}(slambda)$, i don't understand how it should help...
$endgroup$
– Claudio Delfino
Jan 14 at 17:50