Lévy process + scaling property $implies$ Brownian motion












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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 $kin[0,infty)$?










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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


















1












$begingroup$


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 $kin[0,infty)$?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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 $kin[0,infty)$?










share|cite|improve this question











$endgroup$




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






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edited Jan 15 at 20:08









saz

80.9k861127




80.9k861127










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




















  • $begingroup$
    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












1 Answer
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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.






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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.






    share|cite|improve this answer









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      2












      $begingroup$

      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.






      share|cite|improve this answer









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        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 20:05









        sazsaz

        80.9k861127




        80.9k861127






























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