Isonormal Gaussian process

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I am reading Nualart's book The Malliavin Calculus and Related Topics and there is some issues that I am stuck wtih.



Let $H$ be a real separable Hilbert Space with. A stochastic process $W=W(h);;hin H$ defined in a complete probability space $(Omega,mathcalF,P)$ is an isonormal Gaussian process if $W$ is a centered Gaussian family of random variables such that $E(W(h)W(g))=langle h,grangle$ for all $g,hin H.$



The map $hmapsto W(h)$ is linear. Now Nualart said



  1. The mapping $hto W(h)$ provides a linear isometry of $H$ onoto a closed subspace of $L^2$ that will denote by $mathcalH_1.$

Question $1$: Why the map provides a linear isometry ?



Denote, now, by $mathcalG$ the $sigma-$alebra generated by the random variables $W(h);; hin H.$



  1. The set $exp(W(h)),hin H$ form a total subset of $L^2(Omega,mathcalG,P)$

The proof goes as follow, we take $Xin L^2$ such that $E(Xe^W(h))=0$ for all $hin H.$



By linearity of $hto W(h)$ we have $$Ebig( Xexp(sum_i=1^n t_iW(h_i))big)=0.quad (3)$$



This equation says that Laplace transform of the signed measure $Ebig(Xmathbb1_B(W(h_1),W(h_2),ldots,W(h_n))big)$ is identically zero on $BbbR^n.$



Then $E(Xmathbb1_G)=0$ for all $GinmathcalG$ so that $X=0.$



Quesstion $2$: I don't understand why linearity of $hto W(h)$ gives us equation $(3)$ and cannot write down that is the Laplace transform of of the signed measure $Ebig(Xmathbb1_B(W(h_1),W(h_2),ldots,W(h_n))big).$










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    I am reading Nualart's book The Malliavin Calculus and Related Topics and there is some issues that I am stuck wtih.



    Let $H$ be a real separable Hilbert Space with. A stochastic process $W=W(h);;hin H$ defined in a complete probability space $(Omega,mathcalF,P)$ is an isonormal Gaussian process if $W$ is a centered Gaussian family of random variables such that $E(W(h)W(g))=langle h,grangle$ for all $g,hin H.$



    The map $hmapsto W(h)$ is linear. Now Nualart said



    1. The mapping $hto W(h)$ provides a linear isometry of $H$ onoto a closed subspace of $L^2$ that will denote by $mathcalH_1.$

    Question $1$: Why the map provides a linear isometry ?



    Denote, now, by $mathcalG$ the $sigma-$alebra generated by the random variables $W(h);; hin H.$



    1. The set $exp(W(h)),hin H$ form a total subset of $L^2(Omega,mathcalG,P)$

    The proof goes as follow, we take $Xin L^2$ such that $E(Xe^W(h))=0$ for all $hin H.$



    By linearity of $hto W(h)$ we have $$Ebig( Xexp(sum_i=1^n t_iW(h_i))big)=0.quad (3)$$



    This equation says that Laplace transform of the signed measure $Ebig(Xmathbb1_B(W(h_1),W(h_2),ldots,W(h_n))big)$ is identically zero on $BbbR^n.$



    Then $E(Xmathbb1_G)=0$ for all $GinmathcalG$ so that $X=0.$



    Quesstion $2$: I don't understand why linearity of $hto W(h)$ gives us equation $(3)$ and cannot write down that is the Laplace transform of of the signed measure $Ebig(Xmathbb1_B(W(h_1),W(h_2),ldots,W(h_n))big).$










    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am reading Nualart's book The Malliavin Calculus and Related Topics and there is some issues that I am stuck wtih.



      Let $H$ be a real separable Hilbert Space with. A stochastic process $W=W(h);;hin H$ defined in a complete probability space $(Omega,mathcalF,P)$ is an isonormal Gaussian process if $W$ is a centered Gaussian family of random variables such that $E(W(h)W(g))=langle h,grangle$ for all $g,hin H.$



      The map $hmapsto W(h)$ is linear. Now Nualart said



      1. The mapping $hto W(h)$ provides a linear isometry of $H$ onoto a closed subspace of $L^2$ that will denote by $mathcalH_1.$

      Question $1$: Why the map provides a linear isometry ?



      Denote, now, by $mathcalG$ the $sigma-$alebra generated by the random variables $W(h);; hin H.$



      1. The set $exp(W(h)),hin H$ form a total subset of $L^2(Omega,mathcalG,P)$

      The proof goes as follow, we take $Xin L^2$ such that $E(Xe^W(h))=0$ for all $hin H.$



      By linearity of $hto W(h)$ we have $$Ebig( Xexp(sum_i=1^n t_iW(h_i))big)=0.quad (3)$$



      This equation says that Laplace transform of the signed measure $Ebig(Xmathbb1_B(W(h_1),W(h_2),ldots,W(h_n))big)$ is identically zero on $BbbR^n.$



      Then $E(Xmathbb1_G)=0$ for all $GinmathcalG$ so that $X=0.$



      Quesstion $2$: I don't understand why linearity of $hto W(h)$ gives us equation $(3)$ and cannot write down that is the Laplace transform of of the signed measure $Ebig(Xmathbb1_B(W(h_1),W(h_2),ldots,W(h_n))big).$










      share|cite|improve this question













      I am reading Nualart's book The Malliavin Calculus and Related Topics and there is some issues that I am stuck wtih.



      Let $H$ be a real separable Hilbert Space with. A stochastic process $W=W(h);;hin H$ defined in a complete probability space $(Omega,mathcalF,P)$ is an isonormal Gaussian process if $W$ is a centered Gaussian family of random variables such that $E(W(h)W(g))=langle h,grangle$ for all $g,hin H.$



      The map $hmapsto W(h)$ is linear. Now Nualart said



      1. The mapping $hto W(h)$ provides a linear isometry of $H$ onoto a closed subspace of $L^2$ that will denote by $mathcalH_1.$

      Question $1$: Why the map provides a linear isometry ?



      Denote, now, by $mathcalG$ the $sigma-$alebra generated by the random variables $W(h);; hin H.$



      1. The set $exp(W(h)),hin H$ form a total subset of $L^2(Omega,mathcalG,P)$

      The proof goes as follow, we take $Xin L^2$ such that $E(Xe^W(h))=0$ for all $hin H.$



      By linearity of $hto W(h)$ we have $$Ebig( Xexp(sum_i=1^n t_iW(h_i))big)=0.quad (3)$$



      This equation says that Laplace transform of the signed measure $Ebig(Xmathbb1_B(W(h_1),W(h_2),ldots,W(h_n))big)$ is identically zero on $BbbR^n.$



      Then $E(Xmathbb1_G)=0$ for all $GinmathcalG$ so that $X=0.$



      Quesstion $2$: I don't understand why linearity of $hto W(h)$ gives us equation $(3)$ and cannot write down that is the Laplace transform of of the signed measure $Ebig(Xmathbb1_B(W(h_1),W(h_2),ldots,W(h_n))big).$







      stochastic-processes normal-distribution stochastic-analysis






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      asked Sep 4 at 8:57









      TheVie

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          First question: $|W(g)|^2=langle W(g), W(g) rangle =langle g, g rangle =|g|^2$ so $|W(g_1)-W(g_2)|=|W(g_1-g_2)|=|g_1-g_2||$. This proves that $g to W(g)$ is an isometry. To prove (3) all you have to do is write $sum_i=1^n t_iW(h_i)$ as $W(h)$ with $h =sum_i=1^n t_i h_i$. For the last part you need the following fact: suppose $mu(B)=EXI_B(Z)$ for Borel sets $B$ in $mathbb R^n$ where $Z$ is a random vector with $n$ components. Then, for any non-negative measurable function $f$ we have $int f, dmu =EXf(Z)$. To prove this last equation you just have to note that it holds when $f$ is an indiactor function (by definition), hence for all simple functions $f$, hence for all non-negative measurable functions $f$.






          share|cite|improve this answer




















          • Thank you, so what is the closed subspace of $L^2$ for the isometry ? Ok for $(3)$ I didn't understand that we write $h$ as $sum t_ih_i:$ why this decomposition ? Thanks for the last fact.
            – TheVie
            Sep 4 at 9:21










          • You already know that $Xe^W(h)=0$ for all $h$ $(dagger)$. You are given $t_i$'s and $h_i$'s in 3). If you choose $h$ the way I mentioned and apply $(dagger)$ you will get 3).
            – Kavi Rama Murthy
            Sep 4 at 9:24










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          up vote
          2
          down vote













          First question: $|W(g)|^2=langle W(g), W(g) rangle =langle g, g rangle =|g|^2$ so $|W(g_1)-W(g_2)|=|W(g_1-g_2)|=|g_1-g_2||$. This proves that $g to W(g)$ is an isometry. To prove (3) all you have to do is write $sum_i=1^n t_iW(h_i)$ as $W(h)$ with $h =sum_i=1^n t_i h_i$. For the last part you need the following fact: suppose $mu(B)=EXI_B(Z)$ for Borel sets $B$ in $mathbb R^n$ where $Z$ is a random vector with $n$ components. Then, for any non-negative measurable function $f$ we have $int f, dmu =EXf(Z)$. To prove this last equation you just have to note that it holds when $f$ is an indiactor function (by definition), hence for all simple functions $f$, hence for all non-negative measurable functions $f$.






          share|cite|improve this answer




















          • Thank you, so what is the closed subspace of $L^2$ for the isometry ? Ok for $(3)$ I didn't understand that we write $h$ as $sum t_ih_i:$ why this decomposition ? Thanks for the last fact.
            – TheVie
            Sep 4 at 9:21










          • You already know that $Xe^W(h)=0$ for all $h$ $(dagger)$. You are given $t_i$'s and $h_i$'s in 3). If you choose $h$ the way I mentioned and apply $(dagger)$ you will get 3).
            – Kavi Rama Murthy
            Sep 4 at 9:24














          up vote
          2
          down vote













          First question: $|W(g)|^2=langle W(g), W(g) rangle =langle g, g rangle =|g|^2$ so $|W(g_1)-W(g_2)|=|W(g_1-g_2)|=|g_1-g_2||$. This proves that $g to W(g)$ is an isometry. To prove (3) all you have to do is write $sum_i=1^n t_iW(h_i)$ as $W(h)$ with $h =sum_i=1^n t_i h_i$. For the last part you need the following fact: suppose $mu(B)=EXI_B(Z)$ for Borel sets $B$ in $mathbb R^n$ where $Z$ is a random vector with $n$ components. Then, for any non-negative measurable function $f$ we have $int f, dmu =EXf(Z)$. To prove this last equation you just have to note that it holds when $f$ is an indiactor function (by definition), hence for all simple functions $f$, hence for all non-negative measurable functions $f$.






          share|cite|improve this answer




















          • Thank you, so what is the closed subspace of $L^2$ for the isometry ? Ok for $(3)$ I didn't understand that we write $h$ as $sum t_ih_i:$ why this decomposition ? Thanks for the last fact.
            – TheVie
            Sep 4 at 9:21










          • You already know that $Xe^W(h)=0$ for all $h$ $(dagger)$. You are given $t_i$'s and $h_i$'s in 3). If you choose $h$ the way I mentioned and apply $(dagger)$ you will get 3).
            – Kavi Rama Murthy
            Sep 4 at 9:24












          up vote
          2
          down vote










          up vote
          2
          down vote









          First question: $|W(g)|^2=langle W(g), W(g) rangle =langle g, g rangle =|g|^2$ so $|W(g_1)-W(g_2)|=|W(g_1-g_2)|=|g_1-g_2||$. This proves that $g to W(g)$ is an isometry. To prove (3) all you have to do is write $sum_i=1^n t_iW(h_i)$ as $W(h)$ with $h =sum_i=1^n t_i h_i$. For the last part you need the following fact: suppose $mu(B)=EXI_B(Z)$ for Borel sets $B$ in $mathbb R^n$ where $Z$ is a random vector with $n$ components. Then, for any non-negative measurable function $f$ we have $int f, dmu =EXf(Z)$. To prove this last equation you just have to note that it holds when $f$ is an indiactor function (by definition), hence for all simple functions $f$, hence for all non-negative measurable functions $f$.






          share|cite|improve this answer












          First question: $|W(g)|^2=langle W(g), W(g) rangle =langle g, g rangle =|g|^2$ so $|W(g_1)-W(g_2)|=|W(g_1-g_2)|=|g_1-g_2||$. This proves that $g to W(g)$ is an isometry. To prove (3) all you have to do is write $sum_i=1^n t_iW(h_i)$ as $W(h)$ with $h =sum_i=1^n t_i h_i$. For the last part you need the following fact: suppose $mu(B)=EXI_B(Z)$ for Borel sets $B$ in $mathbb R^n$ where $Z$ is a random vector with $n$ components. Then, for any non-negative measurable function $f$ we have $int f, dmu =EXf(Z)$. To prove this last equation you just have to note that it holds when $f$ is an indiactor function (by definition), hence for all simple functions $f$, hence for all non-negative measurable functions $f$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 4 at 9:15









          Kavi Rama Murthy

          26.1k31437




          26.1k31437











          • Thank you, so what is the closed subspace of $L^2$ for the isometry ? Ok for $(3)$ I didn't understand that we write $h$ as $sum t_ih_i:$ why this decomposition ? Thanks for the last fact.
            – TheVie
            Sep 4 at 9:21










          • You already know that $Xe^W(h)=0$ for all $h$ $(dagger)$. You are given $t_i$'s and $h_i$'s in 3). If you choose $h$ the way I mentioned and apply $(dagger)$ you will get 3).
            – Kavi Rama Murthy
            Sep 4 at 9:24
















          • Thank you, so what is the closed subspace of $L^2$ for the isometry ? Ok for $(3)$ I didn't understand that we write $h$ as $sum t_ih_i:$ why this decomposition ? Thanks for the last fact.
            – TheVie
            Sep 4 at 9:21










          • You already know that $Xe^W(h)=0$ for all $h$ $(dagger)$. You are given $t_i$'s and $h_i$'s in 3). If you choose $h$ the way I mentioned and apply $(dagger)$ you will get 3).
            – Kavi Rama Murthy
            Sep 4 at 9:24















          Thank you, so what is the closed subspace of $L^2$ for the isometry ? Ok for $(3)$ I didn't understand that we write $h$ as $sum t_ih_i:$ why this decomposition ? Thanks for the last fact.
          – TheVie
          Sep 4 at 9:21




          Thank you, so what is the closed subspace of $L^2$ for the isometry ? Ok for $(3)$ I didn't understand that we write $h$ as $sum t_ih_i:$ why this decomposition ? Thanks for the last fact.
          – TheVie
          Sep 4 at 9:21












          You already know that $Xe^W(h)=0$ for all $h$ $(dagger)$. You are given $t_i$'s and $h_i$'s in 3). If you choose $h$ the way I mentioned and apply $(dagger)$ you will get 3).
          – Kavi Rama Murthy
          Sep 4 at 9:24




          You already know that $Xe^W(h)=0$ for all $h$ $(dagger)$. You are given $t_i$'s and $h_i$'s in 3). If you choose $h$ the way I mentioned and apply $(dagger)$ you will get 3).
          – Kavi Rama Murthy
          Sep 4 at 9:24

















           

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