Expected value and projection of a normal random variable onto a linear span…?












2












$begingroup$


I just wanted to clarify a part of a proof which used the fact a random variable has zero mean.



Suppose $X, Z_{s_1},dots,Z_{s_n}$ are all jointly normal random variables for all $s_i leq t$ and $n geq 1$.



Define $mathcal{L}(Z,t)$ as set of random variables of the form



$$c_1 Z_{s_1}+ dots + c_n Z_{s_n}$$



for $c_i in mathbb{R}$ and $s_i leq t$, $n geq 1$, along with all their limit points (in $L^2(P)$).



Define $tilde{X} = X - mathcal{P}_L (X)$ where $mathcal{P}_L$ is the projection onto the space $mathcal{L}(Z,t)$.



Then the proof use $mathbb{E}[tilde{X}] = 0$.



My understanding of this, is that the projection $mathcal{P}_L (X)$ coincides with the conditional expectation



$$mathcal{P}_L (X) = E[X|mathcal{F}_L]$$



where $mathcal{F}_L$ is the $sigma$-field generated by the random variables in $mathcal{L}(Z,t)$. And hence by law of total expectation we have



$$mathbb{E}[tilde{X}] = mathbb{E}[X] - mathbb{E}[E[X|mathcal{F}_L]] = mathbb{E}[X]-mathbb{E}[X] =0$$



Is this correct? The proof doesn't really explain, and the relationship between projection and probability spaces is still a bit confusing to me.






geometry probability-theory projective-geometry







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$endgroup$












  • $begingroup$
    What proof is this? What source?
    $endgroup$
    – Zachary Selk
    Dec 21 '18 at 5:02












  • $begingroup$
    It is part of Oksander's SDE book lemma 6.2.2
    $endgroup$
    – Xiaomi
    Dec 21 '18 at 5:11
















2












$begingroup$


I just wanted to clarify a part of a proof which used the fact a random variable has zero mean.



Suppose $X, Z_{s_1},dots,Z_{s_n}$ are all jointly normal random variables for all $s_i leq t$ and $n geq 1$.



Define $mathcal{L}(Z,t)$ as set of random variables of the form



$$c_1 Z_{s_1}+ dots + c_n Z_{s_n}$$



for $c_i in mathbb{R}$ and $s_i leq t$, $n geq 1$, along with all their limit points (in $L^2(P)$).



Define $tilde{X} = X - mathcal{P}_L (X)$ where $mathcal{P}_L$ is the projection onto the space $mathcal{L}(Z,t)$.



Then the proof use $mathbb{E}[tilde{X}] = 0$.



My understanding of this, is that the projection $mathcal{P}_L (X)$ coincides with the conditional expectation



$$mathcal{P}_L (X) = E[X|mathcal{F}_L]$$



where $mathcal{F}_L$ is the $sigma$-field generated by the random variables in $mathcal{L}(Z,t)$. And hence by law of total expectation we have



$$mathbb{E}[tilde{X}] = mathbb{E}[X] - mathbb{E}[E[X|mathcal{F}_L]] = mathbb{E}[X]-mathbb{E}[X] =0$$



Is this correct? The proof doesn't really explain, and the relationship between projection and probability spaces is still a bit confusing to me.






geometry probability-theory projective-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    What proof is this? What source?
    $endgroup$
    – Zachary Selk
    Dec 21 '18 at 5:02












  • $begingroup$
    It is part of Oksander's SDE book lemma 6.2.2
    $endgroup$
    – Xiaomi
    Dec 21 '18 at 5:11














2












2








2





$begingroup$


I just wanted to clarify a part of a proof which used the fact a random variable has zero mean.



Suppose $X, Z_{s_1},dots,Z_{s_n}$ are all jointly normal random variables for all $s_i leq t$ and $n geq 1$.



Define $mathcal{L}(Z,t)$ as set of random variables of the form



$$c_1 Z_{s_1}+ dots + c_n Z_{s_n}$$



for $c_i in mathbb{R}$ and $s_i leq t$, $n geq 1$, along with all their limit points (in $L^2(P)$).



Define $tilde{X} = X - mathcal{P}_L (X)$ where $mathcal{P}_L$ is the projection onto the space $mathcal{L}(Z,t)$.



Then the proof use $mathbb{E}[tilde{X}] = 0$.



My understanding of this, is that the projection $mathcal{P}_L (X)$ coincides with the conditional expectation



$$mathcal{P}_L (X) = E[X|mathcal{F}_L]$$



where $mathcal{F}_L$ is the $sigma$-field generated by the random variables in $mathcal{L}(Z,t)$. And hence by law of total expectation we have



$$mathbb{E}[tilde{X}] = mathbb{E}[X] - mathbb{E}[E[X|mathcal{F}_L]] = mathbb{E}[X]-mathbb{E}[X] =0$$



Is this correct? The proof doesn't really explain, and the relationship between projection and probability spaces is still a bit confusing to me.






geometry probability-theory projective-geometry







share|cite|improve this question











$endgroup$




I just wanted to clarify a part of a proof which used the fact a random variable has zero mean.



Suppose $X, Z_{s_1},dots,Z_{s_n}$ are all jointly normal random variables for all $s_i leq t$ and $n geq 1$.



Define $mathcal{L}(Z,t)$ as set of random variables of the form



$$c_1 Z_{s_1}+ dots + c_n Z_{s_n}$$



for $c_i in mathbb{R}$ and $s_i leq t$, $n geq 1$, along with all their limit points (in $L^2(P)$).



Define $tilde{X} = X - mathcal{P}_L (X)$ where $mathcal{P}_L$ is the projection onto the space $mathcal{L}(Z,t)$.



Then the proof use $mathbb{E}[tilde{X}] = 0$.



My understanding of this, is that the projection $mathcal{P}_L (X)$ coincides with the conditional expectation



$$mathcal{P}_L (X) = E[X|mathcal{F}_L]$$



where $mathcal{F}_L$ is the $sigma$-field generated by the random variables in $mathcal{L}(Z,t)$. And hence by law of total expectation we have



$$mathbb{E}[tilde{X}] = mathbb{E}[X] - mathbb{E}[E[X|mathcal{F}_L]] = mathbb{E}[X]-mathbb{E}[X] =0$$



Is this correct? The proof doesn't really explain, and the relationship between projection and probability spaces is still a bit confusing to me.






geometry probability-theory projective-geometry




geometry probability-theory projective-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 5:59







Xiaomi

















asked Dec 21 '18 at 4:54









XiaomiXiaomi

1,066115




1,066115












  • $begingroup$
    What proof is this? What source?
    $endgroup$
    – Zachary Selk
    Dec 21 '18 at 5:02












  • $begingroup$
    It is part of Oksander's SDE book lemma 6.2.2
    $endgroup$
    – Xiaomi
    Dec 21 '18 at 5:11


















  • $begingroup$
    What proof is this? What source?
    $endgroup$
    – Zachary Selk
    Dec 21 '18 at 5:02












  • $begingroup$
    It is part of Oksander's SDE book lemma 6.2.2
    $endgroup$
    – Xiaomi
    Dec 21 '18 at 5:11
















$begingroup$
What proof is this? What source?
$endgroup$
– Zachary Selk
Dec 21 '18 at 5:02






$begingroup$
What proof is this? What source?
$endgroup$
– Zachary Selk
Dec 21 '18 at 5:02














$begingroup$
It is part of Oksander's SDE book lemma 6.2.2
$endgroup$
– Xiaomi
Dec 21 '18 at 5:11




$begingroup$
It is part of Oksander's SDE book lemma 6.2.2
$endgroup$
– Xiaomi
Dec 21 '18 at 5:11










2 Answers
2






active

oldest

votes


















1












$begingroup$

As @Kavi Rama Murthy pointed out, it holds that
$$
mathcal{L}(Z,t) neq L^2(sigma(Z_s;sleq t)).
$$ But if we assume additionally that $$(X,Z_s;sleq t)$$
is jointly normally distributed, it holds that
$$
P_{mathcal{L}(Z,t) }X = P_{L^2(sigma(Z_s;sleq t))}X.
$$ To see this, notice that if $(X,Y_1,ldots ,Y_n)$ is jointly normal, then
$$
operatorname{Cov}(X,Y_i) = 0,,forall ileq n Leftrightarrow Xtext{ and }(Y_i)_{ileq n} text{ are independent.}
$$ It holds that
$$
operatorname{Cov}(X-P_{mathcal{L}(Z,t) }X, Z_s) = 0,quadforall sleq t,
$$ by the definition of the projection. And since $(X,Z_s;sleq t)$ is jointly normal, so is $(X-P_{mathcal{L}(Z,t) }X,Z_s;sleq t).$ This implies that
$$
X-P_{mathcal{L}(Z,t) }X perp !!! perp (Z_s)_{sleq t}.
$$ Therefore, we have
$$
P_{L^2(sigma(Z_s;sleq t))}[X-P_{mathcal{L}(Z,t) }X]=P_{L^2(sigma(Z_s;sleq t))}X-P_{mathcal{L}(Z,t) }X=0,
$$as desired. In your notation, $mathcal{F}_L=sigma(Z_s;sleq t)$ and $P_{L^2(sigma(Z_s;sleq t))}X = E[X|mathcal{F}_L].$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Your argument is not valid. Conditional expectation is the projection on the space of all random variables measurable w.r.t. $sigma {Z_s:sleq t}$, not the projection on to the vector space spanned by $ {Z_s:sleq t}$. The former space contains many nonlinear functions of $ {Z_s:sleq t}$ like $Z_t^{2}$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      1












      $begingroup$

      As @Kavi Rama Murthy pointed out, it holds that
      $$
      mathcal{L}(Z,t) neq L^2(sigma(Z_s;sleq t)).
      $$       But if we assume additionally that $$(X,Z_s;sleq t)$$
      is jointly normally distributed, it holds that
      $$
      P_{mathcal{L}(Z,t) }X = P_{L^2(sigma(Z_s;sleq t))}X.
      $$       To see this, notice that if $(X,Y_1,ldots ,Y_n)$ is jointly normal, then
      $$
      operatorname{Cov}(X,Y_i) = 0,,forall ileq n Leftrightarrow Xtext{ and }(Y_i)_{ileq n} text{ are independent.}
      $$       It holds that
      $$
      operatorname{Cov}(X-P_{mathcal{L}(Z,t) }X, Z_s) = 0,quadforall sleq t,
      $$       by the definition of the projection. And since $(X,Z_s;sleq t)$ is jointly normal, so is $(X-P_{mathcal{L}(Z,t) }X,Z_s;sleq t).$ This implies that
      $$
      X-P_{mathcal{L}(Z,t) }X perp !!! perp (Z_s)_{sleq t}.
      $$       Therefore, we have
      $$
      P_{L^2(sigma(Z_s;sleq t))}[X-P_{mathcal{L}(Z,t) }X]=P_{L^2(sigma(Z_s;sleq t))}X-P_{mathcal{L}(Z,t) }X=0,
      $$      as desired. In your notation, $mathcal{F}_L=sigma(Z_s;sleq t)$ and $P_{L^2(sigma(Z_s;sleq t))}X = E[X|mathcal{F}_L].$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        As @Kavi Rama Murthy pointed out, it holds that
        $$
        mathcal{L}(Z,t) neq L^2(sigma(Z_s;sleq t)).
        $$         But if we assume additionally that $$(X,Z_s;sleq t)$$
        is jointly normally distributed, it holds that
        $$
        P_{mathcal{L}(Z,t) }X = P_{L^2(sigma(Z_s;sleq t))}X.
        $$         To see this, notice that if $(X,Y_1,ldots ,Y_n)$ is jointly normal, then
        $$
        operatorname{Cov}(X,Y_i) = 0,,forall ileq n Leftrightarrow Xtext{ and }(Y_i)_{ileq n} text{ are independent.}
        $$         It holds that
        $$
        operatorname{Cov}(X-P_{mathcal{L}(Z,t) }X, Z_s) = 0,quadforall sleq t,
        $$         by the definition of the projection. And since $(X,Z_s;sleq t)$ is jointly normal, so is $(X-P_{mathcal{L}(Z,t) }X,Z_s;sleq t).$ This implies that
        $$
        X-P_{mathcal{L}(Z,t) }X perp !!! perp (Z_s)_{sleq t}.
        $$         Therefore, we have
        $$
        P_{L^2(sigma(Z_s;sleq t))}[X-P_{mathcal{L}(Z,t) }X]=P_{L^2(sigma(Z_s;sleq t))}X-P_{mathcal{L}(Z,t) }X=0,
        $$        as desired. In your notation, $mathcal{F}_L=sigma(Z_s;sleq t)$ and $P_{L^2(sigma(Z_s;sleq t))}X = E[X|mathcal{F}_L].$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          As @Kavi Rama Murthy pointed out, it holds that
          $$
          mathcal{L}(Z,t) neq L^2(sigma(Z_s;sleq t)).
          $$           But if we assume additionally that $$(X,Z_s;sleq t)$$
          is jointly normally distributed, it holds that
          $$
          P_{mathcal{L}(Z,t) }X = P_{L^2(sigma(Z_s;sleq t))}X.
          $$           To see this, notice that if $(X,Y_1,ldots ,Y_n)$ is jointly normal, then
          $$
          operatorname{Cov}(X,Y_i) = 0,,forall ileq n Leftrightarrow Xtext{ and }(Y_i)_{ileq n} text{ are independent.}
          $$           It holds that
          $$
          operatorname{Cov}(X-P_{mathcal{L}(Z,t) }X, Z_s) = 0,quadforall sleq t,
          $$           by the definition of the projection. And since $(X,Z_s;sleq t)$ is jointly normal, so is $(X-P_{mathcal{L}(Z,t) }X,Z_s;sleq t).$ This implies that
          $$
          X-P_{mathcal{L}(Z,t) }X perp !!! perp (Z_s)_{sleq t}.
          $$           Therefore, we have
          $$
          P_{L^2(sigma(Z_s;sleq t))}[X-P_{mathcal{L}(Z,t) }X]=P_{L^2(sigma(Z_s;sleq t))}X-P_{mathcal{L}(Z,t) }X=0,
          $$          as desired. In your notation, $mathcal{F}_L=sigma(Z_s;sleq t)$ and $P_{L^2(sigma(Z_s;sleq t))}X = E[X|mathcal{F}_L].$






          share|cite|improve this answer











          $endgroup$



          As @Kavi Rama Murthy pointed out, it holds that
          $$
          mathcal{L}(Z,t) neq L^2(sigma(Z_s;sleq t)).
          $$           But if we assume additionally that $$(X,Z_s;sleq t)$$
          is jointly normally distributed, it holds that
          $$
          P_{mathcal{L}(Z,t) }X = P_{L^2(sigma(Z_s;sleq t))}X.
          $$           To see this, notice that if $(X,Y_1,ldots ,Y_n)$ is jointly normal, then
          $$
          operatorname{Cov}(X,Y_i) = 0,,forall ileq n Leftrightarrow Xtext{ and }(Y_i)_{ileq n} text{ are independent.}
          $$           It holds that
          $$
          operatorname{Cov}(X-P_{mathcal{L}(Z,t) }X, Z_s) = 0,quadforall sleq t,
          $$           by the definition of the projection. And since $(X,Z_s;sleq t)$ is jointly normal, so is $(X-P_{mathcal{L}(Z,t) }X,Z_s;sleq t).$ This implies that
          $$
          X-P_{mathcal{L}(Z,t) }X perp !!! perp (Z_s)_{sleq t}.
          $$           Therefore, we have
          $$
          P_{L^2(sigma(Z_s;sleq t))}[X-P_{mathcal{L}(Z,t) }X]=P_{L^2(sigma(Z_s;sleq t))}X-P_{mathcal{L}(Z,t) }X=0,
          $$          as desired. In your notation, $mathcal{F}_L=sigma(Z_s;sleq t)$ and $P_{L^2(sigma(Z_s;sleq t))}X = E[X|mathcal{F}_L].$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 21 '18 at 6:15

























          answered Dec 21 '18 at 6:08









          SongSong

          16.7k1941




          16.7k1941























              2












              $begingroup$

              Your argument is not valid. Conditional expectation is the projection on the space of all random variables measurable w.r.t. $sigma {Z_s:sleq t}$, not the projection on to the vector space spanned by $ {Z_s:sleq t}$. The former space contains many nonlinear functions of $ {Z_s:sleq t}$ like $Z_t^{2}$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Your argument is not valid. Conditional expectation is the projection on the space of all random variables measurable w.r.t. $sigma {Z_s:sleq t}$, not the projection on to the vector space spanned by $ {Z_s:sleq t}$. The former space contains many nonlinear functions of $ {Z_s:sleq t}$ like $Z_t^{2}$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Your argument is not valid. Conditional expectation is the projection on the space of all random variables measurable w.r.t. $sigma {Z_s:sleq t}$, not the projection on to the vector space spanned by $ {Z_s:sleq t}$. The former space contains many nonlinear functions of $ {Z_s:sleq t}$ like $Z_t^{2}$.






                  share|cite|improve this answer









                  $endgroup$



                  Your argument is not valid. Conditional expectation is the projection on the space of all random variables measurable w.r.t. $sigma {Z_s:sleq t}$, not the projection on to the vector space spanned by $ {Z_s:sleq t}$. The former space contains many nonlinear functions of $ {Z_s:sleq t}$ like $Z_t^{2}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 5:29









                  Kavi Rama MurthyKavi Rama Murthy

                  64.7k42766




                  64.7k42766






























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