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







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












$begingroup$


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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048185%2fexpected-value-and-projection-of-a-normal-random-variable-onto-a-linear-span%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      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











      $endgroup$


















        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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048185%2fexpected-value-and-projection-of-a-normal-random-variable-onto-a-linear-span%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Image
                      0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                      Read more

                      Aardman Animations

                      Image
                      Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                      Read more

                      Are they similar matrix

                      Image
                      0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                      Read more