Covariance in and input-output filter [Stationary Stochastic Processes]












0












$begingroup$



The weakly stationary processes $X_t$, $;t=0$, $,pm$$1$,
$,pm$$2$,$,ldots$ and $Y_t$,$;$ $t=0$, $,pm$$1$, $,pm$$2$,$ldots$
are input and output of a linear filter according to



$Y_t;$$+;0.5Y_{t-1};$$;=;$$X_t$, $quad$ for $t=0$, $,pm$$1$,
$,pm$$2$,$ldots$



The process $X_t,$ has the covariance function $r_X(0)=1$,$,$
$r_X(pm2)=0.2$,$,$ and zero for all other values. Determine the
cross-covariance funtion



$r_{X,Y}(tau)=C[X_t,,Y_{t+tau}]$



for all values of $tau$.




I began by assigning the impulse function h(x) the following values:



$h(0) = 1qquad$
$h(1) = 0.5$



Does it makes sense?






statistics stochastic-processes covariance filters







share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    The weakly stationary processes $X_t$, $;t=0$, $,pm$$1$,
    $,pm$$2$,$,ldots$ and $Y_t$,$;$ $t=0$, $,pm$$1$, $,pm$$2$,$ldots$
    are input and output of a linear filter according to



    $Y_t;$$+;0.5Y_{t-1};$$;=;$$X_t$, $quad$ for $t=0$, $,pm$$1$,
    $,pm$$2$,$ldots$



    The process $X_t,$ has the covariance function $r_X(0)=1$,$,$
    $r_X(pm2)=0.2$,$,$ and zero for all other values. Determine the
    cross-covariance funtion



    $r_{X,Y}(tau)=C[X_t,,Y_{t+tau}]$



    for all values of $tau$.




    I began by assigning the impulse function h(x) the following values:



    $h(0) = 1qquad$
    $h(1) = 0.5$



    Does it makes sense?






    statistics stochastic-processes covariance filters







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      The weakly stationary processes $X_t$, $;t=0$, $,pm$$1$,
      $,pm$$2$,$,ldots$ and $Y_t$,$;$ $t=0$, $,pm$$1$, $,pm$$2$,$ldots$
      are input and output of a linear filter according to



      $Y_t;$$+;0.5Y_{t-1};$$;=;$$X_t$, $quad$ for $t=0$, $,pm$$1$,
      $,pm$$2$,$ldots$



      The process $X_t,$ has the covariance function $r_X(0)=1$,$,$
      $r_X(pm2)=0.2$,$,$ and zero for all other values. Determine the
      cross-covariance funtion



      $r_{X,Y}(tau)=C[X_t,,Y_{t+tau}]$



      for all values of $tau$.




      I began by assigning the impulse function h(x) the following values:



      $h(0) = 1qquad$
      $h(1) = 0.5$



      Does it makes sense?






      statistics stochastic-processes covariance filters







      share|cite|improve this question









      $endgroup$





      The weakly stationary processes $X_t$, $;t=0$, $,pm$$1$,
      $,pm$$2$,$,ldots$ and $Y_t$,$;$ $t=0$, $,pm$$1$, $,pm$$2$,$ldots$
      are input and output of a linear filter according to



      $Y_t;$$+;0.5Y_{t-1};$$;=;$$X_t$, $quad$ for $t=0$, $,pm$$1$,
      $,pm$$2$,$ldots$



      The process $X_t,$ has the covariance function $r_X(0)=1$,$,$
      $r_X(pm2)=0.2$,$,$ and zero for all other values. Determine the
      cross-covariance funtion



      $r_{X,Y}(tau)=C[X_t,,Y_{t+tau}]$



      for all values of $tau$.




      I began by assigning the impulse function h(x) the following values:



      $h(0) = 1qquad$
      $h(1) = 0.5$



      Does it makes sense?






      statistics stochastic-processes covariance filters




      statistics stochastic-processes covariance filters






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 17:24









      LowEnergyLowEnergy

      11




      11






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.



          We can then use the relation
          $$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
          which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
          $$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
            $endgroup$
            – LowEnergy
            Jan 7 at 18:39












          • $begingroup$
            The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
            $endgroup$
            – rzch
            Jan 7 at 18:42












          • $begingroup$
            I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
            $endgroup$
            – LowEnergy
            Jan 7 at 18:52














          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%2f3065249%2fcovariance-in-and-input-output-filter-stationary-stochastic-processes%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.



          We can then use the relation
          $$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
          which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
          $$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
            $endgroup$
            – LowEnergy
            Jan 7 at 18:39












          • $begingroup$
            The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
            $endgroup$
            – rzch
            Jan 7 at 18:42












          • $begingroup$
            I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
            $endgroup$
            – LowEnergy
            Jan 7 at 18:52


















          0












          $begingroup$

          The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.



          We can then use the relation
          $$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
          which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
          $$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
            $endgroup$
            – LowEnergy
            Jan 7 at 18:39












          • $begingroup$
            The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
            $endgroup$
            – rzch
            Jan 7 at 18:42












          • $begingroup$
            I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
            $endgroup$
            – LowEnergy
            Jan 7 at 18:52
















          0












          0








          0





          $begingroup$

          The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.



          We can then use the relation
          $$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
          which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
          $$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$






          share|cite|improve this answer









          $endgroup$



          The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.



          We can then use the relation
          $$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
          which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
          $$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 18:19









          rzchrzch

          1765




          1765












          • $begingroup$
            I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
            $endgroup$
            – LowEnergy
            Jan 7 at 18:39












          • $begingroup$
            The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
            $endgroup$
            – rzch
            Jan 7 at 18:42












          • $begingroup$
            I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
            $endgroup$
            – LowEnergy
            Jan 7 at 18:52




















          • $begingroup$
            I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
            $endgroup$
            – LowEnergy
            Jan 7 at 18:39












          • $begingroup$
            The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
            $endgroup$
            – rzch
            Jan 7 at 18:42












          • $begingroup$
            I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
            $endgroup$
            – LowEnergy
            Jan 7 at 18:52


















          $begingroup$
          I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
          $endgroup$
          – LowEnergy
          Jan 7 at 18:39






          $begingroup$
          I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
          $endgroup$
          – LowEnergy
          Jan 7 at 18:39














          $begingroup$
          The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
          $endgroup$
          – rzch
          Jan 7 at 18:42






          $begingroup$
          The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
          $endgroup$
          – rzch
          Jan 7 at 18:42














          $begingroup$
          I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
          $endgroup$
          – LowEnergy
          Jan 7 at 18:52






          $begingroup$
          I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
          $endgroup$
          – LowEnergy
          Jan 7 at 18:52




















          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%2f3065249%2fcovariance-in-and-input-output-filter-stationary-stochastic-processes%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more