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
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          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













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          1 Answer
          1






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          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




















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