Correlation coefficient of a WSS process












0












$begingroup$


Could someone please tell me why the correlation coefficient of a WSS process is :



$$rho_{XX}(tau) = frac{operatorname{K}_{XX}(tau)}{sigma^2} = frac{operatorname{E}[(X_t - mu)(X_{t+tau} - mu)]}{sigma^2}$$



Instead of :



$$rho_{XX}(tau) = frac{operatorname{K}_{XX}(tau)}{sigma_{t}sigma_{t+tau}} = frac{operatorname{E}[(X_t - mu)(X_{t+tau} - mu)]}{sigma_{t}sigma_{t+tau}} $$



?



Where :$~~~sigma_{t}^2=sigma^2=operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)]$ .



Stated in another way , does a WSS process imply that :



$$operatorname{E}[X^2(t+tau)]=operatorname{E}[X^2(t)]$$ ?



(https://en.wikipedia.org/wiki/Autocovariance)






probability







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Could someone please tell me why the correlation coefficient of a WSS process is :



    $$rho_{XX}(tau) = frac{operatorname{K}_{XX}(tau)}{sigma^2} = frac{operatorname{E}[(X_t - mu)(X_{t+tau} - mu)]}{sigma^2}$$



    Instead of :



    $$rho_{XX}(tau) = frac{operatorname{K}_{XX}(tau)}{sigma_{t}sigma_{t+tau}} = frac{operatorname{E}[(X_t - mu)(X_{t+tau} - mu)]}{sigma_{t}sigma_{t+tau}} $$



    ?



    Where :$~~~sigma_{t}^2=sigma^2=operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)]$ .



    Stated in another way , does a WSS process imply that :



    $$operatorname{E}[X^2(t+tau)]=operatorname{E}[X^2(t)]$$ ?



    (https://en.wikipedia.org/wiki/Autocovariance)






    probability







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Could someone please tell me why the correlation coefficient of a WSS process is :



      $$rho_{XX}(tau) = frac{operatorname{K}_{XX}(tau)}{sigma^2} = frac{operatorname{E}[(X_t - mu)(X_{t+tau} - mu)]}{sigma^2}$$



      Instead of :



      $$rho_{XX}(tau) = frac{operatorname{K}_{XX}(tau)}{sigma_{t}sigma_{t+tau}} = frac{operatorname{E}[(X_t - mu)(X_{t+tau} - mu)]}{sigma_{t}sigma_{t+tau}} $$



      ?



      Where :$~~~sigma_{t}^2=sigma^2=operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)]$ .



      Stated in another way , does a WSS process imply that :



      $$operatorname{E}[X^2(t+tau)]=operatorname{E}[X^2(t)]$$ ?



      (https://en.wikipedia.org/wiki/Autocovariance)






      probability







      share|cite|improve this question









      $endgroup$




      Could someone please tell me why the correlation coefficient of a WSS process is :



      $$rho_{XX}(tau) = frac{operatorname{K}_{XX}(tau)}{sigma^2} = frac{operatorname{E}[(X_t - mu)(X_{t+tau} - mu)]}{sigma^2}$$



      Instead of :



      $$rho_{XX}(tau) = frac{operatorname{K}_{XX}(tau)}{sigma_{t}sigma_{t+tau}} = frac{operatorname{E}[(X_t - mu)(X_{t+tau} - mu)]}{sigma_{t}sigma_{t+tau}} $$



      ?



      Where :$~~~sigma_{t}^2=sigma^2=operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)]$ .



      Stated in another way , does a WSS process imply that :



      $$operatorname{E}[X^2(t+tau)]=operatorname{E}[X^2(t)]$$ ?



      (https://en.wikipedia.org/wiki/Autocovariance)






      probability




      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 16:37









      HilbertHilbert

      1649




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          Weak stationary implies that for all fixed $tau$, the function $tmapsto mathbb Eleft[X_{t+tau}X_tright]$ is constant. In particular, choosing $tau=0$ shows that $mathbb Eleft[X_{t+0}X_tright]=mathbb Eleft[X_t^2right]=mathbb Eleft[X_0^2right]$. The other part of the definition of weakly stationary process implies that the expectation of $X_t$ does not depend on $t$ hence we can safely define $sigma^2:=operatorname{Var}left(X_tright)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I do understand that part but the problem is why they've replaced $sigma_{t}sigma_{t+tau}$ by $sigma^2_{t}$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:21










          • $begingroup$
            Because $sigma_{t+tau}=sigma_{t}$.
            $endgroup$
            – Davide Giraudo
            Jan 4 at 17:24










          • $begingroup$
            This is the very relation that I am trying to prove/understand , how is this equality justified : $$sigma(t+tau) =(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t+tau)])^frac{1}{2}~=~(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t)])^frac{1}{2}~ =? ~(operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)])^frac{1}{2} $$ So basically why : $$operatorname{E}[X^2(t)]~ =~operatorname{E}[X^2(t+tau)]$$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:33













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          $begingroup$

          Weak stationary implies that for all fixed $tau$, the function $tmapsto mathbb Eleft[X_{t+tau}X_tright]$ is constant. In particular, choosing $tau=0$ shows that $mathbb Eleft[X_{t+0}X_tright]=mathbb Eleft[X_t^2right]=mathbb Eleft[X_0^2right]$. The other part of the definition of weakly stationary process implies that the expectation of $X_t$ does not depend on $t$ hence we can safely define $sigma^2:=operatorname{Var}left(X_tright)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I do understand that part but the problem is why they've replaced $sigma_{t}sigma_{t+tau}$ by $sigma^2_{t}$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:21










          • $begingroup$
            Because $sigma_{t+tau}=sigma_{t}$.
            $endgroup$
            – Davide Giraudo
            Jan 4 at 17:24










          • $begingroup$
            This is the very relation that I am trying to prove/understand , how is this equality justified : $$sigma(t+tau) =(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t+tau)])^frac{1}{2}~=~(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t)])^frac{1}{2}~ =? ~(operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)])^frac{1}{2} $$ So basically why : $$operatorname{E}[X^2(t)]~ =~operatorname{E}[X^2(t+tau)]$$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:33


















          0












          $begingroup$

          Weak stationary implies that for all fixed $tau$, the function $tmapsto mathbb Eleft[X_{t+tau}X_tright]$ is constant. In particular, choosing $tau=0$ shows that $mathbb Eleft[X_{t+0}X_tright]=mathbb Eleft[X_t^2right]=mathbb Eleft[X_0^2right]$. The other part of the definition of weakly stationary process implies that the expectation of $X_t$ does not depend on $t$ hence we can safely define $sigma^2:=operatorname{Var}left(X_tright)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I do understand that part but the problem is why they've replaced $sigma_{t}sigma_{t+tau}$ by $sigma^2_{t}$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:21










          • $begingroup$
            Because $sigma_{t+tau}=sigma_{t}$.
            $endgroup$
            – Davide Giraudo
            Jan 4 at 17:24










          • $begingroup$
            This is the very relation that I am trying to prove/understand , how is this equality justified : $$sigma(t+tau) =(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t+tau)])^frac{1}{2}~=~(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t)])^frac{1}{2}~ =? ~(operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)])^frac{1}{2} $$ So basically why : $$operatorname{E}[X^2(t)]~ =~operatorname{E}[X^2(t+tau)]$$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:33
















          0












          0








          0





          $begingroup$

          Weak stationary implies that for all fixed $tau$, the function $tmapsto mathbb Eleft[X_{t+tau}X_tright]$ is constant. In particular, choosing $tau=0$ shows that $mathbb Eleft[X_{t+0}X_tright]=mathbb Eleft[X_t^2right]=mathbb Eleft[X_0^2right]$. The other part of the definition of weakly stationary process implies that the expectation of $X_t$ does not depend on $t$ hence we can safely define $sigma^2:=operatorname{Var}left(X_tright)$.






          share|cite|improve this answer









          $endgroup$



          Weak stationary implies that for all fixed $tau$, the function $tmapsto mathbb Eleft[X_{t+tau}X_tright]$ is constant. In particular, choosing $tau=0$ shows that $mathbb Eleft[X_{t+0}X_tright]=mathbb Eleft[X_t^2right]=mathbb Eleft[X_0^2right]$. The other part of the definition of weakly stationary process implies that the expectation of $X_t$ does not depend on $t$ hence we can safely define $sigma^2:=operatorname{Var}left(X_tright)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 16:50









          Davide GiraudoDavide Giraudo

          127k17154268




          127k17154268












          • $begingroup$
            I do understand that part but the problem is why they've replaced $sigma_{t}sigma_{t+tau}$ by $sigma^2_{t}$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:21










          • $begingroup$
            Because $sigma_{t+tau}=sigma_{t}$.
            $endgroup$
            – Davide Giraudo
            Jan 4 at 17:24










          • $begingroup$
            This is the very relation that I am trying to prove/understand , how is this equality justified : $$sigma(t+tau) =(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t+tau)])^frac{1}{2}~=~(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t)])^frac{1}{2}~ =? ~(operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)])^frac{1}{2} $$ So basically why : $$operatorname{E}[X^2(t)]~ =~operatorname{E}[X^2(t+tau)]$$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:33




















          • $begingroup$
            I do understand that part but the problem is why they've replaced $sigma_{t}sigma_{t+tau}$ by $sigma^2_{t}$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:21










          • $begingroup$
            Because $sigma_{t+tau}=sigma_{t}$.
            $endgroup$
            – Davide Giraudo
            Jan 4 at 17:24










          • $begingroup$
            This is the very relation that I am trying to prove/understand , how is this equality justified : $$sigma(t+tau) =(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t+tau)])^frac{1}{2}~=~(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t)])^frac{1}{2}~ =? ~(operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)])^frac{1}{2} $$ So basically why : $$operatorname{E}[X^2(t)]~ =~operatorname{E}[X^2(t+tau)]$$ ?
            $endgroup$
            – Hilbert
            Jan 4 at 17:33


















          $begingroup$
          I do understand that part but the problem is why they've replaced $sigma_{t}sigma_{t+tau}$ by $sigma^2_{t}$ ?
          $endgroup$
          – Hilbert
          Jan 4 at 17:21




          $begingroup$
          I do understand that part but the problem is why they've replaced $sigma_{t}sigma_{t+tau}$ by $sigma^2_{t}$ ?
          $endgroup$
          – Hilbert
          Jan 4 at 17:21












          $begingroup$
          Because $sigma_{t+tau}=sigma_{t}$.
          $endgroup$
          – Davide Giraudo
          Jan 4 at 17:24




          $begingroup$
          Because $sigma_{t+tau}=sigma_{t}$.
          $endgroup$
          – Davide Giraudo
          Jan 4 at 17:24












          $begingroup$
          This is the very relation that I am trying to prove/understand , how is this equality justified : $$sigma(t+tau) =(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t+tau)])^frac{1}{2}~=~(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t)])^frac{1}{2}~ =? ~(operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)])^frac{1}{2} $$ So basically why : $$operatorname{E}[X^2(t)]~ =~operatorname{E}[X^2(t+tau)]$$ ?
          $endgroup$
          – Hilbert
          Jan 4 at 17:33






          $begingroup$
          This is the very relation that I am trying to prove/understand , how is this equality justified : $$sigma(t+tau) =(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t+tau)])^frac{1}{2}~=~(operatorname{E}[X^2(t+tau)]-operatorname{E}^2[X(t)])^frac{1}{2}~ =? ~(operatorname{E}[X^2(t)]-operatorname{E}^2[X(t)])^frac{1}{2} $$ So basically why : $$operatorname{E}[X^2(t)]~ =~operatorname{E}[X^2(t+tau)]$$ ?
          $endgroup$
          – Hilbert
          Jan 4 at 17:33




















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