Autocorrelation when $m = 0$

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I am trying to understand something with respect to autocorrelation. If the process is iid, I can say:



$R_{XX}[n, n + m] = E[x[n]] times E[x[n + m]] quad textrm{if} quad m ne 0$



But if $m= 0$



We should get:



$R_{XX}[n, n + m] = E[x[n]] times E[x[n]] quad textrm{if} quad m = 0$



That is



$R_{XX}[n, n + m] = (E[x[n]])^2$



But in text books I always see, for m = 0



$R_{XX}[n, n + m] = E[x^2[n]]$



Why is it the latter rather than the former?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am trying to understand something with respect to autocorrelation. If the process is iid, I can say:



    $R_{XX}[n, n + m] = E[x[n]] times E[x[n + m]] quad textrm{if} quad m ne 0$



    But if $m= 0$



    We should get:



    $R_{XX}[n, n + m] = E[x[n]] times E[x[n]] quad textrm{if} quad m = 0$



    That is



    $R_{XX}[n, n + m] = (E[x[n]])^2$



    But in text books I always see, for m = 0



    $R_{XX}[n, n + m] = E[x^2[n]]$



    Why is it the latter rather than the former?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to understand something with respect to autocorrelation. If the process is iid, I can say:



      $R_{XX}[n, n + m] = E[x[n]] times E[x[n + m]] quad textrm{if} quad m ne 0$



      But if $m= 0$



      We should get:



      $R_{XX}[n, n + m] = E[x[n]] times E[x[n]] quad textrm{if} quad m = 0$



      That is



      $R_{XX}[n, n + m] = (E[x[n]])^2$



      But in text books I always see, for m = 0



      $R_{XX}[n, n + m] = E[x^2[n]]$



      Why is it the latter rather than the former?










      share|cite|improve this question











      $endgroup$




      I am trying to understand something with respect to autocorrelation. If the process is iid, I can say:



      $R_{XX}[n, n + m] = E[x[n]] times E[x[n + m]] quad textrm{if} quad m ne 0$



      But if $m= 0$



      We should get:



      $R_{XX}[n, n + m] = E[x[n]] times E[x[n]] quad textrm{if} quad m = 0$



      That is



      $R_{XX}[n, n + m] = (E[x[n]])^2$



      But in text books I always see, for m = 0



      $R_{XX}[n, n + m] = E[x^2[n]]$



      Why is it the latter rather than the former?







      statistics correlation expected-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 5:12









      Andrews

      3831317




      3831317










      asked Dec 9 '18 at 0:24









      Micah MungalMicah Mungal

      83




      83






















          1 Answer
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          0












          $begingroup$

          The general formula is
          $R_{XX}[n, n + m] = E[x[n] x[n + m]],$
          which you can always use.



          Specifically, $(x[n],x[n])$ is not an independent pair so you use the general formula.



          But for i.i.d. the equation reduces to what you have when $mneq 0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok I see, I was looking more at the math itself and didn't realise (x[n],x[n]) is not an independent pair.
            $endgroup$
            – Micah Mungal
            Dec 9 '18 at 0:56











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          1 Answer
          1






          active

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          active

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          0












          $begingroup$

          The general formula is
          $R_{XX}[n, n + m] = E[x[n] x[n + m]],$
          which you can always use.



          Specifically, $(x[n],x[n])$ is not an independent pair so you use the general formula.



          But for i.i.d. the equation reduces to what you have when $mneq 0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok I see, I was looking more at the math itself and didn't realise (x[n],x[n]) is not an independent pair.
            $endgroup$
            – Micah Mungal
            Dec 9 '18 at 0:56
















          0












          $begingroup$

          The general formula is
          $R_{XX}[n, n + m] = E[x[n] x[n + m]],$
          which you can always use.



          Specifically, $(x[n],x[n])$ is not an independent pair so you use the general formula.



          But for i.i.d. the equation reduces to what you have when $mneq 0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok I see, I was looking more at the math itself and didn't realise (x[n],x[n]) is not an independent pair.
            $endgroup$
            – Micah Mungal
            Dec 9 '18 at 0:56














          0












          0








          0





          $begingroup$

          The general formula is
          $R_{XX}[n, n + m] = E[x[n] x[n + m]],$
          which you can always use.



          Specifically, $(x[n],x[n])$ is not an independent pair so you use the general formula.



          But for i.i.d. the equation reduces to what you have when $mneq 0.$






          share|cite|improve this answer









          $endgroup$



          The general formula is
          $R_{XX}[n, n + m] = E[x[n] x[n + m]],$
          which you can always use.



          Specifically, $(x[n],x[n])$ is not an independent pair so you use the general formula.



          But for i.i.d. the equation reduces to what you have when $mneq 0.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 0:45









          kodlukodlu

          3,390716




          3,390716












          • $begingroup$
            Ok I see, I was looking more at the math itself and didn't realise (x[n],x[n]) is not an independent pair.
            $endgroup$
            – Micah Mungal
            Dec 9 '18 at 0:56


















          • $begingroup$
            Ok I see, I was looking more at the math itself and didn't realise (x[n],x[n]) is not an independent pair.
            $endgroup$
            – Micah Mungal
            Dec 9 '18 at 0:56
















          $begingroup$
          Ok I see, I was looking more at the math itself and didn't realise (x[n],x[n]) is not an independent pair.
          $endgroup$
          – Micah Mungal
          Dec 9 '18 at 0:56




          $begingroup$
          Ok I see, I was looking more at the math itself and didn't realise (x[n],x[n]) is not an independent pair.
          $endgroup$
          – Micah Mungal
          Dec 9 '18 at 0:56


















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