Autocorrelation when $m = 0$
$begingroup$
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?
statistics correlation expected-value
edited Dec 9 '18 at 5:12
Andrews
3831317
asked Dec 9 '18 at 0:24
Micah MungalMicah Mungal
83
$endgroup$
add a comment |
$begingroup$
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?
statistics correlation expected-value
edited Dec 9 '18 at 5:12
Andrews
3831317
asked Dec 9 '18 at 0:24
Micah MungalMicah Mungal
83
$endgroup$
add a comment |
$begingroup$
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?
statistics correlation expected-value
edited Dec 9 '18 at 5:12
Andrews
3831317
asked Dec 9 '18 at 0:24
Micah MungalMicah Mungal
83
$endgroup$
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
statistics correlation expected-value
edited Dec 9 '18 at 5:12
Andrews
3831317
asked Dec 9 '18 at 0:24
Micah MungalMicah Mungal
83
edited Dec 9 '18 at 5:12
Andrews
3831317
asked Dec 9 '18 at 0:24
Micah MungalMicah Mungal
83
edited Dec 9 '18 at 5:12
Andrews
3831317
edited Dec 9 '18 at 5:12
Andrews
3831317
edited Dec 9 '18 at 5:12
Andrews
3831317
3831317
asked Dec 9 '18 at 0:24
Micah MungalMicah Mungal
83
asked Dec 9 '18 at 0:24
Micah MungalMicah Mungal
83
asked Dec 9 '18 at 0:24
Micah MungalMicah Mungal
83
83
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
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 $mneq 0.$
answered Dec 9 '18 at 0:45
kodlukodlu
3,390716
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
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oldest
votes
active
oldest
votes
$begingroup$
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 $mneq 0.$
answered Dec 9 '18 at 0:45
kodlukodlu
3,390716
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mneq 0.$
answered Dec 9 '18 at 0:45
kodlukodlu
3,390716
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mneq 0.$
answered Dec 9 '18 at 0:45
kodlukodlu
3,390716
$endgroup$
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 $mneq 0.$
answered Dec 9 '18 at 0:45
kodlukodlu
3,390716
answered Dec 9 '18 at 0:45
kodlukodlu
3,390716
answered Dec 9 '18 at 0:45
kodlukodlu
3,390716
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
add a comment |
$begingroup$
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
add a comment |
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