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  #1 (permalink)  
Old 12-11-2009, 06:54 PM
Frank Chang
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Default squared mean vs. mean squared

Hi group,

I have a (probably very simple) question that I could not think
through. i am hoping that you could please enlighten me a little bit.

Suppose I have two completely uncorrelated signals A and B. We can
assume that they take any form, say, continuous function or discrete
series. For convenience, let's assume they can be indexed. They have
the same length N. The question is, does this relationship hold?

<A> * <B> = <A * B>

Here <> represents mean, i.e. <A> = sum(A)/N.

I have done numerical experiments on this, and it appears that it
hods. I, however, could not come up with an analytical proof. Could
you please help me here?

A related question is that under what condition this relationship
breaks down. I know for sure that when A and B are related, the math
does not hold because <A>^2 != <A^2>. I think an analytical proof for
the previous question can help with this question as well, but I am
not sure.

Thanks a lot!

Regards,
Frank
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  #2 (permalink)  
Old 12-11-2009, 07:10 PM
glen herrmannsfeldt
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Default Re: squared mean vs. mean squared

Frank Chang <etaghtron@gmail.com> wrote:

> I have a (probably very simple) question that I could not think
> through. i am hoping that you could please enlighten me a little bit.


> Suppose I have two completely uncorrelated signals A and B. We can
> assume that they take any form, say, continuous function or discrete
> series. For convenience, let's assume they can be indexed. They have
> the same length N. The question is, does this relationship hold?


> <A> * <B> = <A * B>


> Here <> represents mean, i.e. <A> = sum(A)/N.


Or <A * B> - <A> * <B> = 0

I think you can get from the to the Correlation coefficient,
which is zero in they are uncorrelated.

Start with A and B of length 2, then there is a cross term
that appears in one and not the other. Show that is zero.
Then use induction to show for all N.

Sounds like a good way to me, anyway.

-- glen
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  #3 (permalink)  
Old 12-11-2009, 07:27 PM
Greg Heath
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Default Re: squared mean vs. mean squared

On Dec 11, 1:54*pm, Frank Chang <etaght...@gmail.com> wrote:
> Hi group,
>
> I have a (probably very simple) question that I could not think
> through. i am hoping that you could please enlighten me a little bit.
>
> Suppose I have two completely uncorrelated signals A and B. We can
> assume that they take any form, say, continuous function or discrete
> series. For convenience, let's assume they can be indexed. They have
> the same length N. The question is, does this relationship hold?
>
> <A> * <B> = <A * B>
>
> Here <> represents mean, i.e. <A> = sum(A)/N.
>
> I have done numerical experiments on this, and it appears that it
> hods. I, however, could not come up with an analytical proof. Could
> you please help me here?
>
> A related question is that under what condition this relationship
> breaks down. I know for sure that when A and B are related, the math
> does not hold because <A>^2 != <A^2>. I think an analytical proof for
> the previous question can help with this question as well, but I am
> not sure.
>


If you are going to post in multiple groups, send in only one
post with multiple names in the newgroup line.

See my answer in another newsgroup.

Greg
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  #4 (permalink)  
Old 12-11-2009, 08:16 PM
Tim Wescott
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Default Re: squared mean vs. mean squared

On Fri, 11 Dec 2009 11:27:45 -0800, Greg Heath wrote:

> On Dec 11, 1:54Â*pm, Frank Chang <etaght...@gmail.com> wrote:
>> Hi group,
>>
>> I have a (probably very simple) question that I could not think
>> through. i am hoping that you could please enlighten me a little bit.
>>
>> Suppose I have two completely uncorrelated signals A and B. We can
>> assume that they take any form, say, continuous function or discrete
>> series. For convenience, let's assume they can be indexed. They have
>> the same length N. The question is, does this relationship hold?
>>
>> <A> * <B> = <A * B>
>>
>> Here <> represents mean, i.e. <A> = sum(A)/N.
>>
>> I have done numerical experiments on this, and it appears that it hods.
>> I, however, could not come up with an analytical proof. Could you
>> please help me here?
>>
>> A related question is that under what condition this relationship
>> breaks down. I know for sure that when A and B are related, the math
>> does not hold because <A>^2 != <A^2>. I think an analytical proof for
>> the previous question can help with this question as well, but I am not
>> sure.
>>
>>

> If you are going to post in multiple groups, send in only one post with
> multiple names in the newgroup line.
>
> See my answer in another newsgroup.
>
> Greg


See the Wikipedia article on "cross posting", compare and contrast with
"multiple posting".

--
www.wescottdesign.com
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  #5 (permalink)  
Old 12-11-2009, 08:52 PM
Vladimir Vassilevsky
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Posts: n/a
Default Re: squared mean vs. mean squared



Frank Chang wrote:

> Hi group,
>
> I have a (probably very simple) question that I could not think
> through. i am hoping that you could please enlighten me a little bit.
>
> Suppose I have two completely uncorrelated signals A and B. We can
> assume that they take any form, say, continuous function or discrete
> series. For convenience, let's assume they can be indexed. They have
> the same length N. The question is, does this relationship hold?
>
> <A> * <B> = <A * B>
>
> Here <> represents mean, i.e. <A> = sum(A)/N.
>
> I have done numerical experiments on this, and it appears that it
> hods. I, however, could not come up with an analytical proof. Could
> you please help me here?
>
> A related question is that under what condition this relationship
> breaks down. I know for sure that when A and B are related, the math
> does not hold because <A>^2 != <A^2>. I think an analytical proof for
> the previous question can help with this question as well, but I am
> not sure.


Fun question.

You can always find such i,j,k, so:

sum(A[i]) * sum(B[j]) === sum (A[i]B[i+k])

This holds regardless of A and B.


VLV

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  #6 (permalink)  
Old 12-14-2009, 12:10 AM
dvsarwate
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Posts: n/a
Default Re: squared mean vs. mean squared

On Dec 11, 12:54*pm, Frank Chang <etaght...@gmail.com> wrote:

> Suppose I have two completely uncorrelated signals A and B. We can
> assume that they take any form, say, continuous function or discrete
> series. For convenience, let's assume they can be indexed. They have
> the same length N. The question is, does this relationship hold?
>
> <A> * <B> = <A * B>
>
> Here <> represents mean, i.e. <A> = sum(A)/N.



There are multiple meanings to the phrase "uncorrelated
signals" some of which make the question asked a
"Which came first, the chicken or the egg?" type of
question. In one meaning, signals are said to be
uncorrelated **exactly** when <A> * <B> = <A * B>
holds. So, what does the OP understand by the
phrase "uncorrelated signals" or "I have two
completely uncorrelated signals"? In what sense
are they uncorrelated? What does the rest of the
comp.dsp group understand by this phrase or
these phrases?

--Dilip Nitpick
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  #7 (permalink)  
Old 12-14-2009, 02:37 AM
glen herrmannsfeldt
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Posts: n/a
Default Re: squared mean vs. mean squared

dvsarwate <dvsarwate@gmail.com> wrote:
> On Dec 11, 12:54?pm, Frank Chang <etaght...@gmail.com> wrote:


>> Suppose I have two completely uncorrelated signals A and B.

(snip)

> There are multiple meanings to the phrase "uncorrelated
> signals" some of which make the question asked a
> "Which came first, the chicken or the egg?" type of
> question. In one meaning, signals are said to be
> uncorrelated **exactly** when <A> * <B> = <A * B>
> holds. So, what does the OP understand by the
> phrase "uncorrelated signals" or "I have two
> completely uncorrelated signals"? In what sense
> are they uncorrelated? What does the rest of the
> comp.dsp group understand by this phrase or
> these phrases?


Physics usually uses coherent/incoherent, with the saying:

"For coherent sources add the amplitude, for incoherent sources
add the intensities" (where the "amplitude" includes the appropriate
phase.)

That works in optics, and in general quantum mechanics for the
QM wave function. Otherwise, there is the phrase "statistically
independent" which also describes a similar idea.

-- glen
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