> Hi Peter,
>
> Responses below.
>
> Did you mean
>
> = M^2 + E[XX(t) * XX(t+tau)]?
>
> I'll proceed as if you did.

Yup.

> > So if we are given:
> >
> > E[X(t) X(t+tau)] = A.delta(t)
>
> You mean A delta(tau), right? I'll proceed as if you did.

Yup.

> > doesn't that mean M = 0??
>
> It seems to me that it means
>
> E[XX(t) * XX(t+tau)] = -M^2 (tau != 0)
>
> Why couldn't that happen?

It could, I suppose, if you we'ren't given that XX(t) was IID (or
uncorrelated, as you say below).

> In fact, that makes the point of my paper nicely. If we are ONLY given
> that Rxx(tau) = A delta(tau), then we can't conclude anything about
> the mean. However, if we, in addition, are given that the process is
> uncorrelated sample to sample, then
>
> E[XX(t) * XX(t+tau)] = E[XX(t)] * E[XX(t+tau)]
> = M^2 (since XX is stationary),
>
> from which we may write
>
> M^2 = -M^2
>
> which THEN implies M = 0.

OK, thanks!
Ciao,
Peter K.
--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."

Reply by Randy Yates●February 19, 20062006-02-19

Hi Peter,
Responses below.
p.kootsookos@remove.ieee.org (Peter K.) writes:

> Randy Yates <yates@ieee.org> writes:
>
>> Hey Peter - sorry to take so long to respond - I just noticed this.
>
> NO worries. I'm still getting my head around a response to you and
> Jerry on this thread.
>
>> p.kootsookos@remove.ieee.org (Peter K.) writes:
>>
>> >
>> > You gave the autocorrlation as a constant times a delta function. If
>> > the process had been non-zero mean, it would have been something else.
>>
>> Why?
>>
>> As I wrote in the paper on-line, the autocorrelation of a real process
>> X(t) is
>>
>> Rxx(tau) = E[X(t)X(t+tau)].
>>
>> If X(t) is not uncorrelated, then we don't get to separate this
>> into
>>
>> Rxx(tau) = E[X(t)] * E[X(t+tau)],
>>
>> and so we really can't make any conclusions regarding the mean of
>> that process. We only know (assuming Brown's definition of "white")
>> that E[X(t))X(t+tau)] = 0, and as far as I know, it is possible
>> that this can hold true even when the mean of X(t) is non-zero.
>
> Not as far as I can tell:
>
> X(t) = M + XX(t) where XX(t) is zero mean.
>
> E[X(t) X(t+tau)] = E[(M + XX(t))(M + XX(t+tau))]
> = M^2 + E[M XX(t)] + E[M + XX(t+tau)] + E[XX(t) XX(t+tau)]
> = M^2 + E[XX(t) + XX(t+tau)]

Did you mean
= M^2 + E[XX(t) * XX(t+tau)]?
I'll proceed as if you did.

> So if we are given:
>
> E[X(t) X(t+tau)] = A.delta(t)

You mean A delta(tau), right? I'll proceed as if you did.

> doesn't that mean M = 0??

It seems to me that it means
E[XX(t) * XX(t+tau)] = -M^2 (tau != 0)
Why couldn't that happen?
In fact, that makes the point of my paper nicely. If we are ONLY given
that Rxx(tau) = A delta(tau), then we can't conclude anything about
the mean. However, if we, in addition, are given that the process is
uncorrelated sample to sample, then
E[XX(t) * XX(t+tau)] = E[XX(t)] * E[XX(t+tau)]
= M^2 (since XX is stationary),
from which we may write
M^2 = -M^2
which THEN implies M = 0.
--
% Randy Yates % "She's sweet on Wagner-I think she'd die for Beethoven.
%% Fuquay-Varina, NC % She love the way Puccini lays down a tune, and
%%% 919-577-9882 % Verdi's always creepin' from her room."
%%%% <yates@ieee.org> % "Rockaria", *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Reply by ●February 19, 20062006-02-19

Randy Yates <yates@ieee.org> writes:

> Hey Peter - sorry to take so long to respond - I just noticed this.

NO worries. I'm still getting my head around a response to you and
Jerry on this thread.

> p.kootsookos@remove.ieee.org (Peter K.) writes:
>
> >
> > You gave the autocorrlation as a constant times a delta function. If
> > the process had been non-zero mean, it would have been something else.
>
> Why?
>
> As I wrote in the paper on-line, the autocorrelation of a real process
> X(t) is
>
> Rxx(tau) = E[X(t)X(t+tau)].
>
> If X(t) is not uncorrelated, then we don't get to separate this
> into
>
> Rxx(tau) = E[X(t)] * E[X(t+tau)],
>
> and so we really can't make any conclusions regarding the mean of
> that process. We only know (assuming Brown's definition of "white")
> that E[X(t))X(t+tau)] = 0, and as far as I know, it is possible
> that this can hold true even when the mean of X(t) is non-zero.

Not as far as I can tell:
X(t) = M + XX(t) where XX(t) is zero mean.
E[X(t) X(t+tau)] = E[(M + XX(t))(M + XX(t+tau))]
= M^2 + E[M XX(t)] + E[M + XX(t+tau)] + E[XX(t) XX(t+tau)]
= M^2 + E[XX(t) + XX(t+tau)]
So if we are given:
E[X(t) X(t+tau)] = A.delta(t)
doesn't that mean M = 0??
Ciao,
Peter K.
--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."

Reply by Randy Yates●February 19, 20062006-02-19

Hey Peter - sorry to take so long to respond - I just noticed this.
p.kootsookos@remove.ieee.org (Peter K.) writes:

> Randy Yates <yates@ieee.org> writes:
>
>> p.kootsookos@remove.ieee.org (Peter K.) writes:
>>
>> > Randy Yates <yates@ieee.org> writes:
>> >
>> >>
>> >> I agree. Did I write or state something that required a zero mean?
>> >>
>> >
>> > Just the auto-correlation expression.
>>
>> I'm confused. The definition I've seen of correlation of two
>> WSS random processes W and Z (not necessarily real) is
>>
>> Rwz(tau) = E[W(t)Z'(t+tau)].
>>
>> That definition holds whether the mean is zero or not, right?
>
> True.
>
>> What did I miss?
>
> You gave the autocorrlation as a constant times a delta function. If
> the process had been non-zero mean, it would have been something else.

Why?
As I wrote in the paper on-line, the autocorrelation of a real process
X(t) is
Rxx(tau) = E[X(t)X(t+tau)].
If X(t) is not uncorrelated, then we don't get to separate this
into
Rxx(tau) = E[X(t)] * E[X(t+tau)],
and so we really can't make any conclusions regarding the mean of
that process. We only know (assuming Brown's definition of "white")
that E[X(t))X(t+tau)] = 0, and as far as I know, it is possible
that this can hold true even when the mean of X(t) is non-zero.
--
% Randy Yates % "Though you ride on the wheels of tomorrow,
%% Fuquay-Varina, NC % you still wander the fields of your
%%% 919-577-9882 % sorrow."
%%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

Reply by ●February 16, 20062006-02-16

Randy Yates <yates@ieee.org> writes:

> p.kootsookos@remove.ieee.org (Peter K.) writes:
>
> > Randy Yates <yates@ieee.org> writes:
> >
> >>
> >> I agree. Did I write or state something that required a zero mean?
> >>
> >
> > Just the auto-correlation expression.
>
> I'm confused. The definition I've seen of correlation of two
> WSS random processes W and Z (not necessarily real) is
>
> Rwz(tau) = E[W(t)Z'(t+tau)].
>
> That definition holds whether the mean is zero or not, right?

True.

> What did I miss?

You gave the autocorrlation as a constant times a delta function. If
the process had been non-zero mean, it would have been something else.
I'm too used to seeing autocorrelations of finite duration data, where
if there is a DC offset (non-zero mean), then the (unbiassed)
autocorrelation estimate comes out as something overlaid on a
triangle.
In the case you're looking at, you'd get a constant offset (I think).
Ciao,
Peter K.
--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."

Reply by Randy Yates●February 15, 20062006-02-15

p.kootsookos@remove.ieee.org (Peter K.) writes:

> Randy Yates <yates@ieee.org> writes:
>
>>
>> I agree. Did I write or state something that required a zero mean?
>>
>
> Just the auto-correlation expression.

I'm confused. The definition I've seen of correlation of two
WSS random processes W and Z (not necessarily real) is
Rwz(tau) = E[W(t)Z'(t+tau)].
That definition holds whether the mean is zero or not, right?
What did I miss?
--
% Randy Yates % "The dreamer, the unwoken fool -
%% Fuquay-Varina, NC % in dreams, no pain will kiss the brow..."
%%% 919-577-9882 %
%%%% <yates@ieee.org> % 'Eldorado Overture', *Eldorado*, ELO
http://home.earthlink.net/~yatescr

Reply by ●February 15, 20062006-02-15

Randy Yates <yates@ieee.org> writes:

>
> I agree. Did I write or state something that required a zero mean?
>

Just the auto-correlation expression.
--
"And he sees the vision splendid
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."

Reply by ●February 15, 20062006-02-15

"Randy Yates" <yates@ieee.org> wrote in message
news:m3slql45tc.fsf@ieee.org...

> dvsarwate@ieee.org writes:
>> Can you reveal what you got for E[V(t) * V(t + \tau)]?
>
> I got E[V^2] = T*N0, where the original \phi_{xx}(\tau) was N0 *
> \delta(\tau).

That is certainly the right answer (for \tau > 0). More generally, it is
true that
E[V(t) * V(t + \tau)] = N0*min(t, t+\tau). {V(t)} is called a Wiener
process.

Reply by Randy Yates●February 15, 20062006-02-15

"Peter K." <p.kootsookos@iolfree.ie> writes:

> Randy Yates wrote:
>
>>
>> > It's interesting that you're saying autocorrelation rather than
>> > autocovariance...
>>
>> Why?
>
> Well, because if it was autocovariance, you wouldn't necessarily know
> that the RV is zero mean.

I agree. Did I write or state something that required a zero mean?

> Most definitions of autocovariance I've seen remove the mean before
> calculating the sequence. Autocorreltation definitions that I'm
> used to generally don't mean-correct.

Yes, that's the way I'm seeing them in Proakis, Papoulis, Leon-Garcia,
etc.
--
% Randy Yates % "Maybe one day I'll feel her cold embrace,
%% Fuquay-Varina, NC % and kiss her interface,
%%% 919-577-9882 % til then, I'll leave her alone."
%%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

Reply by Peter K.●February 15, 20062006-02-15

Randy Yates wrote:

>
> > It's interesting that you're saying autocorrelation rather than
> > autocovariance...
>
> Why?

Well, because if it was autocovariance, you wouldn't necessarily know
that the RV is zero mean. Most definitions of autocovariance I've seen
remove the mean before calculating the sequence. Autocorreltation
definitions that I'm used to generally don't mean-correct.
At least that's what I was musing when I wrote the above! :-)
Ciao,
Peter K.