# Simple Question on Random Variables and Random Processes

Started by February 14, 2006
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."


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                %

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."


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

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

> 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."


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
>
>> 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

Randy Yates <yates@ieee.org> writes:

> 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."