Simple Question on Random Variables and Random Processes

Let X(t) be a stationary white noise process with autocorrelation
function R_{xx}(\tau) = N0 * \delta(\tau). Let V be a random variable
determined from X(t) by

  V = \int_{-\infty}^{+\infty} X(t) dt.

How do you compute statistics on V, like E[V^2]?

Yes, this is a homework problem.
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Randy Yates <yates@ieee.org> writes:

> Let X(t) be a stationary white noise process with autocorrelation
> function R_{xx}(\tau) = N0 * \delta(\tau). Let V be a random variable
> determined from X(t) by
>
>   V = \int_{-\infty}^{+\infty} X(t) dt.

Sorry, wrong limits: 

   V = \int_{0}^{T} X(t) dt.
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Randy Yates <yates@ieee.org> writes:

> Randy Yates <yates@ieee.org> writes:
> 
> > Let X(t) be a stationary white noise process with autocorrelation
> > function R_{xx}(\tau) = N0 * \delta(\tau). Let V be a random variable
> > determined from X(t) by
> >
> >   V = \int_{-\infty}^{+\infty} X(t) dt.
> 
> Sorry, wrong limits: 
> 
>    V = \int_{0}^{T} X(t) dt.

Note that integration and expectation are interchangeable:

E[V] = E[\int_{0}^{T} X(t) dt] 
     = \int_{0}^{T} E[X(t)] dt

E[V^2] = E[\int_{0}^{T} X(t) dt \int_{0}^{T} X(s) ds] 
       = \int_{0}^{T} \int_{0}^{T} E[X(t) X(s)] dt ds

You should then be able to use the fact that X is stationary, and the
autocorrelation expression to reduce both further.

Does that help?

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

Ciao,

Peter K.

-- 
"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:
>
>> Randy Yates <yates@ieee.org> writes:
>> 
>> > Let X(t) be a stationary white noise process with autocorrelation
>> > function R_{xx}(\tau) = N0 * \delta(\tau). Let V be a random variable
>> > determined from X(t) by
>> >
>> >   V = \int_{-\infty}^{+\infty} X(t) dt.
>> 
>> Sorry, wrong limits: 
>> 
>>    V = \int_{0}^{T} X(t) dt.
>
> Note that integration and expectation are interchangeable:
>
> E[V] = E[\int_{0}^{T} X(t) dt] 
>      = \int_{0}^{T} E[X(t)] dt
>
> E[V^2] = E[\int_{0}^{T} X(t) dt \int_{0}^{T} X(s) ds] 
>        = \int_{0}^{T} \int_{0}^{T} E[X(t) X(s)] dt ds
>
> You should then be able to use the fact that X is stationary, and the
> autocorrelation expression to reduce both further.
>
> Does that help?

Yes, thanks much, Peter! I've been staring at it for hours! 

I kept on getting hung up on the idea that we can determine a random
variable that is a function of a finite number of random variables
easily enough, but this is a function of an uncountably infinite
number of random variables! Then I got hung up on another viewpoint in
which I modeled this V = V(0), where V(t) is the convolution of a
pulse with X(t), and then using the theorems on output autocorrelation
of a linear system. But that only get's you the autocorrelation,
E[V(t) * V'(t+\tau)] (where "'" denotes conjugate).  What I needed
what E[V(t) * V(t)].

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

Why? 
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Randy Yates wrote:

> Then I got hung up on another viewpoint in
> which I modeled this V = V(0), where V(t) is the convolution of a
> pulse with X(t), and then using the theorems on output autocorrelation
> of a linear system. But that only get's you the autocorrelation,
> E[V(t) * V'(t+\tau)] (where "'" denotes conjugate).  What I needed
> what E[V(t) * V(t)].

I don't understand why conjugation is a problem here.  The original
statement seemed to imply that X(t) was a real process, not a
complex-valued process, and thus conjugation should not matter.
If so, setting \tau = 0 should give the desired answer.

Can you reveal what you got for E[V(t) * V(t + \tau)]?

dvsarwate@ieee.org writes:

> Randy Yates wrote:
>
>> Then I got hung up on another viewpoint in
>> which I modeled this V = V(0), where V(t) is the convolution of a
>> pulse with X(t), and then using the theorems on output autocorrelation
>> of a linear system. But that only get's you the autocorrelation,
>> E[V(t) * V'(t+\tau)] (where "'" denotes conjugate).  What I needed
>> what E[V(t) * V(t)].
>
> I don't understand why conjugation is a problem here.  The original
> statement seemed to imply that X(t) was a real process, not a
> complex-valued process, and thus conjugation should not matter.
> If so, setting \tau = 0 should give the desired answer.

Hi Dilip,

Yes, that would be true. I'm trying to decide if X really is
real. All I know is that it is white, stationary, and 
\phi_{xx}(\tau) = N_0 \delta(\tau). Can you tell from this?

> 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).
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Randy Yates <yates@ieee.org> writes:

> dvsarwate@ieee.org writes:
>
>> Randy Yates wrote:
>>
>>> Then I got hung up on another viewpoint in
>>> which I modeled this V = V(0), where V(t) is the convolution of a
>>> pulse with X(t), and then using the theorems on output autocorrelation
>>> of a linear system. But that only get's you the autocorrelation,
>>> E[V(t) * V'(t+\tau)] (where "'" denotes conjugate).  What I needed
>>> what E[V(t) * V(t)].
>>
>> I don't understand why conjugation is a problem here.  The original
>> statement seemed to imply that X(t) was a real process, not a
>> complex-valued process, and thus conjugation should not matter.
>> If so, setting \tau = 0 should give the desired answer.
>
> Hi Dilip,
>
> Yes, that would be true. I'm trying to decide if X really is
> real. All I know is that it is white, stationary, and 
> \phi_{xx}(\tau) = N_0 \delta(\tau). Can you tell from this?

Dilip et al.,

I also know that it's zero-mean.

This seems to be a nasty problem. It is problem 4.3 b in Proakis, 4th
edition.

First of all, they say z(t) = x(t) + j * y(t) is the (complex) lowpass
equivalent of a bandpass signal in part a). Then they say it's white
in part b), but in the text they state that white noise cannot be
expressed in terms of quadrature components (the x(t) and y(t) are
derived from bandlimited bandpass signals, which white noise is not).

Then I started thinking: is it possible for a complex signal to have a 
real autocorrelation? The answer is "yes." (Note I'm dropping the
LaTeX-ese a bit for convenience.)

Rzz(tau) = E[(x(t) + j*y(t))(x(t+tau) - j*y(t+tau))]
         = Rxx(tau) + Ryy(tau) + j * [Ryx(tau) - Rxy(tau)].

So what we require is that Ryx(tau) = Rxy(tau). 

One case in which that is true is when x(t) = y(t), so that
z(t) = x(t) * (1 + j). 

So then if we interpret z(t) (replace X(t) with z(t) in my original
post) as being complex, then a) we do get different results for
E[V^2] vs. E[V*V'], and b) there is no way to simplify the resulting
equations in x and y since you cannot use the conditions of bandpass
signals which state, e.g.,  Rxx(tau) = Ryy(tau).

Any illumination would be appreciated.
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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.

"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.
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"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.