How can one define Continuous White Noise in a coherent way?
Is there a way to derive it Mathematically?

Specifically, is there a way to define it which will works as the model in Signal Processing yet will obey all Mathematical properties of a Random Process?

  • 1
    $\begingroup$ "is there a difference in the way it is defined in Signal Processing?" Different from what? Perhaps I'm being parochial, but isn't the concept of white noise only defined in signal processing? $\endgroup$
    – TimWescott
    Aug 18, 2023 at 15:54
  • $\begingroup$ Yeah, Royi, I think this question has reached a deadlock. You have one definition in your answer, Tim and others including me have a different one, you're not going to accept these - making the question actually one of "what is the definition, in Royi's opinion", and we usually don't allow opinion-based questions. So, how do you propose we get out of this strange situation? $\endgroup$ Aug 19, 2023 at 9:14
  • $\begingroup$ @MarcusMüller, I won't be accepting mine. I was asking for a coherent definition. I just used my answer to show the issue (Difference). $\endgroup$
    – Royi
    Aug 19, 2023 at 9:17
  • $\begingroup$ @MarcusMüller, By the way, I am pretty sure that all definitions must align. The way I know to align the all is by defining White Noise as a limit of a process noise with certain properties. The only point I made, started in a discussion with Robert, is that the common definition of White Noise in Signal Processing is not complete Mathematically. If you can show me otherwise, you'll get the marked answer. $\endgroup$
    – Royi
    Aug 19, 2023 at 9:27
  • $\begingroup$ @Royi well the problem is that Tim's definition simply through $t_1\ne t_2 \implies E(X_{t_1}\cdot X_{t_2})=0$; is coherent. It doesn't try to do a Fourier transform, and hence doesn't get into trouble when the ACF doesn't exist, or is not transformable, or if the energy diverges. You're arguing in the comments below it that this well-formedness would matter for a definition – but that's not the "coherent" thing to do. You're "inventing" (pointed word, do not take negatively) additional constraints,and then derive properties from that,like the surprising distribution claim! $\endgroup$ Aug 19, 2023 at 9:31

4 Answers 4


In the context of Signal Processing White Noise is usually defined using a single intuitive property: A random process with constant magnitude, for any frequency, of its power spectrum:

If $ w \left( t \right) $ is a white noise then its power spectrum is given by $ {S}_{w} \left( f \right) = {\sigma}^{2} $. In communication it is common to use $ {S}_{w} \left( f \right) = \frac{ {N}_{0} }{2} $.

Using Wiener Khinchin Theorem one would conclude that the Auto Correlation of the process, which is assumed to be a Stationary Process in the wide sense is given by $ {R}_{ww} \left( \tau \right) = {\sigma}^{2} \delta \left( t \right) $.

In the context of Signal Processing we only need the model of Continuous White Noise in order to analyze the output of Linear System with finite frequency support. For this context, the above model is enough.

Yet, if one wants to be rigorous, mathematically, the above model is not enough.
For instance, if one wants the Wiener Khinchin Theorem to hold one must have, arbitrarily small, correlation in the auto correlation function.
So the move from the power density to the auto correlation, as done above, is not justified Mathematically (See for instance "Roy M. Howard - On Defining White Noise").
So the beast, called White Noise, is not plausible both Physically (Which is OK, we are dealing with Math), yet it is not coherent Mathematically, it requires more delicate work. Yet, again, in the context of Signal Processing, this is enough as only deal with it in the context of going through a Linear System with limited bandwidth support.

A more rigorous derivation can be done by defining the (Gaussian) White Noise as the derivative of Wiener Process.
Wiener Process has some properties which makes it interesting:

  1. The difference of 2 samples is Gaussian.
  2. The samples / realization path is continuous (In the almost surely sense).

Then we can define Gaussian White noise as the derivative of the Wiener Process. This will yield a more coherent definition (For instance, it indeed defines a distribution).

Remark I'd be happy if those who -1 will specify why.
I will try to explain the answer.

The answer has the following logic:

  • We want a random process. Specifically a stationary random process (Even if only in the wide sense, there are subtleties to that, but let's skip this).
  • It should obey Wiener Khinchin Theorem we need to be able to analyze it using spectral methods.

A reasonable way to do so is define a random process as following.

  • Let $ v \left( t \right) $ be a stationary random process.
  • For any pair $ {t}_{i} \neq {t}_{j} $ we would like to have $ \mathbb{E} \left[ v \left( {t}_{i} \right) v \left( {t}_{j} \right) \right] = 0 $.

If we can have that, great, we have a White Noise.
It turns out we can't! Why? Read the reference.
It has to do with convergence of a limit and order of integration and the limit operator.

So, what can have?
We can define $ \epsilon > 0 $ arbitrary small for which the auto correlation function is not zero.
It will give us a support, arbitrary large, in the PSD with constant value.
This model of White Noise will obey the needs of Signal Processing, namely for any Linear System with finite bandwidth the properties will be as we know them yet indeed our process will obey the properties of a valid random process, stationarity included.

  • 2
    $\begingroup$ Yet, I've only ever seen the Wiener process defined as the integral of white noise. $\endgroup$
    – TimWescott
    Aug 18, 2023 at 19:21
  • 1
    $\begingroup$ I would, instead, define the Wiener Process (what we Neanderthal electrical engineers call "Brown noise" or "Red noise") as the integral of white noise. And this brown noise has infinite variance, too, (or, at least, grows without bound) if you wait long enough. $\endgroup$ Aug 19, 2023 at 5:27
  • 1
    $\begingroup$ @Royi, this is going to boil down to the same difference in understanding that engineers (and physicists) have with $\delta(t)$ (and treat it as a "function") in comparison to the rigorous mathematical treatment (that makes it into a "distribution" and not a function). We engineers are comfortable with treating the dirac impulse as a special function. If you, somehow, come to a conclusion that "band-unlimited white noise has a variance", then whatever you're talking about is different than what we are talking about. White noise has finite variance only when there is a bandwidth. $\endgroup$ Aug 19, 2023 at 5:33
  • 1
    $\begingroup$ @robertbristow-johnson, Physicists actually don't use the same definition. Again, there are Math laws. If you want the transform from the Auto Correlation to the Power Spectrum, to be Mathematically reasonable there are rules of integration to work by. It has nothing to do with Physicists or Engineers. It has to do with Math. The reason Signal Processing gets away from this is that we don't have a real interest in the Math beast of White Noise, we only care about what happens when it goes through a Linear System with limited bandwidth. $\endgroup$
    – Royi
    Aug 19, 2023 at 5:37
  • 1
    $\begingroup$ @robertbristow-johnson, Also, If you're so sure about your model, I asked you, can you build me a White Noise which has an Exponential distribution? $\endgroup$
    – Royi
    Aug 19, 2023 at 5:43

Well, there's the way that Athanasios Papoulis defines it in Probability, Random Variables, and Stochastic Processes, McGraw-Hill, 1984:

We shall say that a process $\mathbf v(t)$ is white noise if its values $\mathbf v(t_i)$ and $\mathbf v(t_j)$ are uncorrelated for every $t_i$ and $t_j \ne t_i$.

Historically, I'm not sure if this was derived from a spectral point of view (i.e., starting out with a power spectrum and working backwards), or if the concept of "noise that is always different no matter how fine the time interval" came first, and broadband noise came later.

Given that signal processing evolved from pioneer radio technology, and that came from sticking bits of wire and metal plates together first, then identifying inductors, capacitors, and antennas later, I suspect that white noise was first observed and described in terms of spectra, and only later described in the time domain.

If you start with a signal that has a power spectral density $S(\omega) = 1$, then that pretty much defines that it's autocorrelation will be $R(\tau) = \delta(\tau)$. That definition for $R(\tau)$ leads directly* to Papoulis's time-domain definition.

* After, if you're picky about such things, you work through all the infinities involved with trying to apply $\delta(\tau)$ in multiple places.

  • $\begingroup$ The trick is that the property of no correlation is not enough as we need to define a distribution. It is a bit tricky as if you go one path starting with only no correlation you get some "paradoxes". $\endgroup$
    – Royi
    Aug 18, 2023 at 16:46
  • $\begingroup$ @Royi why would you need to define a distribution? You don't! You can have nice gaussian white noise, but just as much uniform distributed. $\endgroup$ Aug 18, 2023 at 18:13
  • 1
    $\begingroup$ @MarcusMüller, Actually you can't :-). This is exactly the point of the question. The model we use in Signal Processing is not coherent Mathematically. There are ways to define it coherently, they require other tools. You may look at my answer and the reference I linked to which shows why you can't have both define the Auto Correlation with Delta and the Power Spectrum as constant and say they are a pair according to Wiener Khinchin Theorem. White Noise is a limit of process. There are some delicate way to work with it which is not a concern for Signal Processing but it is for other fields. $\endgroup$
    – Royi
    Aug 18, 2023 at 18:30
  • 1
    $\begingroup$ The other funny thing is, Tim, is that I got the Papoulis book too. And although he mentions white noise a couple of times (pp 217 and 241), he really stays the hell away from the topic otherwise. That's disappointing. The college references that I am using is A.B. Carlson, Wozencraft and Jacobs, and Van Trees. Papoulis, normally very rigorous, doesn't wanna fuck with it. $\endgroup$ Aug 19, 2023 at 5:48
  • 1
    $\begingroup$ "...difference in understanding that engineers and physical scientists have with mathematicians about $\delta(t)$..." Or if not just that, then the fact that an engineer or physicist takes $\delta(t)$ to be a nice shorthand way of describing something physically impossible and is willing to ignore the 200 pages of exposition one (probably) needs to really nail down what it means, while those 200 pages of exposition are harder for a mathematician to ignore. I suspect that in the case of white noise there's another 200 pages to get from just what we mean by spectrally white to $\delta(t)$. $\endgroup$
    – TimWescott
    Aug 19, 2023 at 14:53

Aaagh, it's 2 in the morning. I'll answer Royi's question sincerely tomorrow.

But it's all gonna boil down to how engineers and physical scientists think about the dirac delta function, $\delta(t)$, versus how the mathematicians think about the same. And this SE is not the same as the math SE. So there is a qualitative difference in the nature of rigor between the two communities.

Strictly speaking, the math guys are right. But it's totally useless in the context of signal processing or even in physical modeling, as per the wonderful quote from Richard Hamming:

Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane.

  • $\begingroup$ Best ideas come late at night :-). I will be reading toughly. Thanks. $\endgroup$
    – Royi
    Aug 19, 2023 at 6:38
  • $\begingroup$ I'm not gonna have anything other than meat and potatoes. $\endgroup$ Aug 19, 2023 at 6:57
  • $\begingroup$ Let $f(x)$ and $g(x)$ be bounded measurable functions on a set of finite measure, $E$. And let $f(x) = g(x)$ "almost everywhere" on the set $x \in E$. Then $$ \int_E f(x) \ \mathrm{d}x = \int_E g(x) \ \mathrm{d}x $$ Now, the problem is that EEs and math people are not going to see this the same. Let $f(x)=\delta(x)$ and $g(x)=0$. And $E=\{x: -\epsilon < x < \epsilon \}$ where $\epsilon>0$. One of them integrals is $1$ and the other is $0$. $\endgroup$ Aug 19, 2023 at 7:06
  • $\begingroup$ You started with "bounded" and then used unbounded function. The delta function is a limit of series of functions. The problem is when you change the orders of the limit and integration without working the delicate Math needed. $\endgroup$
    – Royi
    Aug 19, 2023 at 7:28
  • $\begingroup$ Also, what do you mean by "Now, the problem is that EEs and math people are not going to see this the same". There is one Math to work with. Whether you're EE or Mathematician. The issue is, EE don't need the coherency as they use the Noise only in a single use case where this discrepancy doesn't make a difference. $\endgroup$
    – Royi
    Aug 19, 2023 at 7:30

So, this answer is going to need installments. There needs to be a few chapters to lead up to it. Before getting to the definition of white noise, we gotta begin with stochastic processes (a fancy name for "random processes"), specifically processes that are strictly stationary, more specifically Markov processes.

I am going to make lotsa assumptions at the outset to make my life easier:

  1. This is about finite power signals and not about finite energy signals: $$ 0 \ < \ \lim_{T \to \infty} \frac{1}{T}\int\limits_{-\frac{T}2}^{\frac{T}2} \big|x(t)\big|^2 \ \mathrm{d}t \ < \ \infty $$
  2. So then the appropriate definition of cross-correlation is: $$ R_{xy}(\tau) \triangleq \lim_{T \to \infty} \frac{1}{T}\int\limits_{-\frac{T}2}^{\frac{T}2} x(t+\tau)y^*(t) \ \mathrm{d}t $$
  3. Autocorrelation of $x(t)$ is simply $R_{xx}(\tau)$ and $R_{xx}(0)$ is the mean power of $x(t)$.
  4. $x(t)$ is ergodic in every sense which means that any expressible average or mean w.r.t. time (like those expressed above) can be expressed as a probabilistic average or mean. This is the Expected value (sometimes called "Expectation operator"): $$ \operatorname{E}\Big\{ g(x) \Big\} \triangleq \int\limits_{-\infty}^{\infty} g(\alpha) p_x(\alpha) \ \mathrm{d}\alpha $$ where $p_x(\alpha)$ is the probability density function (p.d.f.) of the random variable $x$ and $g(\cdot)$ is some well-defined function. A random process $x(t)$ sampled at any given time $t$ is a random variable.

So (with another assumption that $x(t)$ and $t$ are both real) the implication that ergodicity has on the autocorrelation is:

$$\begin{align} R_{xx}(\tau) &= \lim_{T \to \infty} \frac{1}{T}\int\limits_{-\frac{T}2}^{\frac{T}2} x(t+\tau)x(t) \ \mathrm{d}t \qquad \qquad & \text{(average w.r.t. time)}\\ \\ &= \operatorname{E}\Big\{ x(t+\tau)x(t) \Big\} \qquad \qquad & \text{(probabilistic average)} \\ \\ &= \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \alpha\beta \ p_{x(t+\tau)x(t)}(\alpha,\beta) \ \mathrm{d}\alpha \, \mathrm{d}\beta\\ \end{align}$$

where $p_{x(t+\tau)x(t)}(\alpha,\beta)$ is the joint p.d.f. for the two random variables $x(t+\tau)$ and $x(t)$.

$ R_{xx}(0) = \operatorname{E}\Big\{ \big|x(t)\big|^2 \Big\}$ is the mean-square of $x(t)$ and, if the mean $\operatorname{E}\Big\{ x(t) \Big\}=0$, the mean-square and variance are equal.

Now let's assume, additionally, that this random process is Gaussian and zero mean:

$$ p_{x(t)}(\alpha) = \frac{1}{\sqrt{2 \pi} \sigma_x} e^{-\frac12 \left(\frac{\alpha}{\sigma_x}\right)^2} $$

Being zero-mean, then the mean-square, $\operatorname{E}\Big\{ \big|x(t)\big|^2 \Big\}$, and variance, $\sigma_x^2$, are the same and larger than zero.

Additionally, we'll assume a Markov process. So the value of the previous states may affect the current state. The value of the state at $x(t)$ may have an effect on the probability of the state at a later time, $x(t+\tau)$.

As a judiciously-arranged example, consider this class of Markov process. Given the known earlier value $x(t)$, the dependent p.d.f. for the later random value $x(t+\tau)$ is:

$$\begin{align} p_{x(t+\tau)}\big(\alpha|x(t)\big) &= \frac{1}{\sqrt{2 \pi \left(\sigma_x^2 - R_{xx}(\tau)\right)}} e^{-\frac12 \frac{\left(\alpha-x(t) \sigma_x^{-2} R_{xx}(\tau)\right)^2}{\sigma_x^2 - R_{xx}(\tau)}} \\ \\ &= \frac{1}{\sqrt{2 \pi} \sigma(\tau)} e^{-\frac12 \left(\frac{\alpha-\mu(\tau)}{\sigma(\tau)}\right)^2} \\ \end{align}$$

where the dependent (on $x(t)$ and $\tau$) mean and variance are:

$$\begin{align} \mu(\tau) &= x(t) \sigma_x^{-2} R_{xx}(\tau) \\ \\ \sigma(\tau) &= \sqrt{\sigma_x^2 - R_{xx}(\tau)} \\ \end{align}$$

I think that the joint p.d.f. is related to the conditional p.d.f. as:

$$\begin{align} p_{x(t+\tau)x(t)}(\alpha,\beta) &= p_{x(t+\tau)}\big(\alpha|\beta \big) \cdot p_{x(t)}(\beta) \\ \\ &= \frac{1}{\sqrt{2 \pi \left(\sigma_x^2 - R_{xx}(\tau)\right)}} e^{-\frac12 \frac{\left(\alpha-\beta\sigma_x^{-2}R_{xx}(\tau)\right)^2}{\sigma_x^2 - R_{xx}(\tau)}} \ \cdot \ \frac{1}{\sqrt{2 \pi} \sigma_x} e^{-\frac12 \left(\frac{\beta}{\sigma_x}\right)^2} \\ \end{align}$$

Then the autocorrelation of this Markov random process is:

$$\begin{align} R_{xx}(\tau) &= \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \alpha\beta \ p_{x(t+\tau)x(t)}(\alpha,\beta) \ \mathrm{d}\alpha \, \mathrm{d}\beta \\ \\ &= \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \alpha\beta \ p_{x(t+\tau)}\big(\alpha|\beta \big) p_{x(t)}(\beta) \ \mathrm{d}\alpha \, \mathrm{d}\beta \\ \\ &= \int\limits_{-\infty}^{\infty} \beta \, p_{x(t)}(\beta) \ \left[ \int\limits_{-\infty}^{\infty} \alpha \ p_{x(t+\tau)}\big(\alpha|\beta \big) \ \mathrm{d}\alpha \right] \ \mathrm{d}\beta \\ \\ &= \int\limits_{-\infty}^{\infty} \beta \, p_{x(t)}(\beta) \ \Big[ \beta \sigma_x^{-2} R_{xx}(\tau) \Big] \ \mathrm{d}\beta \\ \\ &= \left[ \int\limits_{-\infty}^{\infty} \beta^2 \, p_{x(t)}(\beta) \ \mathrm{d}\beta \right] \ \sigma_x^{-2} R_{xx}(\tau) \\ \\ &= \Big[ \sigma_x^2 \Big] \, \sigma_x^{-2} R_{xx}(\tau) \\ \\ &= R_{xx}(\tau) \qquad \qquad \checkmark \\ \end{align}$$

This confirms the choice of dependent mean $\mu(\tau) = x(t) \sigma_x^{-2} R_{xx}(\tau)$.

The choice of dependent variance $\sigma^2(\tau)$ is, perhaps, less critical but we want $\sigma(0)=0$, and when $R_{xx}(\tau)=0$ we want $p_{x(t+\tau)}\big(\alpha|x(t)\big)$ to be independent of $x(t)$ which would mean that $\mu(\tau)=0$ and $\sigma^2(\tau)=\sigma_x^2$. Perhaps, the dependent variance should be $\sigma^2(\tau)=\sigma_x^2-|R_{xx}(\tau)|$. I dunno.

If $R_{xx}(\tau) = \sigma_x^2 \, e^{-|4 \nu \tau|}$ and $\nu > 0$, then this might be the Ornstein-Uhlenbeck process (essentially white noise filtered through a first-order RC low-pass filter) and it might approach Brownian motion (essentially white noise passed through an integrator) as $\nu \to 0$. I would say it approaches white noise as $\nu \to \infty$.

If, instead, $R_{xx}(\tau) = \sigma_x^2 \operatorname{sinc}(2 \nu \tau)$, it might be what I call "bandlimited white noise" (and the bandlimits are $\pm \nu$). Again, send $\nu \to \infty$ and you have a definition for white noise.

So the power spectrum is

$$ S_{xx}(f) \triangleq \mathscr{F}\Big\{ R_{xx}(\tau) \Big\} $$

For the Ornstein-Uhlenbeck process, then $$R_{xx}(\tau) = \sigma_x^2 \, e^{-|4 \nu \tau|}$$ and $$ S_{xx}(f) = \frac{\sigma_x^2}{2\nu} \ \frac{1}{1 + \left( \frac{\pi f}{2 \nu} \right)^2} $$

For the bandlimited white noise process, then $$R_{xx}(\tau) = \sigma_x^2 \operatorname{sinc}(2 \nu \tau)$$ and $$ S_{xx}(f) = \frac{\sigma_x^2}{2\nu} \ \Pi \left( \frac{f}{2\nu} \right) $$

Both of these have $S_{xx}(0) = \frac{\sigma_x^2}{2\nu}$ and an effective noise bandwidth (one-sided) of $\nu$ (two-sided bandwidth is $2\nu$). If the variance $\sigma_x^2$ is fixed to a non-zero value, the value of the power spectrum $S_{xx}(f)$ goes to zero (for all $f$) in the limit as $\nu \to \infty$.

If $S_{xx}(0)$ were to remain constant, say $S_{xx}(0) = \frac{\eta}{2}$, then the variance is $\sigma_x^2 = \eta \nu$ and would increase without bound as the effective noise bandwidth, $\nu$, goes to infinity. Note that, in both processes, as the noise bandwidth $\nu$ increases without bound, the autocorrelation becomes a dirac impulse in the limit.

Ornstein-Uhlenbeck to white noise: $$\begin{align} R_{xx}(\tau) &= \lim_{\nu \to \infty} \eta \nu \, e^{-|4 \nu \tau|} \\ \\ &= \frac{\eta}{2} \delta(\tau) \\ \\ \end{align}$$

bandlimited white noise to white noise: $$\begin{align} R_{xx}(\tau) &= \lim_{\nu \to \infty} \eta \nu \, \operatorname{sinc}(2 \nu \tau) \\ \\ &= \frac{\eta}{2} \delta(\tau) \\ \\ \end{align}$$

And the power spectra are both

$$ S_{xx}(f) = \frac{\eta}{2} \qquad \qquad \forall f \in \mathbb{R} $$

if $S_{xx}(0) = \frac{\eta}{2}$ is held to a constant and bandlimit $\nu$ goes to $\infty$. This is what decorrelates $x(t_1)$ from $x(t_2)$ if $t_1 \ne t_2$, but again, it requires infinite variance, $\sigma_x^2 \to \infty$.


Some definitions so that we can make sure we're all on the same page.

Continuous Fourier transform and inverse: $$\begin{align} \mathscr{F}\Big\{ x(t) \Big\} \triangleq X(f) &= \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \, \mathrm{d}t \\ \\ \mathscr{F}^{-1}\Big\{ X(f) \Big\} \triangleq x(t) &= \int\limits_{-\infty}^{\infty} X(f) \, e^{+j 2 \pi f t} \, \mathrm{d}f \\ \end{align}$$

Rectangular function (sometimes "$\operatorname{rect}(u)$"):

$$\Pi(u) \triangleq \begin{cases} 1 \qquad & \text{ if } |u| < \tfrac12 \\ \tfrac12 \qquad & \text{ if } |u| = \tfrac12 \\ 0 \qquad & \text{ if } |u| > \tfrac12 \\ \end{cases}$$

The inverse Fourier transform of the rectangular function is:

$$ \mathscr{F}^{-1} \left\{ \Pi\left( \tfrac{f}{2\nu} \right) \right\} = 2\nu \, \operatorname{sinc}(2\nu t) $$

The sinc function:

$$\operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} \qquad & \text{ if } u \ne 0 \\ 1 \qquad & \text{ if } u = 0 \\ \end{cases}$$

More definitions and notation:

The probability density function (p.d.f.) of a random variable $x$ is denoted as

$$ p_x(\alpha) \triangleq \lim_{\Delta x \to 0} \frac{1}{\Delta x} \operatorname{P} \Big\{ \alpha-\tfrac12 \Delta x < x < \alpha+\tfrac12\Delta x \Big\} $$

where $\operatorname{P} \Big\{ \alpha-\tfrac12 \Delta x < x < \alpha+\tfrac12\Delta x \Big\}$ is the probability that $x$ exists between $\alpha-\tfrac12 \Delta x$ and $\alpha+\tfrac12 \Delta x$.

The Expected value of some function of single random variable $x$ is

$$ \operatorname{E}\Big\{ g(x) \Big\} \triangleq \int\limits_{-\infty}^{\infty} g(\alpha) \, p_x(\alpha) \ \mathrm{d}\alpha $$

And for a function of two random variables, $x$ and $y$, the Expected value is

$$ \operatorname{E}\Big\{ g(x,y) \Big\} \triangleq \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} g(\alpha,\beta) \, p_{xy}(\alpha,\beta) \ \mathrm{d}\alpha \, \mathrm{d}\beta $$

where $p_{xy}(\alpha,\beta)$ is the joint p.d.f. for random variables $x$ and $y$ and is defined as

$$ p_{xy}(\alpha, \beta) \triangleq \lim_{\Delta x \to 0} \lim_{\Delta y \to 0} \frac{1}{\Delta x} \frac{1}{\Delta y} \ \operatorname{P} \left\{ \begin{matrix} \alpha-\tfrac12 \Delta x < x < \alpha+\tfrac12\Delta x \\ \text{and} \\ \beta-\tfrac12 \Delta y < y < \beta+\tfrac12\Delta y \\ \end{matrix} \right\} $$

  • $\begingroup$ If there's a foul, you get to spell it out. I see no reversal of order. I am yanking multiplicative factors that depend on $\beta$ (which are considered constant w.r.t. $\alpha$) out of the inside integral. $\endgroup$ Aug 20, 2023 at 19:25
  • $\begingroup$ No, @Royi, you spell it out. We got $\LaTeX$ here. You get to spell it out. I'm not looking at the paper. Not at this point. You have to do your due dilligence. $\endgroup$ Aug 20, 2023 at 19:36
  • $\begingroup$ //"I spelled it out."// - - - - That is, what we call in the forensic debate biz, a falsehood. "Denial ain't just a river in Egypt." You get to spell it out, mathematically, or else you're just blowing smoke. I'm done with this now, I'm gonna go canoeing out my backyard here in Vermont. $\endgroup$ Aug 20, 2023 at 19:43
  • 1
    $\begingroup$ No one other than you have said anything about an echo chamber. Are you sure that's the reason you deleted your comments? - - - - As I said originally, even a week ago when I first engaged you about this, White Noise has infinite power. Just as the dirac delta function is unbounded at $\delta(0)$. DC = mean. AC power = variance. DC squared + AC power is total power. Zero mean means that AC power = total power = variance. All growing to infinity without bound as the bandwidth also grows to infinity without bound. And it's not fully "white" if the bandwidth is finite. $\endgroup$ Aug 20, 2023 at 23:41
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    $\begingroup$ Lessee... //"How can one define Continuous White Noise in a coherent way? "// - - - Does my answer do that? $$ $$ //"Is there a way to derive it Mathematically?"// - - - Does my answer do that? $$ $$ //"Specifically, is there a way to define it which will works as the model in Signal Processing yet will obey all Mathematical properties of a a Random Process?"// - - - Does my answer show such a way? $$ $$ @Royi , I picked up the gauntlet that you threw down (to your surprise and chagrin) and you haven't figgered out how to deal with that. Your self-answer is worthless. It's word salad. $\endgroup$ Aug 21, 2023 at 13:12

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