Es seien und b t Prozesse mit weißem Rauschen. Können wir sagen, dass c t = a t + b t notwendigerweise ein Prozess mit weißem Rauschen ist?${a}_{t}$${b}_{t}$${c}_{t}={a}_{t}+{b}_{t}$

**Gaußsches**Rauschen oder weißes Rauschen?

Es seien und b t Prozesse mit weißem Rauschen. Können wir sagen, dass c t = a t + b t notwendigerweise ein Prozess mit weißem Rauschen ist?${a}_{t}$${b}_{t}$${c}_{t}={a}_{t}+{b}_{t}$

1

Was für weißes Rauschen ..?

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Tim
Was ist Ihre Definition von weißem Rauschen?

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Glen_b -Reinstate Monica
Sprechen Sie über weißes **Gaußsches** Rauschen oder weißes Rauschen?

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Mehrdad
Sei . Ist b t ein weißes Rauschen Prozess? Ist b t + a t ? ${b}_{t}=-{a}_{t}$${b}_{t}$${b}_{t}+{a}_{t}$

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user253751
Antworten:

Nein, Sie brauchen mehr (zumindest nach Hayashis Definition von weißem Rauschen). Beispielsweise ist die Summe von zwei *unabhängigen* Weißrauschprozessen Weißrauschen.

Nach Hayashis *Ökonometrie* wird ein stationärer Kovarianzprozess als *weißes Rauschen* definiert, wenn E [ z t ] = 0 und C o v ( z t , z t - j ) = 0 für j ≠ 0 sind .$\{{z}_{t}\}$$\mathrm{E}[{z}_{t}]=0$$\mathrm{C}\mathrm{o}\mathrm{v}({z}_{t},{z}_{t-j})=0$$j\ne 0$

Sei und { b t } ein Prozess mit weißem Rauschen. Definiere c t = a t + b t . Trivialerweise haben wir E [ c t ] = 0 . Überprüfung der Kovarianzbedingung:$\{{a}_{t}\}$$\{{b}_{t}\}$${c}_{t}={a}_{t}+{b}_{t}$$\mathrm{E}[{c}_{t}]=0$

Anwenden von{at}und{b

$$\begin{array}{rl}\mathrm{C}\mathrm{o}\mathrm{v}({c}_{t},{c}_{t-j})& =\mathrm{C}\mathrm{o}\mathrm{v}({a}_{t},{a}_{t-j})+\mathrm{C}\mathrm{o}\mathrm{v}({a}_{t},{b}_{t-j})+\mathrm{C}\mathrm{o}\mathrm{v}({b}_{t},{a}_{t-j})+\mathrm{C}\mathrm{o}\mathrm{v}({b}_{t},{b}_{t-j})\end{array}$$

$\{{a}_{t}\}$ sind weißes Rauschen:
C o v ( c t , c t - j )$\{{b}_{t}\}$$$\begin{array}{rl}\mathrm{C}\mathrm{o}\mathrm{v}({c}_{t},{c}_{t-j})& =\mathrm{C}\mathrm{o}\mathrm{v}({a}_{t},{b}_{t-j})+\mathrm{C}\mathrm{o}\mathrm{v}({b}_{t},{a}_{t-j})\end{array}$$

Ob weißes Rauschen ist, hängt also davon ab, ob C o v ( a t , b t - j ) + C o v ( b t , a t - j ) = 0 für alle j ≠ 0 ist .$\{{c}_{t}\}$$\mathrm{C}\mathrm{o}\mathrm{v}({a}_{t},{b}_{t-j})+\mathrm{C}\mathrm{o}\mathrm{v}({b}_{t},{a}_{t-j})=0$$j\ne 0$

Let $\{{a}_{t}\}$ be white noise. Let ${b}_{t}={a}_{t-1}$. Observe that process $\{{b}_{t}\}$ is also white noise. Let ${c}_{t}={a}_{t}+{b}_{t}$, hence ${c}_{t}={a}_{t}+{a}_{t-1}$, and observe that process $\{{c}_{t}\}$ is not white noise.

Comment to Matthew (add a comment link doesn't work for me): according to more commonly used more rigorous definitions of white noise, even adding two independent white noise sources won't yield true white noise, because the amplitudes are no longer uniform but enveloped.

I'd love to see other, non-econometric, non-economics based definitions of white noise. It's a term that's often thrown around, and I'm not sure how it's used in other fields (or even other definitions used in finance/economics).

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Matthew Gunn
Another example: Let ${d}_{t}=-{a}_{t}$ then ${a}_{t}+{b}_{t}=0$ for all $t$ so not white noise. @MatthewGunn I'd say that the definition in finance would be the same but I don't have a source.

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Bob Jansen
Even simpler than @MatthewGunn's answer,

Consider ${b}_{t}=-{a}_{t}$. Obviously ${c}_{t}\equiv 0$ is not white noise -- it'd be hard to call it any kind of noise.

The broader point is, if we don't know anything about the joint distribution of ${a}_{t}$ and ${b}_{t}$, we won't be able to say what happens when we try and examine objects which depend on both of them. The covariance structure is essential to this end.

Of course, this is exactly the purpose of noise-cancelling headphones! -- to reverse the frequency of external noises and cancel them out -- so, going back to the physical definition of white noise, this sequence is *literal silence*. No noise at all.

0 is a perfectly fine white noise.

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Stig Hemmer
@StigHemmer a usual requirement is that for $j=0$, $\mathrm{Cov}({c}_{t},{c}_{t-j})=\mathrm{Var}({c}_{t})={\sigma}^{2}>0$.

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Therkel
Objection withdrawn.

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Stig Hemmer
@StigHemmer see edit -- it's in fact a very natural definition for 0 *not* to be white noise (in fact it's rather the opposite, by the common definition -- we can exactly predict the value of the sequence given *any* past value)

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MichaelChirico
In electronics, white noise is defined as having a flat frequency spectrum ('white') and being random ('noise'). Noise generally can be contrasted with 'interference', one or more undesired signals being picked up from elsewhere and being added to the signal of interest, and 'distortion', undesired signals being generated from nonlinear processes acting on the signal of interest itself.

While it is possible for two different signals to have correlated parts, and therefore cancel differently at different frequencies or at different times, e.g. completely canceling over a certain band of frequencies or during a certain interval of time, but then not canceling, or even adding constructively over another band of frequencies or during a certain interval of time, the correlation between the two signals presumes a correlation, which is precluded by the presumably random aspect of 'noise', which is what was asked about.

If, indeed, the signals are 'noise' and therefore independent and random, then no such correlations should/would exist, so adding them together will also have a flat frequency spectrum and will therefore also be white.

Also, trivially, if the noises are exactly anti-correlated, then they could cancel to give zero output at all times, which also has a flat frequency spectrum, zero power at all frequencies, which could fall under a sort of degenerate definition of white noise, except that it isn't random and can be perfectly predicted.

Noise in electronics can come from several places. For example, shot noise, arising from the random arrival of electrons in a photocurrent (coming from the random arrival times of photons), and Johnson noise, coming from the Brownian motion of electrons in a resistive element warmer than absolute zero, both produce white noise, although, always with a finite bandwidth at both ends of the spectrum in any real system measured over a finite length of time.

if both white noise sound is traveling in same direction And if their frequency is in phase matched up, then only they get added. But, one thing i am not sure about is after adding up will it remain as white noise or it will become some other type of sound having different frequency.

It seems to me you might be thinking about physical noise rather than statistically? I'm not sure that this answer adds very much - e.g. how can white noise have a single frequency to be matched up? Try looking at a spectrogram of white noise.

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Silverfish
(However, this does appear to be an attempt to answer the question, so reviewers should consider downvoting instead of deletion.)

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Silverfish
The sum of two white noise signals will be white noise if they are *uncorrelated* noise. I also came here from the Hot Network Questions list, not noticing which site it was on. I expect that the statistical definition of white noise is equivalent to the signal-processing definition. About your thought that the two noises will add together occasionally -- yes, they will, but only in certain (random) locations. In other locations, they will subtract. It doesn't stop the result from also being white noise.

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Reinstate Monica
@Justin, *"The sum of two white noise signals will be white noise if they are uncorrelated noise."* - I may be misunderstanding what you mean by "if they are uncorrelated", but under my interpretation your conclusion is wrong. If ${a}_{t}$ is white noise and ${b}_{t}={a}_{t-1}$ (Matthew Gunn's example) then both ${a}_{t}$ and ${b}_{t}$ are white noise, and $\mathrm{c}\mathrm{o}\mathrm{r}({a}_{t},{b}_{t})=0$ for every $t$. Yet, ${c}_{t}={a}_{t}+{b}_{t}$ is not white noise.

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not_bonferroni
@not_bonferroni - Yes, I suppose I was using incorrectly "uncorrelated" to mean "independent".

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Reinstate Monica
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