क्या दो सफेद शोर प्रक्रियाओं का योग जरूरी एक सफेद शोर है?

Question

आइए और सफेद शोर प्रक्रियाएं हैं। क्या हम कह सकते हैं कि एक सफेद शोर प्रक्रिया है?$a_t$$b_t$$c_t=a_t+b_t$

Comments

कैसा सफ़ेद शोर ..?

— टिम

सफेद शोर की आपकी परिभाषा क्या है?

— Glen_b -Reinstate मोनिका

क्या आप सफेद गाऊसी शोर या सफेद शोर के बारे में बात कर रहे हैं ?

— मेहरदाद

आज्ञा देना । क्या एक सफेद शोर प्रक्रिया है? है ? $b_t = -a_t$$b_t$$b_t + a_t$

— user253751

Answer 1

नहीं, आपको अधिक (कम से कम हेयाशी की परिभाषा सफेद शोर के तहत) की आवश्यकता है। उदाहरण के लिए, दो स्वतंत्र का योग सफेद शोर प्रक्रियाओं का सफेद शोर है।

क्यों है और बी टी सफेद के लिए शोर अपर्याप्त एक टी + बी $a_t$$b_t$$a_t+b_t$ सफेद शोर होने के लिए?

हयाशी के अर्थमिति के बाद , एक सहसंयोजी स्थिर प्रक्रिया को सफेद शोर के रूप में परिभाषित किया जाता है यदि E [ z t ] = 0 और C o v ( z t , z t - j ) = 0 के लिए j ≠ $\{z_t\}$$\mathrm{E}[z_t] = 0$$\mathrm{Cov}\left(z_t, z_{t-j} \right) = 0$$j \neq 0$ ।

चलो और { ख टी } सफेद शोर प्रक्रियाओं हो। C t = a t + b t को परिभाषित करें । तुच्छ रूप से हमारे पास E [ c t ] = 0 है । Covariance स्थिति की जाँच:$\{a_t\}$$\{b_t\}$$c_t = a_t + b_t$$\mathrm{E}[c_t] = 0$

उस{at}और{b कोलागू करना

$$\begin{align*} \mathrm{Cov} \left( c_t, c_{t-j} \right) &= \mathrm{Cov} \left( a_t, a_{t-j}\right) + \mathrm{Cov} \left( a_t, b_{t-j}\right) + \mathrm{Cov} \left( b_t, a_{t-j}\right) + \mathrm{Cov} \left( b_t, b_{t-j}\right) \end{align*}$$ $\{a_t\}$ सफेद शोर हैं: सी ओ वी ( सी टी , सी टी - जे$\{b_t\}$ $$\begin{align*} \mathrm{Cov} \left( c_t, c_{t-j} \right) &= \mathrm{Cov} \left( a_t, b_{t-j}\right) + \mathrm{Cov} \left( b_t, a_{t-j}\right) \end{align*}$$

तो क्या सफेद शोर है, चाहे C o v v ( a t , b t - j ) + C o v ( b t , a t - j ) = 0 सभी j ≠ 0 के लिए निर्भर करता है$\{c_t\}$$\mathrm{Cov} \left( a_t, b_{t-j}\right) + \mathrm{Cov} \left( b_t, a_{t-j}\right) = 0$$j\neq 0$ ।

उदाहरण जहां दो सफेद शोर प्रक्रियाओं का योग सफेद शोर नहीं है:

चलो सफेद शोर हो। चलो बी टी = ए टी - 1 । उस प्रक्रिया का निरीक्षण करें { b t } भी सफेद शोर है। चलो c t = a t + b t , इसलिए c t = a t + a t - 1 , और उस प्रक्रिया का निरीक्षण करें { c t } सफेद शोर नहीं है।$\{a_t\}$$b_t = a_{t-1}$$\{b_t\}$$c_t = a_t + b_t$$c_t = a_t + a_{t-1}$$\{c_t\}$

Comments

मैथ्यू पर टिप्पणी करें (एक टिप्पणी लिंक मेरे लिए काम नहीं करता है): सफेद शोर की अधिक सामान्यतः उपयोग की जाने वाली अधिक कठोर परिभाषाओं के अनुसार, यहां तक ​​कि दो स्वतंत्र सफेद शोर स्रोतों को जोड़ने से सही सफेद शोर नहीं होगा, क्योंकि एम्पलीट्यूड अब समान नहीं हैं लेकिन छा गया।

मैं सफेद शोर के अन्य, गैर-अर्थमितीय, गैर-अर्थशास्त्र आधारित परिभाषाओं को देखना पसंद करूंगा। यह एक ऐसा शब्द है जिसे अक्सर चारों ओर फेंक दिया जाता है, और मुझे यकीन नहीं है कि इसका उपयोग अन्य क्षेत्रों में कैसे किया जाता है (या वित्त / अर्थशास्त्र में उपयोग की जाने वाली अन्य परिभाषाएं)।

— मैथ्यू गन

$d_t = -a_t$$a_t + b_t = 0$$t$

Answer 2

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.

---

Addendum:

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.

Comments

0 is a perfectly fine white noise.

— Stig Hemmer

@StigHemmer a usual requirement is that for $j = 0$, $\operatorname{Cov}(c_t,c_{t-j}) = \operatorname{Var}(c_t) = \sigma^2>0$.

— Therkel

Objection withdrawn.

— 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)

— MichaelChirico

Answer 3

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.

Answer 4

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.

Comments

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.

— Silverfish

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

— 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.

— 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 ${\rm cor}(a_t, b_t) = 0$ for every $t$. Yet, $c_t = a_t + b_t$ is not white noise.

— not_bonferroni

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

— Reinstate Monica