Let $a_t$ and $b_t$ be white noise processes. Can we say $c_t=a_t+b_t$ is necessarily a white noise process?

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    $\begingroup$ What kind of white noise..? $\endgroup$
    – Tim
    Oct 9, 2016 at 18:07
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    $\begingroup$ What's your definition of white noise? $\endgroup$
    – Glen_b
    Oct 9, 2016 at 22:53
  • $\begingroup$ Are you talking about white Gaussian noise or white noise? $\endgroup$
    – user541686
    Oct 10, 2016 at 18:42
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    $\begingroup$ Let $b_t = -a_t$. Is $b_t$ a white noise process? Is $b_t + a_t$? $\endgroup$
    – user253751
    Oct 11, 2016 at 10:20

4 Answers 4


No, you need more (at least under Hayashi's definition of white noise). For example, the sum of two independent white noise processes is white noise.

Why is $a_t$ and $b_t$ white noise insufficient for $a_t+b_t$ to be white noise?

Following Hayashi's Econometrics, a covariance stationary process $\{z_t\}$ is defined to be white noise if $\mathrm{E}[z_t] = 0$ and $\mathrm{Cov}\left(z_t, z_{t-j} \right) = 0$ for $j \neq 0$.

Let $\{a_t\}$ and $\{b_t\}$ be white noise processes. Define $c_t = a_t + b_t$. Trivially we have $\mathrm{E}[c_t] = 0$. Checking the covariance condition:

\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*} Applying that $\{a_t\}$ and $\{b_t\}$ are white noise: \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*}

So whether $\{c_t\}$ is white noise depends on whether $\mathrm{Cov} \left( a_t, b_{t-j}\right) + \mathrm{Cov} \left( b_t, a_{t-j}\right) = 0$ for all $j\neq 0$.

Example where sum of two white noise processes is not white noise:

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.

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    $\begingroup$ 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. $\endgroup$
    – user134135
    Oct 10, 2016 at 2:18
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    $\begingroup$ 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). $\endgroup$ Oct 10, 2016 at 3:19
  • $\begingroup$ 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. $\endgroup$
    – Bob Jansen
    Oct 11, 2016 at 13:31

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.

  • $\begingroup$ 0 is a perfectly fine white noise. $\endgroup$ Oct 10, 2016 at 7:10
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    $\begingroup$ @StigHemmer a usual requirement is that for $j = 0$, $\operatorname{Cov}(c_t,c_{t-j}) = \operatorname{Var}(c_t) = \sigma^2>0$. $\endgroup$
    – Therkel
    Oct 10, 2016 at 9:21
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    $\begingroup$ Objection withdrawn. $\endgroup$ Oct 10, 2016 at 9:30
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    $\begingroup$ @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) $\endgroup$ Oct 10, 2016 at 20:15

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.

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    $\begingroup$ 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. $\endgroup$
    – Silverfish
    Oct 10, 2016 at 10:25
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    $\begingroup$ (However, this does appear to be an attempt to answer the question, so reviewers should consider downvoting instead of deletion.) $\endgroup$
    – Silverfish
    Oct 10, 2016 at 10:25
  • $\begingroup$ 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. $\endgroup$
    – Justin
    Oct 10, 2016 at 14:14
  • $\begingroup$ @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. $\endgroup$ Oct 11, 2016 at 14:56
  • $\begingroup$ @not_bonferroni - Yes, I suppose I was using incorrectly "uncorrelated" to mean "independent". $\endgroup$
    – Justin
    Oct 11, 2016 at 17:53

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