Discrete white noise

Question

I have a binary time series $\{X(n), n= 0,1,2,\cdots\}$, $X(n)\in \{0,1\}$ that can be written as: $$X(n)=S(n)+\epsilon(n)$$ Where $S(n)$ is a random stationary binary signal and $\epsilon(n)$ is a zero-mean binary white noise process (uncorrelated with $S(n)$). My question is what is the definition of a binary (or discrete) white noise?

Answer 1

Assuming that the $S(n)$ are also binary (taking on values in $\{0,1\}$) as are the $X(n)$, then I suspect that the $\epsilon(n)$ are also meant to be taking on values in $\{0,1\}$ and that $+$ in $X(n) = S(n) + \epsilon(n)$ is intended to be a modulo 2 sum or Exclusive-OR sum which might be better written as $$X(n) = S(n)\oplus \epsilon(n)\tag{1}$$ or, in real-number arithmetic as $$X(n) = S(n) + \epsilon(n) - 2S(n)\epsilon(n).\tag{2}$$

The model for this white noise process is an IID sequence of Bernoulli random variables with parameter $p$ that are independent of the $S(n)$ series.

Yes, I know that you said uncorrelated, but uncorrelated Bernoulli random variables are also independent random variables.

Readers of stats.SE and time-series books will undoubtedly be horrified at the nonlinear equation $(2)$, but this model is quite commonly used in the communications and information theory literature under the name Binary Symmetric Channel with crossover probability $p$ and readers of dsp.SE will be quite familiar with the model.

Answer 2

Discrete white noise definition is quite similar to continuous noise definition, meaning that it has mean zero (constant), and its variance is also constant (nonzero), and there's no autocorrelation:

$$E[\varepsilon(n)]=0$$ $$Var[\varepsilon(n)]=\sigma^2$$ $$E[\varepsilon(n)\varepsilon(n-k)]=0,\space k>0 $$

Example: $\varepsilon(n)\sim Pois(\lambda)-\lambda$, you have: $$E[\varepsilon(n)]=0$$ $$Var[\varepsilon(n)]=\lambda$$ $$E[\varepsilon(n)\varepsilon(n-k)]=0,\space k>0 $$

This is used widely in finance to model so called jump processes as opposed to diffusion, which is a usual Brownian motion. Also, look up Levy and Poisson processes. This one is interesting that it has some similarities to Brownian motion since Poisson distribution like Gaussian is a stable distribution. So accumulation of this particular noise remains to be Poisson!