As is shown in wikipedia: Click [here] (http://en.wikipedia.org/wiki/White_noise#Mathematical_definitions)

A random vector (that is, a partially indeterminate process that produces vectors of real numbers) is said to be a white noise vector or white random vector if its components each have a probability distribution with zero mean and finite variance, and are statistically independent.

Here comes my question, does the vector is still white noise, if the variances of the components are not the same any more?

Thank you very much.


2 Answers 2


The components are pairwise independent so they are also pairwise uncorrelated. This implies that the covariance matrix associated with a white random vector is diagonal, i.e.:

$$\mathbf C_x=diag(\lambda_1,\lambda_2,\dots,\lambda_n)$$

If the uncorrelated random variables are further identically distributed, i.e., possess identical statistical characteristics then the random vector is said to be an independent and identically distributed random vector. In this case the covariance matrix becomes an identity matrix:

$$\mathbf C_x=\sigma^2 \mathbf I$$

here $\sigma$ is the common standard deviation of the components. A random vector is said to be weakly white if the components are just statistically uncorrelated. Henceforth when we refer to a white random vector it will mean white in the weak sense.

So if I understand your question correctly the clue is that you need to distinguish if there are random characteristics involved.

See also reference for details >>> here


On the context of random processes, a process $X$ with an auto correlation function $R_X(t,s)=\sigma^2\delta(t-s)$ is considered a (wide sense) stationary white process. However, if $R_X(t,s)=\sigma^2(t)\delta(t-s)$ then the process is still considered white, though not stationary. Translating this logic to random vectors, I guess the answer to your question is yes.


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