It could seem an easy question and without any doubts it is but I'm trying to calculate the variance of white Gaussian noise without any result.

The power spectral density (PSD) of additive white Gaussian noise (AWGN) is $\frac{N_0}{2}$ while the autocorrelation is $\frac{N_0}{2}\delta(\tau)$, so variance is infinite?

  • $\begingroup$ Isn't the noise power the variance of the noise voltage? One could also ask about the variance (or standard deviation) of the power measured over a specific time interval. I think the central limit theorem would describe the relationship between the duration the measurement time and the variance of the results. $\endgroup$
    – user17323
    Commented Sep 9, 2015 at 14:22

3 Answers 3


White Gaussian noise in the continuous-time case is not what is called a second-order process (meaning $E[X^2(t)]$ is finite) and so, yes, the variance is infinite. Fortunately, we can never observe a white noise process (whether Gaussian or not) in nature; it is only observable through some kind of device, e.g. a (BIBO-stable) linear filter with transfer function $H(f)$ in which case what you get is a stationary Gaussian process with power spectral density $\frac{N_0}{2}|H(f)|^2$ and finite variance $$\sigma^2 = \int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,\mathrm df.$$

More than what you probably want to know about white Gaussian noise can be found in the Appendix of this lecture note of mine.

  • 2
    $\begingroup$ The curious thing about this for me is that the $\sigma^2$ parameter that is used as the "variance" of the Gaussian distribution of $x(t)$ is not the variance of the sequence. As you say, it's because $E[x^2(t)]$ is infinite. Thanks for the clear explanation! $\endgroup$
    – Peter K.
    Commented Apr 13, 2013 at 0:29
  • 10
    $\begingroup$ @PeterK. There is a difference between the notions of white Gaussian noise for discrete time and continuous time. If a discrete-time process is considered as samples from a continuous-time process, then, taking into consideration that the sampler is a device with a finite bandwidth, we get a sequence of independent Gaussian random variables of common variance $\sigma^2$ which is what you have in your answer. If your $Y[n]$ is $$Y[n]=\int_{(n-1)T}^{nT}X(t)\,\mathrm dt$$ where $X(t)$ is the OP's AWGN, then $\sigma_{Y[n]}^2=\frac{N_0}{2}T$, not $\frac{N_0}{2}$ as you have it (except if $T=1$). $\endgroup$ Commented Apr 13, 2013 at 1:22
  • 1
    $\begingroup$ @DilipSarwate I read your interesting appendix. But you say " One should not, however, infer that the random variables in the WGN process are themselves Gaussian random variables". I did not fully understand this. If the random variables aren't Gaussian (and this seems reasonable to me since they have infinite variance), why is the process named Gaussian? $\endgroup$ Commented Jul 4, 2017 at 7:04
  • 1
    $\begingroup$ @Surferonthefall Try writing down the probability density function $f_{X(t)}(x)$ of the alleged Gaussian random variables in the white Gaussian noise process $\{X(t)\colon -\infty < t < \infty\}$. The density function has value $0$ for all $x$. How can $X(t)$ be viewed as a Gaussian random variable? As I said repeatedly in the document you read, one should not look too closely at the random variables in a white noise process $\{X(t)\colon -\infty < t < \infty\}$. The process is a mythical one and it is defined by what it produces at the output of linear filter, not by anything else. $\endgroup$ Commented Jul 4, 2017 at 14:21
  • 2
    $\begingroup$ Sorry, that should have read ".... take the limit as $\sigma \to \infty$" not as $\sigma \to 0$. $\endgroup$ Commented Jul 4, 2017 at 21:27

Suppose we have a discrete-time sequence $x[t]$ which is stationary, zero mean, white noise with variance $\sigma^2$. Then the autocorrelation of $x$ is: $$ \begin{array} RR_{xx}[\tau] &=& E\left[ x[t] x[t+\tau] \right]\\ &=& \left \{ \begin{array} EE \left[ x[t]^2 \right], {\rm if\ }\tau=0 \\ 0, {\rm otherwise} \end{array} \right. \\ &=& \sigma^2 \delta[\tau] \end{array} $$ where $\delta[\tau]$ is the Kronecker delta.

So, that implies that $\sigma^2 = \frac{N_0}{2}$.


Yes it is: unless you take into account that infinite power is hard to come by in these post big-bang times. Actually all white noise processes end up in a physical implementation that has a capacitance and thus limits on the effective bandwidth. Consider the (reasonable) arguments leading to Johnson R noise: they would produce infinite energy; except there are always bandwidth limits in implementation. A similar situation applies at the opposite end: 1/F noise. Yes some processes fit 1/f noise very well over a long time; I have measured them. But in the end you are constrained by physical laws.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.