I'm an electrical engineering student studying statistics.

I saw in an article that White Noise variance = power = rms^2.

Could you explain how this expression is derived or recommend reference materials?

And in the case of discrete white noise, does the sampling frequency affect the variance a lot?

  • $\begingroup$ Hi! Can you add the definitions of "Variance", "RMS" and "Power" to your question? That would give us a good start at understanding what you need help with. As it is now, your question is rather broad! $\endgroup$ Commented Sep 6, 2022 at 15:51

2 Answers 2


The relationship between variance and rms value follows directly from the definition of variance. For a continuous random variable, it is defined as:

$$ \sigma^2 = E\left(\left(X - \mu\right)^2\right) $$

Where $\mu$ is the mean of the random variable $X$. Noise processes are typically modeled as zero-mean, so this simplifies to:

$$ \sigma^2 = E\left(X^2\right) $$

So the variance $\sigma^2$ is defined as the expected value, or the mean, of $X^2$.

How does this relate to $X$'s RMS value? RMS stands for "root mean squared", or "the square root of the mean of the square" of the random variable. By inspection, you can see the relationship you asked about:

$$ \text{RMS} = \sqrt{E(X^2)} = \sqrt{\sigma^2} = \sigma $$

$\sigma$ is also known as the standard deviation of the random variable.

  • $\begingroup$ Thank you for your reply. Could you also explain the relationship between Variance and Power? $\endgroup$
    – Jiwon
    Commented Sep 7, 2022 at 7:07

I saw the article variance = power = rms^2 in White noise.

That's true for almost all mean-free signals, not just white noise. For simplicity, we assume a discrete real signal, $x[n]$ of length $N$ and just take a look at the definitions.

$$P = \frac{1}{N}\sum_{n=0}^{N-1} x[n].^2 $$

$$x_{RMS} = \sqrt{\frac{1}{N}\sum_{n=0}^{N-1} x[n].^2} $$

$$\sigma^2_x = \frac{1}{N-1}\sum_{n=0}^{N-1} (x[n]-\mu_x).^2 $$

From the definition it's obvious that $x_{RMS}^2 = P$. Since $x[n]$ is mean free, $\mu_x = 0$ we get $\sigma^2_x = \frac{N}{N-1} P$. For sufficiently large $N$ the difference becomes trivial.

  • $\begingroup$ Your definition of $\sigma_x^2$ is the definition for the canonical estimated variance of a sample set -- if it were my writing and I were being careful, I'd write $\hat \sigma_x^2$, not $\sigma_x^2$. Technically $\sigma_x^2 = E\{(x - \mu_x)^2\}$, and is a function of the probability density of $x$, not of any sample set of $x$. $\endgroup$
    – TimWescott
    Commented Sep 6, 2022 at 20:51

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