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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?
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.
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.
