RMS averaging in spectrum analyzers
A straightforward way of estimating the PSD is to compute magnitude squared DFT (a.k.a. periodogram).

Suppose we have an $N$ point sequence $x(n)$ and we would like to estimate its power spectral density (PSD) $P_{xx}(f)$.

Method 1: Periodogram 

We compute the magnitude squared DFT. No averaging.
$$
\hat{P}_{xx}(f) = \frac{1}{N}\left |  \sum_{i=0}^{N-1} x(n) e^{-j2\pi fn/N} \right |^2
$$

Method 2: Periodogram Averaging

We instead use $M$ different realizations of the $N$-point sequence $x(n)$ giving us $M$ sequences $x_i(n)$, $0\leq i\leq M-1$. We compute the periodogram estimate $\hat{P}_{x_ix_i}(f)$ of each $x_i$ and average them to get the final PSD estimate:
$$
\hat{P}(f) = \frac{1}{M}\sum_{i=0}^{M-1} \hat{P}_{x_ix_i}(f).
$$

It can be shown that the variance of this averaged periodogram estimate is lower:

$$
\mbox{Var}[\hat{P}(f)] = \frac{1}{M} \mbox{Var}[\hat{P}_{xx}(f)].
$$

Note that it is a little unfair to compare Methods 1 and 2 because Method 2 uses more data ($MN$ data samples, instead of just $N$). If all you have is the original $N$ point sequence $x(n)$, you end up making a tradeoff between PSD estimator's variance and frequency resolution.

The PSD of the underlying signal is what it is. A computational routine will produce an estimate of this PSD. Whether a PSD estimate is good or bad will depend on your definition of good or bad. One commonly used metric variance. In that sense, periodogram averaging estimate $\hat{P}$ is a better estimator of the PSD $P_{xx}$ than $\hat{P}_{xx}$.

cf. Bartlett Method, Welch Method

Reference: Digital Signal Processing, Proakis and Manolakis, Ch. 14 (4th ed.)