The link is not freely available. And your question is not entirely clear. I will guess you are trying to estimate an unknown
Suppose $\theta$ is the unknown parameter, and you have an
estimator $T$ of $\theta$ based on $n$ observations.
We say that $T$ is an unbiased estimator of $\theta$ if $E(T) = \theta.$
If you have two unbiased estimators $T_1$ and $T_2,$ then
the estimator with the smaller standard deviation is considered
to be better because it is more likely to be near the correct
Two simple examples: (1) If the data are from a normal
distribution with unknown mean $\mu$, then the sample mean
$\bar X$ and the sample median $\tilde X$ of the $n$
observations are both unbiased. In this case, the sample
mean is considered the better estimator because it has smaller
standard deviation (or variance).
(2) If the data are from a population distributed $Unif(0, \theta),$
then twice the mean and $(n+1)/n$ times the maximum are both
unbiased, and the latter is preferred because it has the smaller
However, if estimators are biased, then the variance or SD
is no longer an optimal guide. They can be in error because
of the bias or because of sampling variability and it is
difficult compare such estimators using the SD or variance alone.
For biased estimators one reasonable criterion for 'goodness'
is to have a small mean squared error. One can show that
$$MSE_\theta (T) = E[(T-\theta)^2] = V(T) + [b_\theta(T)]^2,$$
where $[b_\theta (T)]^2 = [E(T)-\theta]^2$ is the square of the
In example (2) above, the maximum is biased unless it is
multiplied by $(n+1)/n.$ However, the maximum has smaller MSE
than double the mean, and so the maximum is considered the
better estimator according to the MSE criterion.