Here there are some formula named, mean square error, balanced relative error etc.

I have been developing an experiment with two methods which of both methods estimate a same size set of values.

I mean at the end of experiments i have same sized;

n real value,

n first experiment result,

n second experiment result,


I need to discuss the methods those i suggested and express some opinion about one of them is more successful on estimation.

First question, does comparison of methods' standard deviation give an idea about their success of estimation?

At the linked page there are some formula, do you think they are appropriate to use one of them in my situation, if so which of them?


The link is not freely available. And your question is not entirely clear. I will guess you are trying to estimate an unknown parameter.

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 value $\theta.$

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 standard deviation.

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 bias.

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.


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