# What does the SEE measure?

Depending on context, I've seen the term "Standard Error of Estimate" used for both the standard deviation and for the variance of both the standard error and of the residuals. Is SEE a well-defined term that is often misused, or is it a casual concept that can be applied to different situations as need be?

Frustratingly, there does not seem to be a page on wolfram.com, which is my usual go-to place when I find contradictory information online. If there is a better resource to be had, I would much appreciate knowing about it. I am learning statistics from an online course where there is little facility for asking such questions.

• You are always welcome to ask questions here. – gung - Reinstate Monica Nov 30 '14 at 2:18

An estimator in the most general sense is a (deterministic) mathematical function of all possible potential data. That is, if the data from an experiment will be considered as an ordered $n$-tuple $(x_1, x_2, \ldots, x_n)$, then an estimator is a function

$$t:\mathbb{R}^n \to \mathbb{R}$$

which produces a number $t(x_1, x_2, \ldots, x_n)$ (the estimate) for each possible element $(x_1, x_2, \ldots, x_n)$ of $\mathbb{R}^n$ (the data).

When the data are modeled as the outcome of a (vector-valued) random variable $(X_1, X_2, \ldots, X_n)$, then

$$T = t(X_1, X_2, \ldots, X_n)$$

is itself a random variable (by virtue of the randomness of its argument) provided $t$ is measurable (which is an important technical condition that for conceptual purposes may be ignored). (For more about measurability, see the "afterward" in the preceding link.)

The standard error of estimate of $t$ is the standard deviation (SD) of $T$.

A far-reaching example is to take a random sample $(X_1, X_2, \ldots, X_n)$ (independently, with replacement) of a population having mean $\mu$ and standard deviation $\sigma$ which are unknown. In order to estimate $\mu$ one might use the sample mean,

$$t(x_1, x_2, \ldots, x_n) = \frac{1}{n}(x_1 + x_2 + \cdots + x_n).$$

Probability theory, as supported by the assumptions about the sample, tells us that $t$ estimates the mean in the sense that

$$\mathbb{E}(T) = \mu$$

and it provides a measures of how closely it tends to estimate the mean, such as

$$\text{SD}(T) = \frac{\sigma}{\sqrt{n}}.$$

This value is the standard error of the [estimate of the] mean, where the term "mean" describes $T$ (the mean of the sample), not $\mu$ (the mean of the population)! The fact that it is also a standard deviation (of the random variable $T$) may lead to misunderstandings among those who assume that "standard deviation" applies only to the population.

• Thank you very much! This is an outstanding explanation, slightly above my level of understanding but close enough that I can google the concepts that I am missing. Thank you whuber! – dotancohen Nov 30 '14 at 14:01