# Justification for removing μ out of brackets in definitions of average of deviations and variance

Here are three equations for some basic parameters.

Average of deviations:

$$\lim_{N\to \infty} \bar{d} = \lim_{N\to \infty} \left[\frac{1}{N}\sum(x_i-\mu) \right]= \lim_{N\to \infty} \left(\frac{1}{N}\sum x_i\right) - \mu = 0$$

Variance: $$\sigma^2 \equiv \lim_{N\to \infty} \left[\frac{1}{N}\sum(x_i-\mu)^2 \right]= \lim_{N\to \infty} \left(\frac{1}{N}\sum x_i^2\right) - \mu^2$$

Variance of a continuous distribution: $$\sigma^2 = \int_{-\infty}^\infty(x-\mu)^2p(x)dx = \int_{-\infty}^\infty x^2p(x)dx - \mu^2$$

In all of the above cases, the mean $$\mu$$ has been taken out of the brackets $$()$$ in what seems to be a violation of algebra laws. What is the justification for that?

Intuitively I think that because $$\mu$$ is a constant, it is unrelated to all of the other things that are going on in there, but I can't fully understand it yet.

You maybe need to consider some intermediate steps to understand what is going on, as these are more or less just the consequences algebraic equalities: Let us use $${\bar{x}_N}$$ for the sample mean of $$x_1,x_2,\ldots,x_N$$. As @MartijnWeterings mentioned, under the right conditions $${\bar{x}_N} \to \mu$$.

In the first case we have

\begin{align}\frac{1}{N} \sum_i (x_i - \mu) &= \frac{1}{N}\left[ \left(\sum_i x_i \right) - \left(\sum_i \mu \right)\right] \\ &= \underbrace{\frac{1}{N} \left(\sum_i x_i \right)}_{={\bar{x}_N}} - \underbrace{\frac{1}{N}\underbrace{\left(\sum_i \mu \right)}_{=N\mu} }_{=\mu} \\ &= {\bar{x}_N} - \mu \to 0 \end{align}

In the first equation I use that the sum commutes, in the second equation I use the distributive law.

You can expand the other two cases with similar intermediate steps, so it is not just a matter of "taking $$\mu$$ out of the brackets".

In the second case consider following steps:

\begin{align} \frac{1}{N} \sum (x_i - \mu)^2 &= \frac{1}{N} \sum (x_i^2 - 2x_i\mu + \mu^2) \\ &= \left(\frac1N\sum x_i^2 \right) - \left(\frac1N\sum 2x_i \mu\right) + \underbrace{\left(\frac1N\sum \mu^2 \right)}_{=\mu^2} \\ &= \left(\frac1N\sum x_i^2 \right) - 2\mu\underbrace{\left(\frac1N\sum x_i \right)}_{=\bar{x}_N} + \underbrace{\left(\frac1N\sum \mu^2 \right)}_{=\mu^2} \\ &\to \lim_{N\to\infty} \left(\frac1N\sum x_i^2 \right) - \mu^2 \end{align}

The third case is just the continuous analogue of the second one and you can use the exact same steps.

• There is a slight complication that has slipped. You have more exactly $\frac{1}{N} \left( \sum_i x_i \right) = \bar{x}$, ie. the sample mean. It is only equal to the popuation mean $\mu$ in the limit $N \to \infty$ (and if the right conditions for the law of large numbers are met). – Sextus Empiricus Apr 3 at 13:35
• @MartijnWeterings Thanks for clarifying I have mistaken $\mu$ for $\bar x$ indeed. – flawr Apr 3 at 13:50
• It is also a bit confusing due to the third case, which doesn't seem to be about the sample but uses the population distribution density $p(x)$. And "average of deviations" is a bit ambiguous (although the formula, with terms like $x_i$, looks like a sample to me). Is it the average of residuals, the average of error-terms, etc. ? – Sextus Empiricus Apr 3 at 13:57