I am reading this paper. At the first part it says $\langle g_n \rangle=0$ and $\langle g_{n} \rangle ^2=1$. I know that $\langle x \rangle$ is used in physics to refer to the mean. So, the second assertation is related to the mean of $g_{n}^2$? I don't think so. I think it should be the variance of $g_n$. Is that right?

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    $\begingroup$ If the mean of $x$ is 0, then the variance of $x$ simplifies to the mean of $x^2$. Here I am setting aside, as the author is presumably is also doing, any question of a divisor $n - 1$ for a sample estimator. See also e,g, stats.stackexchange.com/questions/187620/… $\endgroup$
    – Nick Cox
    Commented Feb 21, 2016 at 17:43
  • $\begingroup$ I think that the author would have referred to the variance, since that, as you have observed, the variance of $x$ is the mean of $x^2$ if the mean of $x$ is 0. Thank you. $\endgroup$
    – foolcool
    Commented Feb 21, 2016 at 17:53
  • $\begingroup$ I didn't read the paper. Perhaps by "would" you mean "should"? $\endgroup$
    – Nick Cox
    Commented Feb 21, 2016 at 18:07

1 Answer 1


The mean of $g_n^2$ would be $\langle g_n^2 \rangle$ – which, on checking, is what the paper actually says. $\langle g_n\rangle^2$ would refer to the square of the mean of $g_n$.

This means the same thing as $\mathbb E[g_n^2]$. Note that, as Nick pointed out, $$\mathrm{Var}[g_n] = \mathbb E[g_n^2] - \mathbb E[g_n]^2 = \mathbb E[g_n^2],$$ since $\mathbb E[g_n] = 0$, so you're correct that in this case it refers to the variance.

(Sometimes $\langle \cdot \rangle$ is used to refer to the sample mean, but in this paper it's clear that it means the expectation.)

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    $\begingroup$ I've always heard the angle brackets described as "physicists notation". I believe it is widely used in statistical mechanics. $\endgroup$ Commented Feb 21, 2016 at 18:40

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