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I am reading the original 1908 paper "The probable error of a mean" from William Sealy Gosset (Student pseudonym) where the Student T Distribution was first derived.

On section 1 (at the end of page 2 in the original paper) the author starts with the following sentence: "Samples of n individuals are drawn out of a population distributed normally, to find an equation which shall represent the frequency of the standard deviations of these samples. If s be the standard deviation found from a sample x1 x2 . . . xn (all these being measured from the mean of the population), then $s^2=\frac{S(x_1^2)}{n}-\left(\frac{S(x_1)}{n}\right)^2=\frac{S(x_1^2)}{n}-\frac{S(x_1^2)}{n^2}-\frac{2S(x_1x_2)}{n^2}$."

The S in the formula above is not previously defined...it may represent a summation (as I guess) or the sample variance. Also it is not clear to me whether x1, x2 etc are single individuals or samples of several individuals...

You see I am pretty confused about the formula above, so I'm not able to follow the (apparently elementary) mathematical derivations in the Section 1 until I don't get this preliminary identity.

Any help would be hugely appreciated.

Brian Moehring
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Andrea Andrea
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  • $S(x_1)$ is what we would write $\sum_{i=1}^n x_i$ nowadays. Similarly, $S(x_1^2)=\sum_{i=1}^n x_i^2$, and (this is the tricky one) $S(x_1x_2)=\sum_{i=1}^{n-1}\sum_{j=1+1}^n x_i x_j$, which we would also write as $\sum_{i<j}x_i x_j$. – kimchi lover Aug 28 '23 at 22:33
  • Many thanks Kimchi lover. So the author of the paper is computing the sample variance about the sample mean: $\frac{\sum_{i=1}^n (x_i-\frac{\sum_{i=1}^nx_i}{n})^2}{n}$ . In this case S(x1 x2) seems to be simply the double sum $\sum_{i=1}^n\sum_{j=1}^nx_ix_j$. I never found this notation for summations elsewhere. – Andrea Andrea Aug 29 '23 at 17:41
  • Thank you @Brian Moehring for correcting the formula. – Andrea Andrea Aug 29 '23 at 19:25
  • @AndreaAndrea You're welcome (honestly, I just didn't want the question derailed by a typo). It seems unlikely that $S(x_1x_2)$ relates to your double sum, though, unless you intend that one of $S(x_1) = \sum_{i=1}^n x_i$ or $S(x_1^2) = \sum_{i=1}^n x_i^2$ is an incorrect interpretation of the notation. Once we have those, we immediately find $S(x_1x_2) = \sum_{1\leq i < j \leq n} x_ix_j$. – Brian Moehring Aug 29 '23 at 19:40
  • @BrianMoehring and kimchilover ...you are perfectly right. I had a momentary eclipse in my brain :-) Sorry, and thanks again – Andrea Andrea Aug 29 '23 at 20:19

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