# Difference of sums or sum of differences this is the question

If you have a bunch of paired measurements, let's say you measured peoples weights before and after holidays (AND you are assuming these are samples from an underlaying Gaussian distribution) what is the correct thing to do, to calculate the difference in weight between before and after holidays?! Difference of averages or average of differences?

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Hint: Suppose you have just two measurements $(x_1, y_1) = (144, 148)$ and $(x_2, y_2) = (110, 120)$ for two people #$1$ and #$2$, What is the difference of the averages? What is the average of the differences? Generalize to the formula that you would use to compute the "difference of averages" and the "average of differences". Use the Gaussian assumption if you like – Dilip Sarwate Dec 1 '11 at 21:59
Both the difference of means and the mean of differences are well-defined calculations, and epsilonhalbe shows they are the same. It can be interesting to check the variances as well. It is possible that the variance before and after the holidays is the same, while the variance in the weight change of each subject is zero. This would be the case if every individual gained or lost the same amount of weight. You need to think about what you want to learn and be clear about what you have done. – Ross Millikan Dec 1 '11 at 22:32
if my answer was helpful, please accept it - or tell me what it lacks to be a good answer – epsilonhalbe Dec 3 '11 at 14:33

If you have $x_1,…,x_n$ sample points (before holidays) and $y_1,…,y_n$ afterwards. If you use the usual arithmetic mean $\overline{x}:=\frac{x_1+…+x_n}{n}$ and $\overline{y}:=\frac{y_1+…+y_n}{n}$ then you have $\overline{x}-\overline{y}=\frac{x_1+…+x_n}{n}-\frac{y_1+…+y_n}{n}=\frac{x_1-y_1+…+x_n-y_n}{n}=\overline{x-y}.$
$\overline{x}:=\sqrt[n]{x_1…x_n}$ and $\sqrt[n]{y_1…y_n}$ then $\overline{x}-\overline{y}\neq\overline{x-y}$ where the latter is defined as $\overline{x-y}:=\sqrt[n]{(x_1-y_1)…(x_n-y_n)}.$