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Wonder if anyone can help.

I have a set of lists of numbers, around 300 lists in the set, each list of around 200 numbers. What I wish to calculate is the "relative stability" of each list.

For example:

List A: 100,101,103,99,98 - the range is v small - so stable.
List B: 0.3, 0.1, -0.2, 0.1 - again, v small range, so stable.
List C: 0.00003, 0.00002, 0.00007, 0.00008 - stable.

Now, I could use standard deviation - but the values returned by the standard deviation will be relative to the values within each list. So std for list C would be tiny in comparison to std for list A - and therefore numerically not give me a comparable measure of stability/volatility enabling me to meaningfully ask: if list A more or less stable than list C?

So, I wondered if anyone had any suggestions for a measure that will be comparable across such lists?

Many thanks for any help. R

Martin Evans
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Roland Dunn
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  • Your lists B and C are not "stable" at all. Their variation coefficients are 238% and 48% (vs 1.7% for A). it seems that you have a wrong notion of stability. –  Jul 03 '17 at 22:12

1 Answers1

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You could use the standard deviation of the list divided by the mean of the list.

Those measures have the same units so their quotient will be a pure number, without a unit. This scales the variability (standard deviation) to the size of the numbers (mean).

The main difficulty with this is for lists that have both positive and negative numbers, like your List B. The mean could end up being an order of magnitude less that the numbers, exaggerating the stability measure. Worse, the mean could end up being zero, making the measure undefined. I cannot think of any correction that would work well in all cases. The "stability" of a list with both positive and negative numbers is very doubtful and would depend on the context, so I doubt that any general stability measure would work well in all such cases. You would need a variety for different situations.

Rory Daulton
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  • Thanks to both for your answers. Variation coefficient is probably what I'm looking for. Thanks v much. All v helpful. – Roland Dunn Jul 04 '17 at 08:53