Re: time series with binary data


You're asking for an index that weights periods of not working?

If the number of days in the gaps where it doesn't work are g(1), g(2),
.... g(k)
what about something like index = 1 - [Sum_i
{g(i)^(1+m)}/90^(1+m)}^{1/(1+m)}

where m is some small positive number (between 0 and say 3, most likely
between 0 and 1).

for example, if we have two gaps of 4 days each and take m=1, the index
is
1 - sqrt[(4^2 + 4^2)/90^2] = 0.937

If we have 7 two day gaps, and m=1 index = 1 - sqrt[(7*2^2)/90^2] =
0.941

(so two 4-day gaps is slightly worse than 7 two-day gaps)

Larger m penalises long gaps more.

If you give more of an idea of what should produce some of the
intermediate values, more could be done to help you construct a
function.

Does this help?

Glen


You can tune m so as to give the sort of ordering you ask for in the
single four day gap vs every third day.


There are many easy-to-write-down functions that might suit your
purpose. It might be easier to make suggestions if you gave an example
of what you'd like to come out somewhere near 0.5, and maybe 0.1, 0.25,
0.75 and 0.9.

.

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