When dividing a timeseries by its mean value so that its mean becomes 1, does the resulting data still have a unit or is it unitless?

  • 5
    $\begingroup$ Unitless$\displaystyle$ $\endgroup$ – Henry Sep 15 '14 at 18:19

Since the series and the mean are measured in the same units, the ratio is unit-free.

Adding or subtracting two things in the same units leaves you with the sum or difference in the same units again. But ratios are in the ratio of the units.

If the numerator is in dollars and the denominator is in weeks, you have the ratio in units of dollars per week.

Imagine the units in your series were dollars, or people, or nanometers. The mean is in the same units, so the units of the ratio would be dollars per dollar, or people per person or nanometers per nanometer, any of which cancel out - the result is unit-free.

Similarly when dividing by a standard deviation, mean deviation, interquartile range, median absolute deviation from the median, which are all in the original units ... ratios of anything measured in the same units will cancel out to give a unitless result. So coefficient of variation (sd/mean) for example, is dimensionless, as is moment-based skewness (it's a ratio of cubed-units: third central moment / sd^3).

See the wikipedia article relating to dimensionless quantities:

Dimensionless quantities are often defined as products or ratios of quantities that are not dimensionless, but whose dimensions cancel out

Several sub-sections of the Wikipedia article on dimensional analysis are also relevant.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.