| VARIANCE in IDL [message #14412] |
Tue, 23 February 1999 00:00 |
ashmall
Messages: 14
Registered: October 1998
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Junior Member |
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Dear All,
I have a question regarding the variance as calculated by IDL - I expect to
get thoroughly flamed by some statistician types but I'm keen to know if I'm
wrong!
I always thought the definition of variance was the mean of the squares of the
differences from the mean, i.e.:
VARIANCE = { SUM [ (x - mean_x)^2 ] } / N
and this is what I *thought* I was getting from IDL - it wasn't until I was
testing a prog to calculate the means and variances of rows and columns of an
array that I spotted that IDL's variance has N-1 as the denominator:
VARIANCE = { SUM [ (x - mean_x)^2 ] } / N-1
Now I realise the latter ( let's call it Var(n-1) ) is the best estimate of
the variance of the overall population, if my data is a sample from that
population, but that's not what I want (or expect) from the variance function.
More worrying is the fact that this isn't mentioned in any way in the on-line
help for the VARIANCE function (although the equation does appear in the help
on the MOMENT function). Perhaps a keyword to the function would be in order
so you could select if you wanted "population estimate" or "sample" variance
at the very least.
A simple example is given calculating Var(n) and Var(n-1) on the numbers
1,2,3,4,5. The mean is obviously 3 but I would say the variance is 2.0
(Var(n)), not 2.5 as given by IDL (Var(n-1)).
I'd be interested to hear if my definition of variance is correct and whether
other people made the same assumption regarding variance as myself.
Incidentally, I use IDL 5.1.1.
Thanks,
Justin
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