3
$\begingroup$

Consider a data set $X$ made up of smaller subsets: $X=A \cup B \cup C$, with $A,B,C$ disjoint data sets. Eg: $A=\{1.0, -1.0, 0\}$, $B=\{5.0, -7.0, 2.0\}$, $C=\{1.5, -5.0, 8.0\}$

Is is possible to transform (scale, shift) $A,B,C$ such that each is individually standardised to have mean=0, and std dev=1.0, and so that $X$ also has mean=0, std dev=1.0?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ You might get some strange answers to this question because it conflates standardized (having zero mean and unit standard deviation) with normal. People will point out that $X$ itself cannot possibly be normally distributed (for one thing, it's a discrete finite set of values) and that if you mean for $X$ to become approximately normally distributed, even achieving that will generally be impossible via only changes of scale and location. So what are you really asking--is your question about standardization or is it truly about normal distributions? $\endgroup$
    – whuber ♦
    Commented May 25, 2014 at 1:53

1 Answer 1

Reset to default
2
$\begingroup$

The answer is both 'no, you can't' and 'well, yes, sort of' -- it depends on how strict you want to be about things.

Once you standardize the subgroups there are no degrees of freedom left in linear transformations to do anything about the overall variable, so you're stuck with whatever that achieves. However, the situation isn't necessarily bad, it depends on what you need.

The mean is no issue, that works simply by standardizing the subgroups.

It cannot be achieved exactly for the usual Bessel-corrected standard deviation (the $n-1$ denominator version). It will only approximately apply - but in large samples, the approximation is very good. In small samples like your example with three groups of 3, it's not so good.

It's not possible in general to simultaneously standardize the subgroups and the whole sample, because making the individual subgroup standard deviations exactly $1$ relies on dividing by $n_i-1$ (the degrees of freedom).

This means that the sums of squared deviations (from $0$, since they're already standardized) for each subgroup is $n_i-1$, so when you sum the squared deviations for all standardized subgroups concatenated into one single group, their sum of squares is $N-g$, where $g$ is the number of groups and $N=\sum_{i=1}^g n_i$.

This means for the overall variable, the usual Bessel-corrected standard deviation will be $\sqrt{\frac{N-g}{N-1}}\approx 1-\frac{g}{2N}$. In your problem that's $\sqrt{\frac{9-3}{9-1}}=\sqrt{0.75}\approx 0.866$ (the $1-\frac{g}{2N}$ approximation gives $0.833$). That's a fair way short of 1.

If you used the $n$-denominator version of standard deviation for both the subgroups and the overall variable, then it works "automatically" - you get the overall variable standardized for free.

So if you want both standardized you must settle for it being only approximate in not-too-small samples, or you must settle for defining standard deviation as the $n$-denominator form.

Edit: it seems from your comments that your sample sizes are very large (on the order of 200 per subgroup or so) so the effect will be very small. For most people the small difference from 1 will be of little consequence in that situation.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ My actual dataset is about 20,000 and i would break it up into less than 100 subsets, so that gives me a bessel-corrected std deviation of >0.9975. Should I edit my question to reflect this? $\endgroup$
    – Frames Catherine White
    Commented May 25, 2014 at 2:49
  • $\begingroup$ No need to edit the question; I think the discussion of your question in the small sample case is broadly useful even if it doesn't relate to your needs. I'll add an edit that relates to your comment. $\endgroup$
    – Glen_b
    Commented May 25, 2014 at 2:51
  • $\begingroup$ Is the n-denominator form the regular, high-school stats form? I can't find any reference to what the n-denominator form is on w quick google. $\endgroup$
    – Frames Catherine White
    Commented May 25, 2014 at 2:56
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Glen_b
    Commented May 25, 2014 at 2:59
  • $\begingroup$ "The sums of squared deviations (from 0, since they're already standardized) for each subgroup is $n_i−1$, so when you sum the squared deviations for all standardized subgroups concatenated into one single group, their sum of squares is $N−g$." Is this correct? The (squared) deviations from the mean of a standardised subset are likely to be much smaller than the deviations from the mean of the whole data set — consider the subset $\{-1, -1, -1\}$ of an otherwise standardised data set. You can't just add them to get the squared deviations for the entire set. $\endgroup$
    – sjy
    Commented May 30, 2014 at 6:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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