Return to Answer

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 form 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.

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.

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 $\sqrt{\frac{N-g}{N-1}}$. In your problem that's $\sqrt{\frac{9-3}{9-1}}=\sqrt{0.75}\approx 0.866$. 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 form 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.

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 form 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.

The answer is both 'no, you can't' and 'yes, but only 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 yours, it's not so good.

It's not possible in general to apply a set of linear transformations that simultaneously standardizes 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 $\sqrt{\frac{N-g}{N-1}}$. In your problem that's $\sqrt{\frac{9-3}{9-1}}=\sqrt{0.75}\approx 0.866$. 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 form 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.

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 $\sqrt{\frac{N-g}{N-1}}$. In your problem that's $\sqrt{\frac{9-3}{9-1}}=\sqrt{0.75}\approx 0.866$. 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 form 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.