Symbol to indicate normalized or standardized variables

Is there a symbol to indicate that variables have been standardized?

For example, if I have 2 different scoring functions Score1 and Score2. Let's say I want to form a combo score and show that the scores have been standardized, e.g.,

$\text{ComboScore1} = \frac{\text{std. Score1} + \text{std. Score2}}{2}$

two differentiate from a flavor of a ComboScore where the individual scores have not been standardized

$\text{ComboScore2} = \frac{\text{Score1} + \text{Score2}}{2}$

• You are right, I removed the tag
– user39663
Oct 13, 2014 at 2:48
• How are you standardizing exactly? Are these standardized for sample mean and standard deviation, or in some other way? Oct 13, 2014 at 2:50

Standards can vary from one application area to another.

Speaking for myself, I'd probably do something this:

Let $z_1 = \frac{\text{score}_1-\text{?}}{\text{??}}$ (depending on how you're standardizing, I'd fill in the missing parts differently) and similarly for $z_2$. Then

$\text{ComboScore}_1 = \frac{z_1 + z_2}{2}$

Which is to say I'd explicitly (algebraically) define the z-scores, then explicitly algebraically define the combo score.

• Yes, I would mention in the prepending text something like "after standardization (z-score normalization) of the individual scores, the combo score is calculated as follows [ComboScore equation]." That's a good idea, I will just add also something like $Score1_z = \frac{Score1_i - \bar{Score1}}{\sigma_{Score1}}$
– user39663
Oct 13, 2014 at 2:52
• Conventially, $\overline{x}$ indicates a sample mean of $x_i$'s, while $\sigma_x$ indicates a population standard deviation. If you know the population s.d. how is it you don't know the mean? Oct 13, 2014 at 2:55
• I am confused now, where did I say that I don't know the mean?
– user39663
Oct 13, 2014 at 2:58
• The fact that you use a sample mean to standardize rather than the population mean implies you don't know the population mean. If you knew the population parameters, you'd use $\frac{x_i-\mu_x}{\sigma_x}$. If you only had sample quantities you'd used $\frac{x_i-\bar x}{s_x}$. You mixed the two (sample mean, population sd), which leads to the question. Oct 13, 2014 at 2:59
• The answer to that should be fairly clear from the fact that I used $s_x$ for the sample standard deviation of $x$ in my previous comment... it would be $s_{\text{score}_1}$. The use of $s$ for sample standard deviations is pretty conventional. See here for example. Oct 13, 2014 at 3:06