Timeline for Combining Z Scores by Weighted Average. Sanity Check Please?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 30, 2019 at 14:13 | vote | accept | Alexander | ||
| Apr 27, 2019 at 18:02 | comment | added | COOLSerdash | This answer seems to imply that standardization means that the resulting variables are following a standard normal distribution. But standardization only means that the resulting variables will have mean 0 and standard deviation of 1. Standardization will not make non-normal variables into normal ones. | |
| Apr 27, 2019 at 17:54 | history | edited | rpatel | CC BY-SA 4.0 |
fixed square root in denominator for Zw formula
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| Apr 24, 2019 at 22:23 | comment | added | Marouen | Would that still hold if A and B are random vectors? | |
| May 28, 2018 at 15:23 | comment | added | rpatel | It makes sense to weigh an attribute twice the other in a linear combination if it conforms to the goals of your project. That said, it is fine for you to use the original method, as long as you acknowledge that the variable would then have a normal distribution with mean 0 and variance as written above (instead of the standard normal, with mean 0 and variance 1) | |
| May 28, 2018 at 14:50 | comment | added | Alexander | Does this mean that the weighting intuition if "attribute 1 is twice as important as attribute 2, so attribute 1 will have a weight of two and attribute 2 will have a weight of one" is incorrect if calculated this way? Perhaps I should just do a weighted average of the cumulative percentages | |
| May 28, 2018 at 14:46 | comment | added | Alexander | I just ran a test case. Let vector $Z = (-1, 2, 1, 0, -1)$ representing 5 input data's Z-Scores. Let vector $W = (0.2, 0.2, 0.2, 0.2, 0.2)$ represent the weights for these data, equally weighting all data. The cumulative percent for vector $Z$ are $(42\%, 65\%, 58\%, 50\%, 42\%)$ with a mean of $57.9\%$. With the original method, I get weighted average of $0.2$, which translates to a cumulative percent of $52.7\%$. Not the same but close. However if I use your method to obtain $Z_W$, the denominator of $Z_W$ is also $0.2$, which makes $Z_W = 1$. As a result, the cumulative percent is $84.1\%$ | |
| May 28, 2018 at 13:40 | comment | added | Alexander | Got it. Does the sanity check for the following statement hold as well? "If I believe attribute 1 is twice as important as attribute 2, so attribute 1 will have a weight of two and attribute 2 will have a weight of one. I do the same for all observations." | |
| May 28, 2018 at 13:31 | comment | added | rpatel | Exactly correct - read more about linear combinations of normal RVs here: en.wikipedia.org/wiki/… | |
| May 28, 2018 at 12:40 | comment | added | Alexander | Thank you so much in finding an error in my reasoning. Currently reading and analyzing this further. "To make inferences like "a value of W (the weight-averaged Z-scores) =1 is greater than ~84% of observations" would involve having to standardize by dividing W by its standard deviation." I'm just doing a sanity check on that too. The statistic $Z_W$ follows a normal distribution because it is a sum of two normal distributions $Z_A$ and $Z_B$, scaled by constants. Is that correct as well? | |
| May 28, 2018 at 12:21 | history | answered | rpatel | CC BY-SA 4.0 |