Timeline for Why do we calculate pooled standard deviations by using variances?
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| May 11, 2017 at 0:46 | comment | added | Glen_b | But the variance does something almost as simple as adding, which is why my second sentence is deliberately phrased precisely the way it is. I don't want to add something like what you say there because I disagree with the equivalence it draws between variance and standard deviation when $|\rho|$ is neither $0$ or $1$. Variance is simpler than standard deviation in those cases, and (outside some limited special cases) it's unique among all the ways of measuring spread-outed-ness in that simplicity. | |
| May 11, 2017 at 0:37 | comment | added | user795305 | Yeah, I completely agree. Having $|\rho| = 1$ gives that the random variables are almost surely proportional. This case appears to be much, much less interesting. I'm just trying to point out that I think that an answer to the question should include this. My understanding of your answer is that (simply put and very roughly) "independence gives variances adding, and this is widely useful (seen via t test)" However, a similar argument can't be made for having $|\rho|=1$ since that case is uninteresting. The argument concludes by then noticing that neither the variance nor sd add in other cases. | |
| May 10, 2017 at 23:22 | comment | added | Glen_b | @Ben See my links to the Wikipedia page on basic properties of variance, which shows the non independence case. Our models are often independence models, but it's always true that $\text{Var}(aX+bY)=a^2\sigma^2_X+2ab\sigma_X\sigma_Y\rho_{X,Y}+b^2\sigma^2_Y$. There's no correspondingly-as-simple result for standard deviations (at least that's not simpler still by working in terms of variances), except when $\rho=\pm 1$. How often are we dealing with that case, compared to dealing with $\rho\neq \pm 1$? | |
| May 10, 2017 at 22:25 | comment | added | user795305 | If $\mathrm{cor}(X,Y) = 1$, then $\mathrm{sd}(X+Y) = \mathrm{sd}(X) + \mathrm{sd}(Y)$. In light of this, I think it's fair to say that just because it's true that variances add when $X, Y$ have correlation $0$, doesn't automatically mean it's why we do it. We have to explain why purely uncorrelated random variables are more common to come across (or more useful?) than perfectly correlated ones, I think. | |
| May 10, 2017 at 22:17 | history | edited | Glen_b | CC BY-SA 3.0 |
clarification about distinction between equal-variance version and Welch
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| May 10, 2017 at 16:10 | vote | accept | gabryll | ||
| May 9, 2017 at 10:03 | history | answered | Glen_b | CC BY-SA 3.0 |