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Suppose one has three quantities $X$, $X_1$ and $X_2$, such that $X = X_1X_2$. Since percentages uncertainties of products just add up we have: $$\frac{\delta X}{X} = \frac{\delta X_1}{X_1} + \frac{\delta X_2}{X_2}$$

therefore: $$\frac{\delta X_1}{X_1} = \frac{\delta X}{X} - \frac{\delta X_2}{X_2}$$

However, since $X_1 = \frac{X}{X_2}$, and percentage uncertainties of ratios also add up: $$\frac{\delta X_1}{X_1} = \frac{\delta X}{X} + \frac{\delta X_2}{X_2}$$

Which directly contradicts the formula given above. How can one resolve this paradox?

Here's what I have thought about to try to resolve the above paradox so far: Let $X, X_1, X_2$ be random variables such that $X = X_1X_2$, if the first and second moments of $X, X_1$ and $X_2$ exist, then: $$\frac{Var[X]}{(E[X_1]E[X_2])^2} = \frac{Var[X_1]}{E[X_1]^2} + \frac{Var[X_2]}{E[X_2]^2} + \frac{Var[X_1]Var[X_2]}{(E[X_1]E[X_2])^2}$$

In practice the third term is negligibly small, also errors in experiments are usually uncorrelated so we have: $$\frac{Var[X]}{E[X]^2} \approx \frac{Var[X_1]}{E[X_1]^2} + \frac{Var[X_2]}{E[X_2]^2}$$

However, it seems to me that the formula for the ratio of two random variables has to be a lot more complicated in general, so I couldn't derive it. So that's where I got stuck.

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However, since $X_1 = \frac{X}{X_2}$, and percentage uncertainties of ratios also add up: $$\frac{\delta X_1}{X_1} = \frac{\delta X}{X} + \frac{\delta X_2}{X_2}$$

Where did you get this formula?

Even for multiplication you have to understand that the "error propagation" equations are not exact. They are approximate, order of magnitude, back of an envelope kind of relations. You may look at "propagation of uncertainties" subject in Google to see how they come up with these relations. In any case, your statement about the propagation of errors in division is unusual at best.

Applying the taylor expansion approach from the second line, you can get easily $$\frac{\delta X_1}{X_1} = \frac{\delta X}{X} - \frac{\delta X_2}{X_2}$$, which is compatible with your base case of $X_1X_2$. This is how it's done. $$\left(\frac{x}{y}\right)'=\frac{x'}{y}-\frac{xy'}{y^2}$$ Divide both sides by $\frac{x}{y}$ and get: $$\frac{\left(\frac{x}{y}\right)'}{\frac{x}{y}} = \frac{x'}{y}\frac{y}{x}-\frac{xy'}{y^2}\frac{y}{x}=\frac{x'}{x}-\frac{y'}{y}$$ If you call $z=\frac{x}{y}$ then you get $$\frac{z'}{z}=\frac{x'}{x}-\frac{y'}{y}$$. Q.E.D.

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  • $\begingroup$ the issue is that I didn't "get," it from anywhere. Essentially, we are told some "error propagation," formulae and that's it, no rhyme nor reason whatsoever. It would be great if you could include the derivation of the propagation of errors formula for division, if it is valid, stating under which conditions it is valid. $\endgroup$ Commented Apr 15, 2015 at 14:15
  • $\begingroup$ Also, I get that they are approximations, in fact for multiplication that is the reason why I used an approximately equals sign, instead of an equals sign. $\endgroup$ Commented Apr 15, 2015 at 14:20
  • $\begingroup$ @11Kilobytes take a look at this section in the link I gave earlier. E.g. using rtaylor expansion, $(\frac{x}{y})'\frac{y}{x}=\frac{x'}{x}-\frac{y'}{y}$, i.e. in your formula for division the sign should be minus, not plus, which resolves your "paradox" $\endgroup$
    – Aksakal
    Commented Apr 15, 2015 at 14:23
  • $\begingroup$ K. I accepted your answer +Aksakal. Oh well, I can't wait until college where I can (hopefully) take a serious statistics course. :D $\endgroup$ Commented Apr 15, 2015 at 14:25
  • $\begingroup$ @11Kilobytes I admire your attitude to a college. I learned this trick not in statistics course but in one of the physics related courses, probably in the physics lab class when learning to process experimental data. $\endgroup$
    – Aksakal
    Commented Apr 15, 2015 at 14:28

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