# Difference between these two equations to calculate uncertainty when multiplying

I have seen these 2 equations for how to calculate the uncertainty when 2 numbers (that have uncertainty) are multiplied.

$$(A\pm a)\times(B\pm b)=(A\times B)\pm \left[\left(\frac{a}{A}\cdot100\right)+\left(\frac{b}{B}\cdot100\right)\right]\%$$

With a very beautiful proof here.

But I have also seen this more "correct" formula:

$$\frac{\sigma_f}{f} = \sqrt{\frac{\sigma_A^2}{A^2} + \frac{\sigma_B^2}{B^2} + 2 \frac{\sigma_{AB}}{AB}}.$$

I have seen the proof of this error propagation formula, but unfortunately, my physics/maths knowledge is stuck at a high school level so I don't really understand it (I regrettably did not study STEM in University).

But my main question is, why is the first equation a mere approximation? What's wrong with the proof linked here? It looks quite valid to me. Why is the proper/more accurate error propagation formula different?

The first equation is not an approximation, it's a conditional equation.

The condition is that the errors, $$a$$ and $$b$$, are uncorrelated.

It might be simpler to first work it out with addition. Suppose I measure 2 quantities--lengths, call them lengths, $$A, B$$ relative to some origin, with $$N$$ samples each. Suppose they have gaussian random error with standard deviations $$\sigma_A, \sigma_B$$.

So our set of measurements is:

$$A_j, B_j;\ \ \ \ \ j=(1, \ldots, N)$$

which means and uncertainties are:

$$\bar A \pm \frac{\sigma_A}{\sqrt N} \equiv \bar A \pm \sigma_{\bar A}$$

$$\bar B \pm \frac{\sigma_B}{\sqrt N}\equiv \bar B \pm \sigma_{\bar B}$$

We want to know the separation, $$L=A-B$$, between $$A$$ and $$B$$.

Our measurement is:

$$\langle L \rangle = (\bar A - \bar B) \pm \sigma_L$$

If the Gaussian random errors are uncorrelated, meaning:

$$\langle a_i b_i \rangle \rightarrow 0;\ \ \ \ N\rightarrow \infty$$

where the $$a_i, b_i$$ are the individual measurement uncertainties.

In our model:

$$\langle a_i b_i \rangle = \sigma_A\sigma_B \langle n^{(a)}_i n^{(b)}_i \rangle$$

where the $$n^{(a,b)}_i$$ are uncorrelated normal Gaussian random numbers, and the expectation of their product is indeed zero. Hence:

$$\sigma^2_L = \sigma_{\bar A}^2 + \sigma_{\bar B}^2$$

If you're learning this, this is a really good time to use your favorite software (python) and just generate some distributions and and look at the standard deviations of difference, correlations, and convergence with $$N$$.

Note that if $$n^{(a)}_i = n^{(b)}_i$$ (complete correlation), they cancel in the $$L_i = A_i - B_i$$ so the estimated error is zero.

Now you may say: but what if the measured difference in $$L_j$$ is not constant for different $$j$$? Well, then your error model is wrong and you get results that can't be interpreted properly.

Suppose you generate 3 sets of random numbers ($$m^{(a)}_i,m^{(b)}_i, n'_i$$) and share one to model correlated random errors, i.e.:

$$n^{(a)}_i = \frac{m^{(a)}_i + C n'_i}{\sqrt{1 + C^2}}$$
$$n^{(b)}_i = \frac{m^{(b)}_i + C n'_i}{\sqrt{1 + C^2}}$$

then you will find that:

$$\sigma^2_L = \sigma^2_{\bar A} + \sigma^2_{\bar B} -2C_{AB} \sigma_{\bar A}\sigma_{\bar B}$$

where

$$C_{AB} = \frac{C^2}{1+C^2}$$

is the coefficient-of-correlation between $$a_i, b_i$$, and has this form because I made a poor choice of definitions.

I strongly recommend using your favorite software (python, numpy) to go investigate. If you (scatter) plot $$b_i$$ vs $$a_i$$ you get a circle/tilted ellipse/line for un-/partial/total correlations. No eclipse pun intended. (and by circle I mean disk with a Gaussian density, and like-wise for the ellipse, but the circle may be an ellipse (though not tilted) if $$\sigma_A \ne \sigma_B$$).