Say you had 3 independent variables - A, B, and C
with mean values and associated variances (the actual quantities are just an example):
$\mu_A=1$; $\sigma^2_A=0.01$
$\mu_B=2$; $\sigma^2_B=0.02$
$\mu_C=2$; $\sigma^2_C=0.03$
and then you construct two ratios: $R_{AB}$ and $R_{AC}$ - which should have the same mean value.
$R_{AB} = \frac{A}{B}$; $R_{AC} = \frac{A}{C}$;
with mean values and associated variances (Formula for variance of a ratio (Wikipedia) - Section: Example formulae):
$\mu_{AB} = \frac{\mu_{A}}{\mu_{B}}$; $\sigma^2_{AB} = \left(\frac{\mu_{A}}{\mu_{B}}\right)^2 \left[ \left(\frac{\sigma_A}{\mu_A}\right)^2 + \left(\frac{\sigma_B}{\mu_B}\right)^2 \right]$
$\mu_{AC} = \frac{\mu_{A}}{\mu_{C}}$; $\sigma^2_{AC} = \left(\frac{\mu_{A}}{\mu_{C}}\right)^2 \left[ \left(\frac{\sigma_A}{\mu_A}\right)^2 + \left(\frac{\sigma_C}{\mu_C}\right)^2 \right]$
how would you find a linear combination of these two ratios that has a minimum possible variance?
$R = w_0R_{AB} + w_1R_{AC}$, where $w_0+w_1=1$
I initially used the inverse-variance weighting method to find the weights: Inverse-variance weighting (Wikipedia) but then realized I couldn't because the two ratios $R_{AB}$ and $R_{AC}$ are not independent and have a non-zero covariance due to their common numerator.