If I have a function $$f(r,V,B)=\frac {2V}{r^2B^2} = 2Vr^{-2}B^{-2}$$ what is the propagation of error? If I use the power rules and multiplication rules described here http://www.fas.harvard.edu/~scphys/nsta/error_propagation.pdf and http://www.physics.pomona.edu/sixideas/labs/LRM/LR09.pdf

I get: $$ \sigma_f = \mid f\mid*\sqrt{\frac{4}{r^2}(\sigma_r)^2 + \frac {(\sigma_V)^2}{V^2}+\frac{4}{B^2}(\sigma_B)^2} $$

Is this correct?


Your answer is correct. The underlying theory of propogation is alluded to in the footnote of the first link you provide.

Generally, if you assume that the errors in your measurements (say of, r, V, B to use your example) are unbiased with variance $\sigma_r^2,\sigma_V^2,\sigma_B^2$, respectivley, then a given function of these variables, $f(r,V,B)$, will be affected by the uncertainty in the underlying variables through its first derivative (NOTE: this is only valid if the errors are assumed to be "small", so that the function is well approximated by a line over the expected range of error):

From calculus, the total derivative of $f$ is: $df = \frac{\partial f}{\partial r} dr + \frac{\partial f}{\partial V}dV +\frac{\partial f}{\partial B}dB$.

Now, we re assuming that $dr, dV, dB$ are small, independent, random errors, with $E[dr]=E[dV]=E[dB]=0$ and variances $\sigma_r^2,\sigma_V^2,\sigma_B^2$.

From the linearity of expectation, we get the usual variance formula:

$\sigma_f^2 = \left(\frac{\partial f}{\partial r}\right)^2\sigma_r^2 + \left(\frac{\partial f}{\partial V}\right)^2 \sigma_V^2 +\left(\frac{\partial f}{\partial B}\right)^2 \sigma_B^2$

Therefore, the standard deviation of $f$ is just the square root of the above. If the errors are assumed to be normally distributed, then $f$ is normally distributed about the estiamted value with the above variance.

  • $\begingroup$ I understand the usual variance formula that you provided. However, isn't the propagated error if I use the partial of f with respect to r/V/B different than if I use the power law rules for uncertainty in the first link that I provide? (i.e. (df/dr)^2 does not equal 4/r^2) $\endgroup$ – abalabazn Mar 24 '14 at 15:08
  • $\begingroup$ @user136816 That is true, but you pulled out $f$ in your standard deviation equation. Re-integrate the $f$ into the square root and you will get the proper result. $\endgroup$ – user76844 Mar 24 '14 at 15:12
  • $\begingroup$ I understand now. Many thanks. $\endgroup$ – abalabazn Mar 24 '14 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.