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I am trying to understand how to combine uncertainties when they are dependent and independent from each other.

Using this formula :

$$\delta z = \sqrt {\Biggl(\dfrac{\partial f}{\partial x} \delta x\Biggr)^2+\Biggl(\dfrac{\partial f}{\partial y} \delta y\Biggr)^2+2\Biggl(\dfrac{\partial f}{\partial x}\cdot \dfrac{\partial f}{\partial y}\Biggr)\text{cov}(x,y)}$$

Intuitively, if the covariance between the two is zero, the last term will disappear and the equation just becomes the square root of sums for combining uncertainty.

My question is does $\delta z$ then need to be divided by 2 to get the final uncertainty value.(i.e. divide by $N$)

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    $\begingroup$ No. The square root takes care of that. The book(s) by Bevington is an excellent reference source. $\endgroup$
    – Carl Witthoft
    Commented Feb 10, 2014 at 14:53
  • $\begingroup$ @CarlWitthoft: Do you think you might be able to expand your comment into an answer? $\endgroup$
    – Kyle Kanos
    Commented Feb 11, 2014 at 1:33
  • $\begingroup$ @KyleKanos I'll take a whack at it, but it's pretty dang trivial if you just write out the derivative expansion. $\endgroup$
    – Carl Witthoft
    Commented Feb 11, 2014 at 1:56
  • $\begingroup$ @CarlWitthoft: I'd say your comment plus a link to the Bevington book would suffice by my standards. I just hate unanswered questions when a decent one is in the comments. $\endgroup$
    – Kyle Kanos
    Commented Feb 11, 2014 at 1:57
  • $\begingroup$ @Kyle -- how's this answer? $\endgroup$
    – Carl Witthoft
    Commented Feb 11, 2014 at 2:12

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ok, here goes... Direct from "Data Reduction and Error Analysis for the Physical Sciences," Bevington, McGraw-Hill, Chapter Four, when $ z = f(x,y) $ then since $ \sigma_z^2 = \Sigma(x_j-<x>)^2 $ and $\Delta z = \Delta x *\frac{\delta z}{\delta x} + \Delta y * \frac{\delta z}{\delta y}$ , some equation hacking leads to

$ \sigma_z^2 = \sigma_x^2 * (\frac{\delta z}{\delta x})^2 + \sigma_y^2 * (\frac{\delta z}{\delta y} )^2 $+ covariance term

Take the square root of both sides and Bob's your uncle.

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  • $\begingroup$ Looks good to me :D $\endgroup$
    – Kyle Kanos
    Commented Feb 11, 2014 at 2:12
  • $\begingroup$ Thank you! I'm trying to understand how to reduce the uncertainty if I measure with two different devices. Lets say they are totally independent, then using the equation above and combining the two uncertainties, why does the uncertainty not reduce when compared to only using one device. $\endgroup$
    – user2165209
    Commented Feb 12, 2014 at 10:04
  • $\begingroup$ @user2165209 That's a completely different question. What you described here is a repeated measurement of a single quantity. The measurements are independent but the variable is the same. In this case, you're heading towards a "Chi-square" test, so take a look at the literature on that. $\endgroup$
    – Carl Witthoft
    Commented Feb 12, 2014 at 12:26

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