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Context: The variance of a sum of independent random walks is a sum of their variances: $\sigma^2 = \sigma_1^2 + \sigma_2^2$. In case of a dependent random walks with bivariate normal distribution it will be $\sigma^2 = \sigma_1^2 + \sigma_2^2 + 2 \rho \sigma_1 \sigma_2$.

Question: Is there any known class of dependent random walk processes for which the reciprocal of variance is a sum of reciprocals of variances of each random walk process: $1/\sigma^2 = 1/\sigma_1^2 + 1/\sigma_2^2$ ? Can one infer any useful statistical information about such processes, given the relation holds true?

Remark: The relation $1/\sigma_{total} = 1/\sigma_{diffusion-controlled-process} + 1/\sigma_{advection-controlled-process}$ follows from the coupling theory of eddy dispersion [1] (formula 2.10-8): the total dispersion of a chromatographic peak is due to diffusional-based and advectional-base mechanisms. This relation describes a corpus of the experimental data quite well, however Giddings derived the theory with a lot of assumptions and to me it seems more like an empirical correlation, which lacks rigorous statistical reasoning. I am trying to reverse engineer it and find out what kind of a relationship between two processes may lead to this peculiar rule for the sum of the variances.

[1] Giddings, J. C. (1965). Dynamics of Chromatography, Part 1: Principles and Theory. New York, NY, USA: Marcel Dekker.

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    $\begingroup$ It is difficult to see how the relation you quote for chromotographic peaks could be true: it would imply that somehow diffusion and advection together create less dispersion than either one separately. Incidentally, the harmonic mean of any positive numbers $x$ and $y$ is $2/(1/x+1/y)$. $\endgroup$ – whuber Sep 1 '14 at 19:24
  • $\begingroup$ Such an equation would suggest a very particular kind of (possibly quite strong, presumably nonlinear) negative dependence. $\endgroup$ – Glen_b -Reinstate Monica Sep 1 '14 at 22:52
  • $\begingroup$ @whuber Indeed the harmonic mean does not fit here. Thank you for pointing that out. I will change the wording now. $\endgroup$ – Anton Daneyko Sep 2 '14 at 11:51
  • $\begingroup$ @whuber I will try to address the rest of your concerns in several steps. First: I am sorry for misleading you -- the subscript diffusion is not correct (I have changed it now). I did not mean diffusion per se, rather I meant a process, whose dynamics is controlled by diffusion. The derivation of this relationship by Giddings is quite lengthy and probably is not relevant for Cross Validated audience. If you are interested I might lay out the derivation of Giddings somewhere else and point you to it, so you can follow his reasoning. $\endgroup$ – Anton Daneyko Sep 2 '14 at 12:07
  • $\begingroup$ @whuber Second: There has been an argument in the chromatographic community which started in the end of the fifites [1] between the theory which stated that the variances sum up and the one that yields (or rather assumes) the sum of reciprocals of variances. The experimental and simulation data of numerous authors including yours truly [2] corroborate the latter model. I suspect it actually stemmed from the attempt to describe the experimental data. [1] nature.com/nature/journal/v184/n4683/abs/184357a0.html [2] pubs.acs.org/doi/abs/10.1021/ac200424p $\endgroup$ – Anton Daneyko Sep 2 '14 at 12:09
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From

$$1/\sigma^2 = 1/\sigma_1^2 + 1/\sigma_2^2$$ we get

$$\sigma^2 = \frac {\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2}$$

Consider the process

$$Z = c_1X_1+c_2X_2$$ where $c_1, c_2$ are constants. If $X_1$ and $X_2$ are independent, we want to have

$$\text{Var}(Z) = \sigma^2 \Rightarrow c_1^2\sigma_1^2+c_2^2\sigma_2^2 = \frac {\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2}$$

$$\Rightarrow \frac {1}{\sigma_2^2}c_1^2+\frac {1}{\sigma_1^2}c_2^2 = \frac {1}{\sigma_1^2+\sigma_2^2}$$

There are many pairs of $\{c_1,c_2\}$ that satisfy this equation, so there are many linear combinations of the two independent random walks that can satisfy the initial equation.

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