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I am trying to figure out an equation from the following paper by Cadena and Kovak (2016): http://pubs.aeaweb.org/doi/pdfplus/10.1257/app.20140095 on pages 264 and 265. I don't think the context is necessarily important for the purposes here, but the math definitely is.

They define a variable with a dot over it as: $\dot{x} = \log(x_1) - \log(x_0)$

Defining $\gamma_{ic}^{t_0} = \frac{L_{ic}^{t_0}}{L_c^{t_0}}$, they then state this:

$\dot{L}_c = \sum_i \gamma_{ic}^{t_0} \dot{L}_{ic} ~~~~~~~~~~ [1]$

(I assume that $L_c^{t_0} = \sum_i L_{ic}^{t_0},$ but I am not positive on this.)

I can't figure out why equation [1] holds. I made the toy example below, but it fails, likely indicating that I am misunderstanding something.

$L_c^1 = 13 ~~~~~~~ L_c^0 = 6$

$L_{1c}^1 = 6 ~~~~~~~ L_{1c}^0 = 3, ~~~~~~~ \gamma_{1c}^{t_0} = 3/6 = 1/2$

$L_{2c}^1 = 3 ~~~~~~~ L_{2c}^0 = 2, ~~~~~~~ \gamma_{2c}^{t_0} = 2/6 = 1/3$

$L_{3c}^1 = 4 ~~~~~~~ L_{3c}^0 = 1, ~~~~~~~ \gamma_{3c}^{t_0} = 1/6$

Then, trying to follow formula [1] and remembering that $\log(b) - \log(a) = \log(\frac{b}{a})$, I do:

$\frac{1}{2} \times \log(\frac{6}{3}) + \frac{1}{3} \times \log(\frac{3}{2}) + \frac{1}{6} \times \log(\frac{4}{1}) = .3089 \neq .336 = \log(13) - \log(6)$

What am I missing?

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    $\begingroup$ When people use logs in formulae like these, they are basically creating instantaneous approximations to what are really discrete values. To see this, consider replacing the formula for $\dot{x}$ with a finite approximation: $x_1 - x_0$. If you do that throughout, you'll see the math works. But it takes time for an appreciable delta to actually occur. Now shrink that time to 0 and you wind up with logs (rates of change) instead of finite differences. It's akin to working with derivatives rather than finite differences. $\endgroup$
    – jbowman
    Dec 26, 2016 at 2:05
  • $\begingroup$ I still don't get it. When I replace $\dot{x}$ with $x_1 - x_0$ and re-do the math, I get $13 - 6 = 7$ on the one hand and $(1/2)*(3) + (1/3)*(1) + (1/6)*(3) = (7/3)$ on the other. I think that equation 1 should not have weights in this case. $\endgroup$
    – bill999
    Jan 12, 2017 at 22:25

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