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I've just encountered the following problem. Define the following geometric mean $$Y_t = \left( \prod_{k \neq i} X^k_t \right)^{1/n} $$ where $i = 1,2,...,n$ and $t$ means time. My question is how to take following differentiation $$\frac{dY_t}{Y_t}?$$ I tried several times but got stuck. I appreciate any hint or answer. Many thanks!

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  • $\begingroup$ $\frac{dY_t}{Y_t}$? What is this notation supposed to mean? Also, show your attempts $\endgroup$ – Brevan Ellefsen May 27 at 12:41
  • $\begingroup$ If I understand correctly, I believe that $\frac{dY_t}{Y_t}$ means the differentiation of $Y_t$ devided by $Y_t$. I wrote $\prod_{k \neq i} X^k_t = X^1_t X^2_t ... X^{i-1}_t X^{i+1}_t ... X^n_t$ and then tried to get $\frac{dY_t}{Y_t}$ but getting overwhelmed since there are many variables. $\endgroup$ – Min Haw May 27 at 12:50
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$$\log Y_t=\frac1n\log\left(\prod_{k\ne i} X_t^k\right)=\frac1n\sum_{k\ne i}\log X_t^k$$

and

$$\frac{dY_t}{Y_t}=\frac1n\sum_{k\ne i}\frac{d X_t^k}{ X_t^k}.$$

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  • $\begingroup$ Thanks for your response. Do you mean $d \log Y_t = \frac{dY_t}{Y_t}$? I am not sure whether we can treat $Y_t$ as a normal variable in this case or not. $\endgroup$ – Min Haw May 27 at 13:14
  • $\begingroup$ @MinHaw: yes I am. $\endgroup$ – Yves Daoust May 27 at 13:50
  • $\begingroup$ @ Yves Daoust: I got it. Thanks! $\endgroup$ – Min Haw May 28 at 1:21

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