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I've been looking for the formula of the rolling geometric mean, which could be calculated iteratively, i.e $G(x)=f(G(x-1))$. This is needed for the computationally efficient spreadsheets.

Gregory Grant
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xelvet
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1 Answers1

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$$G_{n} = \bigg(\prod_{i=1}^n x_i\bigg)^{1/n} = \bigg(x_n\prod_{i=1}^{n-1} x_i\bigg)^{1/n} = x_n^{1/n}\bigg(\prod_{i=1}^{n-1} x_i\bigg)^{1/n} = \left(x_n\left(G_{n-1}\right)^{n-1}\right)^{\frac 1n}$$

jameselmore
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  • Please could you explain this in plain English? How does the formula work? Does it solve the issue of underflow/overflow in computation (for large products)? – Aalawlx Apr 11 '18 at 17:44