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I'm encountering the following minimization problem in my research:

$$\hat b = \underset{b}{\arg\min} \sum_i^n \left( \log \frac{a_i}{b} \right)^2$$

I could iteratively optimize, but I think that there should be a closed-form solution. Intuitively I think the solution should be the geometric mean, i.e.

$$\hat b = \prod_i^n (a_i)^{\frac{1}{n}}$$

but I'm struggling to prove it to myself. Is there a good way to go about proving or disproving this? Thanks for your help.

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    $\begingroup$ You are minimizing the quadratic univariate function $\sum_i(\log(a_i) - \beta)^2$ where $\beta=\log(b)$. When it is written in the form $n(\beta-r)^2 + C$ for constants $r, C$, the unique minimum obviously is at $\beta=r$ because squared values are non-negative. You don't even need to do the algebra: just directly check your guess that $r = \log(\hat b)$. $\endgroup$
    – whuber ♦
    Commented Feb 19, 2015 at 19:24

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Just take the derivative with respect to b. This yields $$ \sum_i 2 \log(a_i/b)*(-1/b) = 0 $$

Ignoring the $b\to\infty$ solution (which, if you work it out, gives a max rather than a min), the remaining solution is $$ \sum_i \log(a_i/b) = \log\left(\frac{\prod_i a_i}{\prod_i b}\right)=\log\left(\frac{\prod_i a_i}{b^n}\right) = 0 $$

$$ \frac{\prod_i a_i}{b^n} = 1 $$

$$ b = \left(\prod_i a_i \right)^{1/n} $$

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  • $\begingroup$ Ahhh of course. Thanks. (Will accept when the 5 mins are up) $\endgroup$
    – rhombidodecahedron
    Commented Feb 19, 2015 at 18:30
  • $\begingroup$ whuber's comment shows the easy way to get there. $\endgroup$
    – Glen_b
    Commented Feb 20, 2015 at 2:22

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