Minimize squared relative error

The Endeavour 2025-06-21

Suppose you have a list of positive data points y₁, y₂, …, y_n and you wanted to find a value α that minimizes the squared distances to each of the y‘s.

$\sum_{i=1}^n (y_i - \alpha)^2$

Then the solution is to take α to be the mean of the y‘s:

$\alpha = \frac{1}{n} \sum_{i=1}^n y_i$

This result is well known [1]. The following variation is not well known.

Suppose now that you want to choose α to minimize the squared relative distances to each of the y‘s. That is, you want to minimize the following.

$\sum_{i=1}^n \left( \frac{y_i - \alpha}{\alpha} \right)^2$

The value of alpha this expression is the contraharmonic mean of the y‘s [2].

$\alpha = \frac{\sum_{i=1}^n y_i^2}{\sum_{i=1}^n y_i}$

[1] Aristotle says in the Nichomachean Ethics “The mean is in a sense an extreme.” This is literally true: the mean minimizes the sum of the squared errors.

[2] E. F. Beckenbach. A Class of Mean Value Functions. The American Mathematical Monthly. Vol. 57, No. 1 (Jan., 1950), pp. 1–6

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Minimize squared relative error

The Endeavour 2025-06-21

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