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Consider we want to compare algorithm A and algorithm A* (a modified version of algorithm A). To evaluate them, both are executed on multiple datasets (D1, D2, ...) and their performance is measured by a reasonable metric M.

Is the relation for improvement of algorithm A* w.r.t algorithm A, with the following formula a standard effect size to compare their performance on dataset $D_i$?

$\frac{M(A*, D_i)-M(A, D_i)}{M(A, D_i)}$

For example let $M(A, D_i)$ be 1000, $M(A*, D_i)$ be 1050. Then:

$\frac{M(A*, D_i)-M(A, D_i)}{M(A, D_i)} = \frac{1050-1000}{1000} = 5%$

Does this simple formula have a name in literature? Can this be considered a standard effect size?

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I would not consider this as an estimate of a standardized effect size, because the term "standardized" means, you express the effect in terms of the standard deviation of the measurements: $\frac{\mu_1-\mu_0}{\sigma}$.

In your case, the denominator is not a standard deviation.

If I would like to invent a name for your performance measure, I would call it e.g. the "relative improvement".

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  • $\begingroup$ In my field (Empirical Software Engineering), few papers have used something like Cohens d (en.wikiversity.org/wiki/Cohen%27s_d) and many papers have used this "relative improvement". How can this be explained? $\endgroup$ – Mostafa Mahdieh Nov 1 '19 at 11:59
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    $\begingroup$ Cohens d can be used to get a sense about the relevance of an effect in a population. How large is the effect in relation to the variation, that can be observed across the population. Silly purely invented example: maybe an apple a day raises the mean IQ about half a point, but it is not really relevant, considering the IQ variation across the population. With $100 \times$ "relative improvement", you ask another question: How many percent points is the new method better, than the old one. The variation of your $M$ is not your primary interest. $\endgroup$ – ghlavin Nov 1 '19 at 12:08

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