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EI(x)=E ⁣[max⁡(f(x)−f+, 0)]\mathrm{EI}(x) = \mathbb{E}\!\left[\max(f(x) - f^{+},\ 0)\right]

Why this formula appears here

The acquisition function must trade off predicted value against uncertainty, not just rank by predicted value. A pure exploitation policy — always synthesise the top-ranked candidate — collapses quickly onto whatever region of chemical or materials space the initial training data already covered well, and stops finding anything the model did not already believe. The standard fix is an acquisition function such as expected improvement, which for a candidate x with predictive mean μ(x)\mu(x) , predictive standard deviation σ(x)\sigma(x) , and current best observed value f+f^{+} is EI(x)=E ⁣[max⁡(f(x)−f+, 0)]\mathrm{EI}(x) = \mathbb{E}\!\left[\max(f(x) - f^{+},\ 0)\right]. a quantity that rewards both a high predicted mean and high predictive uncertainty. In practice this…

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EI(x)=E ⁣[max⁡(f(x)−f+, 0)],\mathrm{EI}(x) = \mathbb{E}\!\left[\max(f(x) - f^{+},\ 0)\right],

Equation 5 · AI for Science

AI for Science and Medicine in Practice: An Advanced Technical Guide

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

The acquisition function must trade off predicted value against uncertainty, not just rank by predicted value. A pure exploitation policy — always synthesise the top-ranked candidate — collapses quickly onto whatever region of chemical or materials space the initial training data already covered well, and stops finding anything the model did not already believe. The standard fix is an acquisition function such as expected improvement, which for a candidate x with predictive mean μ(x)\mu(x) , predictive standard deviation σ(x)\sigma(x) , and current best observed value f+f^{+} is EI(x)=E ⁣[max⁡(f(x)−f+, 0)]\mathrm{EI}(x) = \mathbb{E}\!\left[\max(f(x) - f^{+},\ 0)\right]. a quantity that rewards both a high predicted mean and high predictive uncertainty. In practice this…

Meanings in this article

  • E\mathbb{E}: The expected value operator: the probability-weighted average of the quantity inside its brackets.
  • f+f^{+}: the current best observed value.
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