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So I am learning Bayesian Optimization and came across expected improvement.

My question is are we searching for the point in the Gaussian Process model whose expected value (determined by mean and confidence) shall be decreased the most if sampled at that point? So is the starting criteria is to take the lowest point in GP and from there determine what is the next point that whose expected value is lowest then any other point in the GP?

How do we intuitively quantify expected improvement distribution $\phi$ in the graph attached?

enter image description here

enter image description here

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My question is are we searching for the point in the Gaussian Process model whose expected value (determined by mean and confidence) shall be decreased the most if sampled at that point?

No. At any iteration, you've observed some inputs, and one of those inputs ($x^*$) is the current optimum with a function value $f(x^*)$.

In expected improvement, what we want to do is calculate, for every possible input, how much its function value can be expected to improve over our current optimum. This is expressed in your post by the equation:

$$I(x) = max(f^* - Y, 0)$$

I think it's clearer to write this as:

$$I(x) = max(f(x^*) - f(x), 0)$$

In words, this means that the improvement for any input $x$ is how much better lower its function value f(x) is than the current lowest function value found $f(x^*)$. If $f(x)$ is greater than $f(x^*)$, then there's no improvement, so $I(x) = 0$.

Under a GP posterior, $f(x)$ is a random variable, which means that $I(x)$ is also a random variable, and so we want to calculate the expected value of $I(x)$. We do this for every possible $x$, and pick the one that gives the greatest expected improvement. After observing that point, we add its function value to our GP posterior and repeat.

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  • $\begingroup$ This makes sense but why is $ϕ$ in the start of integration defined as $e^2/2$ and why the positive infinity in the integral is changed to the $(f(x^∗)−μ)/σ$ so graphically $ϕ$ is only the distribution part that is lower then f* value i.e. $ϕ$ doesn't represent full distribution? $\endgroup$ – GENIVI-LEARNER Sep 12 '19 at 12:01
  • $\begingroup$ So they define $\epsilon \sim \mathcal{N}(0, 1)$ (the standard normal) and $\phi$ as the pdf of $\epsilon$. The standard normal pdf is $e^{-\epsilon^2/2}$. $\endgroup$ – Kevin Yang Sep 13 '19 at 16:17
  • $\begingroup$ The positive infinity in the integral gets changed because $I(x) = 0$ whenever $\mu > f(x^*)$. $\endgroup$ – Kevin Yang Sep 13 '19 at 16:19
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As far as I know, in practice, the first observations in Bayesian optimization are random before the Gaussian processes take over. After the initial observations, the expected improvement can be calculated for a data point x. By doing so, we want to select the value for x that is expected to improve the results of our objective function the most.

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