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I'm working on a function maximisation problem, witch is related to a non-linear regression problem.

The problem is as follows:

The target variable is $y$ and is continuous. The data set consists of some features $x_1,\dots,x_N$, $y$ and a parameter $p$, which is a continuous and bounded. The parameter $p$ is a feature that I can tune for new samples.

The objective is to, given a new data point (i.e., a set of fetaures $x_1, \dots, x_N$), find the parameter $p$ that will yield maximum $y$.

This is problem is somehow similar to some reinforcement learning problems. New data points are being sampled at constant time intervals, for each new data point ($x_1, \dots, x_N$) I get to decide a parameter $p$ and then I get a feedback $y$. For each new data point, I want to decide the best $p$ so that $y$ is maximized.

My approach to solve this problem is as follows:

  1. Use non-linear regression techniques to estimate $y$ as a function of $p$ and $x_1, \dots, x_N$ (i.e, regression-$y$).

  2. Evaluate regression-$y$ over all possible (well just a subset using global optimization techinques) values of $p$ in order to find the one that yields higher $y$.

  3. Construct a second data set with $x_1, \dots, x_N$ and $p^*$, where $p^*$ is the parameter that yielded higher $y$ in regression-$y$.

  4. Use another non-linear regression to estimate $p^*$ as a function of $x_1, \dots, x_N$ (i.e., regression-$p$).

  5. Use the regression-$p$ to estimate the best $p$ for each new data point.

I'm wondering if I could use another procedure that estimates the best $p^*$.

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    $\begingroup$ How are you going to construct a second data set? Is not your data fixed? $\endgroup$
    – generic_user
    Commented Feb 5, 2018 at 18:25
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    $\begingroup$ @generic_user the values of $x_1$ through $x_N$ are fixed, but $p$ is an adjustable parameter. I haven't thought hard about this, but it seems to be a situation in which $p$ would always end up at one of its bounds. $\endgroup$
    – EdM
    Commented Feb 5, 2018 at 18:54
  • $\begingroup$ Adjusting $p$ to maximize $y$ is trivial. Instead, do you want to choose a combination of $x$'s to maximize $y$ given $p$? That'd be a more typical problem, given an estimate of $p$ that minimizes some loss function. $\endgroup$
    – generic_user
    Commented Feb 5, 2018 at 19:04
  • $\begingroup$ @generic_user Adjusting $p$ to maximize $y$ is exactly what I want. For each set of features x's I can decide the value of $p$ and then I observe an $y$. $y$ is what I want to maximize. $\endgroup$
    – PolBM
    Commented Feb 5, 2018 at 21:08
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    $\begingroup$ If you are using $y = \beta_0 + \beta_1 x_1 + ...+ \beta_N x_N + \beta_{N+1} p + \varepsilon$, then $p=\infty$ gives you the maximal possible $y=\infty$... If you want to opitmize a function, why not directly use optimization algorithm that optimizes your desired objective? $\endgroup$
    – Tim
    Commented Feb 5, 2018 at 22:55

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