In a typical regression set up, we want to maximize the expected reward(or minimize the expected loss). Empirically, we are maximizing the average return over all samples. However, in reality wouldn't we want to maximize the total reward/revenue over a year for example?
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2$\begingroup$ Expected total reward = expected average reward $\times$ number of attempts. If making more attempts with a smaller return per attempt makes sense to you, go right ahead, but if you can do that, why not make the same number of attempts with a better reward per attempt? $\endgroup$– Glen_bCommented Jan 19, 2023 at 2:44
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In calculus, finding the values that maximize a function $f(\theta)$ is the same as finding the values that maximize $af(\theta)$ for any positive $a$ (such as the reciprocal of a sample size). Consequently, maximizing the mean reward yields the same result as maximizing the sum of the rewards. We typically do not care about the maximum or minimum values achieved, just where they occur. (Rather, we often do care about those values, just for different reasons.)
$$ \underset{\theta\in\Theta}{\arg\max} \left\{f(\theta)\right\} = \underset{\theta\in\Theta}{\arg\max}\left\{ \frac{1}{n}f(\theta)\right\} $$
Ditto for minimization.
In fact, $\arg\min$ remains unchanged upon applying a strictly increasing function, and $\arg\max$ remains unchanged upon applying a strictly decreasing function.
Concretely, if we do a linear regression and estimate the parameter vector $\beta$ using ordinary least squares, we get the same estimate $\hat\beta$ whether we minimize $\overset{n}{\underset{i=1}{\sum}}\left(y_i-\hat y_i\right)^2$, $\dfrac{1}{n}\overset{n}{\underset{i=1}{\sum}}\left(y_i-\hat y_i\right)^2$, or $\sqrt{\dfrac{1}{n} \overset{n}{\underset{i=1}{\sum}}\left(y_i-\hat y_i\right)^2 }$, except maybe for some numerical issues when you do the math on a computer.
