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Let $\mathcal X$ be a finite set and $f$ a function from $\mathcal X$ to $\mathbb R^+$. Basic probabilistic method says that if I can find a probability distribution on $\mathcal X$ and show that $E[f(X)]\leq \epsilon_1$ (where $X$ is a random variable with that probability distribution), then there exists a $x_0\in\mathcal X$ such that $f(x_0)\leq \epsilon_1$.

My question is: suppose I have an additional objective function $g:\mathcal X\mapsto\mathbb R$. How could I use the probabilistic method to show that there exists a $x_0\in\mathcal X$, such that $f(x_0)\leq \epsilon_1$ and $f(x_0)\leq \epsilon_2$.

If I use the basic methods twice, meaning that if I could show $E[f(X)]\leq \epsilon_1$ and $E[g(X)]\leq \epsilon_2$, it does not imply that there is one $x_0$ that satisfies both inequalities.

One way to solve this is to show $E[f(X)+g(X)]\leq \min\{\epsilon_1, \epsilon_2\}$. But I suspect that there is a more general and more elegant solution to this problem.

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    $\begingroup$ If $f,g$ are nonnegative, you could do something like showing that $E\left[\frac{f(X)}{\epsilon_1} + \frac{g(X)}{\epsilon_2}\right] \le 1$. $\endgroup$ Commented Mar 28, 2019 at 5:17
  • $\begingroup$ If $E[f(X)]=\varepsilon_1$, $E[g(X)]=\varepsilon_2$ and $E[(f(X)-\varepsilon_1)(g(X)-\varepsilon_2)]=0$, then there exists a $x_0$ such that either $f(x_0) < \varepsilon_1$ and $g(x_0) \leq \varepsilon_2$ or $f(x_0)>\varepsilon_1$ and $g(x_0) > \varepsilon_2$. $\endgroup$ Commented Mar 28, 2019 at 9:07
  • $\begingroup$ Only a guess, could be wrong. $\endgroup$ Commented Mar 28, 2019 at 9:14
  • $\begingroup$ @NateEldredge, yes I am thinking of nonnegative function. Your idea sounds good, thanks! $\endgroup$ Commented Mar 29, 2019 at 1:04
  • $\begingroup$ I guess my idea is actually equivalent to your original one, since by rescaling $f,g$, without loss of generality we can assume $\epsilon_1 = \epsilon_2 = 1$. $\endgroup$ Commented Mar 29, 2019 at 3:02

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