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(Re-posted from StackOverflow as suggested)

I have the following problem.

The functions $f(x),g(x)$ are defined as $$ f(x) = \begin{cases} f_1(x) & 0 \leq x \leq 10, \\ f_2(x) & 10 < x \leq 20, \\ 0 & \text{otherwise}, \end{cases} \qquad g(x) = \begin{cases} g_1(x) & 0 \leq x \leq 5, \\ g_2(x) & 5 < x \leq 20, \\ 0 & \text{otherwise}, \end{cases} $$ In addition, we require the constraints $$ \int_0^{20} f(x) dx \geq K, \quad \int_0^{20} g(x) dx \geq Q, \quad f(x)+g(x) \leq R \text{ for all $x$}. $$ where $K,Q,R$ are parameters.

I assume there is quite some elaborate theory behind it, and was wondering if anybody could point me in the right direction to devise an algorithm that can generate $f_1(x), f_2(x), g_1(x), g_2(x)$?

I would like to add that for a given $K$ and $Q$, the interest is to keep $R$ as low as possible.

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  • $\begingroup$ Do you mean that when $x \in [0, 10)$, $f(x) = f_1(x)$ and when $x \in [10, 20]$, $f(x) = f_2(x)$? What sort of restriction is that? $\endgroup$ Apr 5, 2014 at 16:26
  • $\begingroup$ Yeah, that's what I mean, f1(x)!=f2(x) $\endgroup$
    – MrD
    Apr 5, 2014 at 16:28
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    $\begingroup$ I can take any function $f$ and define $f_1(x) = f(x)\cdot 1_{[0, 10)}(x)$ where $1_{[...]}$ is an indicator function. My point is that dividing $f$ into two functions adds nothing to the problem. $\endgroup$ Apr 5, 2014 at 16:33
  • $\begingroup$ I may have gotten the syntax wrong, but I hope what I mean is still clear? $\endgroup$
    – MrD
    Apr 5, 2014 at 16:34
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    $\begingroup$ So is there actually a point to $f_1, f_2$ or not? $\endgroup$ Apr 5, 2014 at 16:36

1 Answer 1

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Let $X$ be a random variable distributed uniformly over $[0,20]$. Your constraints imply $$ \mathbb{E}[f(X)] \geq \frac{K}{20}, \qquad \mathbb{E}[g(X)] \geq \frac{Q}{20}. $$ We conclude that $$ \mathbb{E}[f(X)+g(X)] \geq \frac{K+Q}{20}, $$ and so there is some point $x \in [0,20]$ such that $f(x) + g(x) \geq (K+Q)/20$. In particular, $R \geq (K+Q)/20$. This bound is tight, as shown by the functions $$ f(x) = \frac{K}{20} \mathbf{1}_{x \in [0,20]}, \qquad g(x) = \frac{Q}{20} \mathbf{1}_{x \in [0,20]}. $$

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  • $\begingroup$ Hi Yuval, thanks for taking the time to answer and format my post. I'm afraid I don't fully understand your solution though. Why are you saying that my constraints inply that the expected values of f and g are k/20 and q/20 respectively? $\endgroup$
    – MrD
    Apr 5, 2014 at 17:43
  • $\begingroup$ You have a constraint on $\int_0^{20} f(x) dx$, and this translates to a constraint on $\mathbb{E}[f(X)]$ since the density of $X$ is $(1/20)\mathbf{1}_{x \in [0,20]}$. $\endgroup$ Apr 5, 2014 at 18:22
  • $\begingroup$ Hi Yuval, I'm sorry, I still don't understand how the solution guarantees that both functions are less than R for all given X. $\endgroup$
    – MrD
    Apr 6, 2014 at 13:53
  • $\begingroup$ This solution only works for $R \geq (K+Q)/20$, in which case I hope you can answer this yourself. The first part shows that there is no solution if $R < (K+Q)/20$. $\endgroup$ Apr 6, 2014 at 14:15

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