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The following question, somewhat edited here, was asked and then closed at The best bound of the integral of a nondecreasing real function in a closed interval.

Let $F\colon[0,1]\to[0,1]$ be a nondecreasing function. For $t\in[0,1]$, let $$G(t):=\int_t^1 F(x)\,dx.\tag{1}$$ Take any $a,b,c$ such that $0\le a<b<c<1$. Find the best upper and lower bounds on $G(b)$ in terms of $a,b,c,G(a),G(c)$.


I think the answer to this question is not altogether trivial, and it might be useful in some research in probability. Indeed, changing, if necessary, values of the function $F$ on an at most countable set and extending the resulting function to $\mathbb R$, without loss of generality we may assume that $F$ is the cumulative distribution function (cdf) of a random variable (r.v.) $X$ with values in $[0,1]$, and then for all $t\in[0,1]$ $$G(t)=\int_t^1 F(x)\,dx=\int_t^1 P(X\le x)\,dx=\int_t^1E1(X\le x)\,dx=E\int_t^1 1(X\le x)\,dx=E\int_0^1 dx\,1(x\ge t)1(X\le x)=E\int_0^1 dx\,1(x\ge t\vee X)=E(1-t\vee X)=1-E(t\vee X),$$ where $u\vee v:=\max(u,v)$. So, the question can be restated as the one about the best upper and lower bounds on $H(b)$ in terms of $a,b,c,H(a),H(c)$, where $H(t):=E(t\vee X)$.

Therefore, an answer will be given below.

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1 Answer 1

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$\newcommand\G{\mathscr G}$From the probabilistic interpretation of the question, it follows that without loss of generality the function $F$ in (1) is right-continuous on $[0,1)$. It follows that the function $G$ is nonincreasing, with the nonincreasing on $[0,1)$ right derivative $G'_+=-F$. So, $G$ is in the (convex) set -- say $\G$ -- of all functions $g\colon[0,1]\to[0,1]$ such that $g$ is nonincreasing and concave, $g(1)=0$, and $g(t)\le1-t$ for all $t\in[0,1]$.

Vice versa, take any $g\in\G$. Then (1) holds with $g$ in place of $G$ and some nondecreasing function $f\colon[0,1]\to[0,1]$ in place of $F$. Indeed, let $f(t):=-g'_+(t)$ for $t\in[0,1)$, with $f(1):=f(1-)$. Then $f$ is nonnegative and nondecreasing, and (1) holds with $g$ in place of $G$. Moreover, then $$f(1-)=\lim_{t\uparrow1}\frac{g(t)-g(1)}{1-t}=\lim_{t\uparrow1}\frac{g(t)}{1-t}\le1,$$ by the conditions $g(1)=0$ and $g(t)\le1-t$ for all $t\in[0,1]$. So, $f(t)\le1$ for $t\in[0,1]$; that is, we have a function $f\colon[0,1]\to[0,1]$.

Thus, the problem can be restated as follows: Take any $a,b,c$ such that $0\le a<b<c<1$. Find $\max\{G(b)\colon G\in\G_{A,C}\}$ and $\min\{G(b)\colon G\in\G_{A,C}\}$ for all real $A$ and $C$ such that $$\G_{A,C}:=\{G\in\G\colon G(a)=A,G(c)=C\}\ne\emptyset.$$ Also find all real $A$ and $C$ such that $\G_{A,C}\ne\emptyset$.

Note that $\G_{A,C}\ne\emptyset$ iff the function $h\colon\{a,c,1\}\to\mathbb R$ defined by the conditions $h(a):=A$, $h(c):=C$, and $h(1):=0$ satisfies the following conditions: $h$ is nonincreasing and concave, and $h(t)\le1-t$ for all $t\in\{a,c,1\}$ -- indeed, then $h$ can be extended to $[0,1]$ by linear interpolation, with $h(0):=h(a)=A$, and the resulting extended function will be in $\G$. So, $\G_{A,C}\ne\emptyset$ iff $$0\le A\le1-a\quad\text{and}\quad \frac{1-c}{1-a}\,A\le C\le A\wedge(1-c),\tag{*}$$ where $u\wedge v:=\min(u,v)$. In what follows, assume that (*) holds, so that $\G_{A,C}\ne\emptyset$.

Similarly using linear interpolation, we can now see that $\max\{G(b)\colon G\in\G_{A,C}\}$ equals the maximum of $h(b)$ over all functions $h\colon\{a,b,c,1\}\to\mathbb R$ such $h(a)=A$, $h(c)=C$, $h(1)=0$, $h$ is nonincreasing and concave, and $h(t)\le1-t$ for all $t\in\{a,b,c,1\}$. It follows that $$\max\{G(b)\colon G\in\G_{A,C}\}=A\wedge\Big(\frac{1-b}{1-c}\,C\Big).$$ Similarly, $$\min\{G(b)\colon G\in\G_{A,C}\}=A+\frac{C-A}{c-a}\,(b-a).$$

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  • $\begingroup$ @losif Pinelis Many thanks for your detailed analysis. Due to the requirement of $C\le A$, the last formula, i.e. the minimum of $G(b)$, can be directly replaced by $A+\frac{C-A}{c-a}(b-a)$. $\endgroup$
    – RyanChan
    Nov 24, 2020 at 8:12
  • $\begingroup$ @RyanChen : I am glad this was of help. Thank you also for the suggested simplification, now done. $\endgroup$ Nov 24, 2020 at 14:16
  • $\begingroup$ @AlexandreEremenko : In your answer, now deleted, you said "this question is too easy for this site". It was not quite easy for me to come up with this answer, perhaps because of an unfortunate turn of events. First, as explained in the question on this page, I reduced the problem to that of finding the exact upper and lower bounds on $E(b\vee X)$ given $E(a\vee X)$ and $E(c\vee X)$. $\endgroup$ Nov 24, 2020 at 14:41
  • $\begingroup$ Previous comment continued: Then I tried to used the fact that such bounds are attained -- given the three linear restrictions on the distribution of $X$: $E(a\vee X)=A$, $E(c\vee X)=B$, and $\int P(X\in dx)=1$ -- when $X$ takes at most three values, and then feed this to Mathematica, which did not result in anything good. Only after that the simple ideas that worked (some of which you expressed in two sentences in your answer) occurred to me. Still, there remained a number of details to work on. $\endgroup$ Nov 24, 2020 at 14:42
  • $\begingroup$ Previous comment continued: Overall, I believe the answer is not significantly simpler or significantly more detailed than an average MO answer. $\endgroup$ Nov 24, 2020 at 14:42

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