Behaviour of a certain function

Question

While reading a paper on game theory, I´ve found myself stuck with the following problem:

Given the function $$ f(x,a,p)\left\{ \begin{array}{ll} \frac{1-\left(\frac{1-p}{p}\right)^x}{1-\left(\frac{1-p}{p}\right)^a}&\;\;\; \hbox{if }p\neq\frac{1}{2} \\ \frac{x}{a}&\;\;\;\hbox{if }p=\frac{1}{2} \end{array} \right. $$ where $a\in\mathbb{N}$, $x\in\{0,1,\ldots,a\}$, and $p\in(0,1)$,

show that for fixed $x$ and $a$, $f(x,a,p)$ increases from 0 to 1 as $p$ increases from 0 to 1.

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I have tried, at least, a couple of ways: first, directly consider $p_1, p_2\in(0,1)$ with $p_1<p_2$ with the object of trying to prove that $f(x,a,p_1)<f(x,a,p_2)$. The second try was to consider the derivative with respect to $p$ and try to prove that it is positive all through the interval $(0,1)$. Neither one nor the other did I get the desired result.

Answer 1

I just wanted to provide another way of deriving that $$ \phi(u) = \frac{u^x-1}{u^a-1},\quad u>0,\ 0< x\leq a $$ is a decreasing function of $u$, without the expansion as in the answer of @jwhite.

We know $$ \phi'(u) = \frac{xu^{x-1}(u^a-1) - au^{a-1}(u^x -1)}{(u^a-1)^2}, $$ and we want to show the top $\leq 0$.

Use the generalized mean value theorem, we have $$ \frac{u^a-1}{u^x-1} = \frac{a\zeta^{a-1}}{x\zeta^{x-1}}=\frac{a}{x}\zeta^{a-x} $$ for some $\zeta$ between 1 and $u$. (This is even true when $u=1$ in the limit sense.) There are two cases.

If $u\in (0, 1)$, then $\zeta>u$, $u^x-1\leq 0$, and we have $$ \text{top} = {(u^x-1)}( a\zeta^{a-x} u^{x-1} - au^{a-1})\leq 0. $$

Similarly for $u\in (1, \infty)$. We are done.

Answer 2

The case $x=a$, we have $f(a,a,p)=1$ for any $p\in (0,1)$. Also, $[(1-p)/p]^x=1$ for $x=0$, so $f(0,a,p)=0$ for any $p\in (0,1)$. So the result is not true for all $x$. It is true for integers $x$ with $0<x<a$.

Let $u=(1-p)/p=p^{-1}-1$. Note that as $p\to 0^+$, $u\to +\infty$. As $p\to 1^-$, $u\to 0$. It is easy to see from the expression $u=p^{-1}-1$ that $u$ is strictly decreasing.

Recall the formula for a finite geometric series: $\sum_{n=0}^N u^n = \frac{1-u^{N+1}}{1-u}$, assuming $u\neq 1$. Then $$1-u^x=(1-u)\sum_{n=0}^{x-1}u^n$$ and $$1-u^a = (1-u)\sum_{n=0}^{a-1}u^n.$$ The first of these two is where $0<x$ becomes necessary. Taking the ratio and cancelling the $1-u$ gives $$\frac{1-u^x}{1-u^a}=\frac{\sum_{n=0}^{x-1}u^n}{\sum_{n=0}^{a-1}u^n}.$$ We can see that this equals $x/a$ when $u=1$ (we have cancelled the problematic $1-u$ factor), so this gives us a continuous expression for $f(x,a,p)$ in terms of $x,a,u$, valid for all $0<u<\infty$.

The function $\frac{\sum_{n=0}^{x-1}u^n}{\sum_{n=0}^{a-1}u^n}$ is continuous on $[1,\infty)$, equals $1$ at $u=0$ ($p\to 1^-$), and tends to $0$ as $u\to\infty$, since the dominating term in the denominator is $u^{a-1}$ and the leading term in the denominator is $u^{x-1}$. This is where $x<a$ becomes necessary. So this gives the correct values at the endpoints. We last need to know that this is a strictly decreasing function of $u$, so that it becomes a strictly increasing function of $p$. I'm going to change notation a little to approach this.

Fix $0<\theta<\infty$ and positive numbers $A,A',B,B'$ such that $A'/A<B'/B$. Then $A'/A<\frac{A'+B'}{A+B}<B'/B$. To see this, let $t=A/(A+B)\in (0,1)$, so $1-t=B/(A+B)$. Then $$\frac{A'+B'}{A+B}=\frac{A}{A+B}\cdot \frac{A'}{A}+\frac{B}{A+B}\cdot \frac{B'}{B}=t\cdot\frac{A'}{A}+(1-t)\frac{B'}{B}.$$ This is an average of $A'/A$ and $B'/B$ with weights $t,1-t$, and therefore lies strictly between the averaged values $A'/A$, and $B'/B$.

Now consider two polynomial functions $g,h:(0,\infty)\to \mathbb{R}$ such that $g'(u)/g(u)<h'(u)/h(u)$ for some $u\in (0,\infty)$. By the previous paragraph, $$[\log(g+h)]'(u)=\frac{g'(u)+'h(u)}{g(u)+h(u)}<h'(u)/h(u)=[\log(h)]'(u).$$ If this is true for all $u\in (0,\infty)$, then for any $0\leqslant a<b<\infty$, we have $$m_{g+h}:=\int_a^b [\log(g+y)]'(u)du<\int_a^b [\log(h)]'(u)du=:m_h.$$ Similarly, we have $$m_g:=\int_a^b [\log(g)]'(u)du <m_{g+h}.$$ Let $M_{g+h}=e^{m_{g+h}}$, and let $M_g$, $M_h$ be defined similarly. Note that $$\log(g+h)(b)-\log(g+h)(a)=M_{g+h},$$ so that, raising $e$ to both sides, we get $$(g+h)(b)=(g+h)(a)M_{g+h},$$ $$g(b)=g(a)M_g,$$ and $$h(b)=h(a)M_h.$$ Then we have $$\frac{g(b)}{(g+h)(b)} = \frac{g(a)M_g}{(g+h)(a)M_{g+h}}=\frac{M_g}{M_{g+h}}\cdot \frac{g(a)}{(g+h)(a)}.$$ Since $M_g/M_{g+h}<1$, we see that $g/(g+h)$ is strictly decreasing.

Now, for integers $0\leqslant l\leqslant m$ and $d\in \mathbb{N}$, let $g(u)=\sum_{n=l}^{m}u^n$ and $h(u)=\sum_{n=m+1}^{m+d}u^n$. Then for $u\in (0,\infty)$, $$g'(u)/g(u)=\frac{\sum_{n=l}^{m}nu^{n-1}}{\sum_{n=l}^{m}u^n} < \frac{(m+1)\sum_{n=l}^{m}u^{n-1}}{\sum_{n=l}^{m}u^n}=(m+1)/u\leqslant \frac{\sum_{n=m+1}^{m+d}n u^{n-1}}{\sum_{n=m+1}^{m+d}u^n}.$$ Therefore $$\frac{\sum_{n=l}^m u^n}{\sum_{n=l}^{m+d}u^n}$$ is strictly decreasing for $d\in \mathbb{N}$.

Apply this with $l=0$, $m=x-1$, and $d=a-x$. Then the preceding argument applies when $0<x<a$, since it requires $m>0$ and $d>0$.