3
$\begingroup$

I try to show that $$r(k,n,\delta)=\frac{k\cdot(\delta^k-\delta^n)}{(n-k)\cdot(1-\delta^k)}$$ is decreasing in $k$ for all $0<k<n$ and $\delta \in (0,1)$. I checked the claim for numerous parameters and it seems to hold.

The first derivative with resepect to $k$ is negative if $$-\ln \delta < \frac{n\cdot (1-\delta^k)\cdot (\delta^k-\delta^{n})}{(1-\delta^n)\cdot \delta^k \cdot k \cdot (n-k)}.$$ I do not know how to show that the above inequality holds. Thanks for your ideas!

$\endgroup$
2
  • $\begingroup$ Just to confirm, $k$ can be negative? $\endgroup$
    – orlp
    Commented Sep 18, 2018 at 6:39
  • $\begingroup$ No, $k$ is positive. $\endgroup$
    – user583877
    Commented Sep 18, 2018 at 7:40

1 Answer 1

1
$\begingroup$

The claim is true.

Let us first define $F,G$ and $k_0$ as follows :

$$\begin{align}F:&=r'(n-k)^2(1-\delta^k)^2=n(\delta^k-\delta^n)(1-\delta^k)+k\delta^k\ln\delta(n-k)(1-\delta^n)\\\\ G:&=\frac{F'}{\delta^k\ln\delta}=2(n-k)+k\ln\delta(n-k)(1-\delta^n)+2k\delta^n-2n\delta^k\\\\ G'&=-2+\ln\delta(1-\delta^n)(n-2k)+2\delta^n-2n\delta^k\ln\delta\\\\ G''&=-2\ln\delta(1-\delta^n+n\delta^k\ln\delta)\\\\ G'''&=-2n(\ln\delta)^3\delta^k\\\\ k_0:&=\log_{\delta}\left(\frac{1-\delta^n}{n(-\ln\delta)}\right)\quad\text{which is the only $k$ such that $G''=0$}\end{align}$$

Using the following facts (the proofs for lemmas are written at the end of the answer):

  • $k_0\gt 0\quad$ (Lemma 1)
  • $k_0\lt n\quad$ (Lemma 2)
  • $G''$ is increasing since $G'''\gt 0$
  • $\displaystyle\lim_{k\to 0^+}G'=\displaystyle\lim_{k\to n^-}G'\gt 0\quad$ (Lemma 3)
  • $G'(k_0)\lt 0\quad$ (Lemma 4)
  • $\displaystyle\lim_{k\to 0^+}F=\displaystyle\lim_{k\to n^-}F=0$

$\qquad$enter image description here

we get that $F\lt 0,$ i.e. $r'\lt 0$ as claimed.


Finally, let us prove the lemmas.

Lemma 1 : $k_0\gt 0$

Proof : $k_0\gt 0$ is equivalent to $\delta^n-1-n\ln\delta\gt 0$. Let $H(n)$ be the LHS. Then, $H'(n)=-\ln\delta(1-\delta^n)\gt 0$, so $H(n)$ is increasing with $\displaystyle\lim_{n\to 0^+}H(n)=0$ from which $H(n)\gt 0$ follows.$\quad\square$

Lemma 2 : $k_0\lt n$

Proof : $k_0\lt n$ is equivalent to $1-\delta^n+n\delta^n\ln\delta\gt 0$. Let $I(n)$ be the LHS. Then, $I'(n)=n\delta^n(\ln\delta)^2$ which is positive. So, $I(n)$ is increasing with $\displaystyle\lim_{n\to 0^+}I(n)=0$ from which $I(n)\gt 0$ follows.$\quad\square$

Lemma 3 : $\displaystyle\lim_{k\to 0^+}G'=\displaystyle\lim_{k\to n^-}G'\gt 0$

Proof : $\displaystyle\lim_{k\to 0^+}G'=\displaystyle\lim_{k\to n^-}G'=-2+n\ln\delta(1-\delta^n)+2\delta^n-2n\ln\delta$. Let this be $J(n)$. Then, $J'(n)=\ln\delta(-1+\delta^n-n\delta^n\ln\delta),J''(n)=-n\delta^n(\ln\delta)^3\gt 0$. So, $J'(n)$ is increasing with $\displaystyle\lim_{n\to 0^+}J'(n)=0$. So $I'(n)$ is positive from which we have that $J(n)$ is increasing with $\displaystyle\lim_{n\to 0^+}J(n)=0$. It follows that $J(n)\gt 0$.$\quad\square$

Lemma 4 : $G'(k_0)\lt 0$

Proof : $G'(k_0)=\ln\delta(1-\delta^n)(n-2k_0)$, so $G'(k_0)\lt 0$ is equivalent to $n-2k_0\gt 0$, i.e. $1-\delta^n+n\delta^{\frac n2}\ln\delta\gt 0$. Let $L(n)$ be the LHS. Then, $L'(n)=-\delta^{\frac n2}\ln\delta(1-\delta^{\frac n2}-\frac n2\ln\delta)$ which is positive. So, $L(n)$ is increasing with $\displaystyle\lim_{n\to 0^+}L(n)=0$ from which $L(n)\gt 0$ follows.$\quad\square$

$\endgroup$

You must log in to answer this question.