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I have been trying to solve the following function is non-increasing (non-decreasing) with respect $\theta$ where $\theta \in (0,1)$ (resp. $\theta \in (-1,0)$) \begin{equation} f(\theta)= \frac{h(t,\beta)}{h(\beta, \beta)} = \frac{1-t+\frac{\varphi_{\theta}(t)}{\varphi_{\theta}^{\prime}(\beta)}}{1-\beta+\frac{\varphi_{\theta}(\beta)}{\varphi_{\theta}^{\prime}(\beta)}}=\frac{1-t+\frac{\beta (1-\theta(1-\beta))}{\theta-1}\log ( \frac{1-\theta(1-t)}{t})}{1-\beta+\frac{\beta (1-\theta(1-\beta))}{\theta-1}\log ( \frac{1-\theta(1-\beta)}{\beta})} \end{equation} where $0<\beta<t<1$. The function $\varphi_{\theta}(t)$ is the generator of the Ali-Mikhail-Haq copula, \begin{equation} \varphi_{\theta}(t)=\log \left( \frac{1-\theta(1-t)}{t}\right)\\ \varphi_{\theta}^{\prime}(\beta)=\frac{\theta-1}{\beta(1-\theta(1-\beta))} \end{equation}

However, $f^{\prime}(\theta)$ seems to get me nowhere. I would appreciate any hint. Thank you so much.

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  • $\begingroup$ Sorry for that. $f(\theta)$ depends on $t,\beta$ and $\theta$ through the generator $\varphi_{\theta}(t)$. I will edit the question. Thank you. $\endgroup$
    – MSquared
    Commented Oct 8 at 15:53

1 Answer 1

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$\newcommand{\p}{\partial}$Assuming that $\log$ here stands for $\ln$, write $s,b,t$ in place of $t,\beta,\theta$, respectively, so that $0<b<s<1$ and $t\in(-1,0)\cup(0,1)$; the latter conditions on $s,b,t$ will be always assumed by default.

The task is to show that \begin{equation*} f(t):=f(t;b,s):=\frac{F(t)}{G(t)} \end{equation*} is decreasing in $t\in(0,1)$ and increasing in $t\in(-1,0)$, where \begin{equation*} F(t):=F(t;b,s):=\frac{(s-1)(1-t)}{1-(1-b) t}+b \ln \frac{1-(1-s) t}{s}, \end{equation*} \begin{equation*} G(t):=G(t;b):=\frac{(b-1)(1-t)}{1-(1-b) t}+b \ln \frac{1-(1-b) t}{b}. \end{equation*}

Proof: Key here are the so-called l'Hospital-type rules for monotonicity.

To use them, note first that \begin{equation*} G'(t)=\frac{(1-b)^2 b t}{(1-(1-b) t)^2}, \end{equation*} which is of the same sign as $t$. So, $G$ is increasing in $t\in(0,1)$ and decreasing in $t\in(-1,0)$. Also, $G(1)=0$. So, \begin{equation*} G<0, G'>0 \text{ on }(0,1). \tag{10}\label{10} \end{equation*} Also, the derivative of $G(-1)/b$ in $b$ is $\frac{4 (1-b)^2}{(2-b)^2 b^2}>0$, so that $G(-1)=G(-1;b)/b$ is increasing in $b\in(0,1)$, to $G(-1;1)/1=0$, so that $G(-1)<0$. So, \begin{equation*} G<0, G'<0 \text{ on }(-1,0). \tag{20}\label{20} \end{equation*}

Consider now the "derivative ratio" (multiplied by the positive factor $\frac{(1-b)^2}{1-s}$ not depending on $t$) \begin{equation*} \rho(t):=\frac{F'(t)}{G'(t)} \frac{(1-b)^2}{1-s} =\frac{1 - 2 b + s - (1- b)^2 t}{1-(1-s) t}, \end{equation*} which is a simple rational expression. Note that \begin{equation*} \rho'(t)=-\frac{(b-s)^2}{(1-(1-s) t)^2}<0. \tag{30}\label{30} \end{equation*} Also, $F(1)=G(1)=0$. So, by the special l'Hospital-type rule for monotonicity given by Proposition 4.1 in the linked paper and in view of \eqref{30} and \eqref{10}, $f=F/G$ is decreasing on $(0,1)$.

It is somewhat harder to show that $f=F/G$ is increasing on $(-1,0)$. Here we have to use a so-called general l'Hospital-type rule for monotonicity, namely the one given in line 2 of Table 1.1 in the linked paper, which is applicable here in view of \eqref{30} and \eqref{20}. According to this rule, there is some $c\in[-1,0]$ such that $f=F/G$ is increasing on $(-1,c)$ and decreasing on $(c,0)$.

So, it remains to show that $f=F/G$ is increasing in a left neighborhood (l.n.) of $0$, so that $c=0$ and $f=F/G$ is increasing on the entire interval $(-1,0)$.

So, it is enough to show that \begin{equation*} \begin{aligned} L(b,s)&:=\frac{(1 - b + b \ln b)^2}{b^2}\lim_{t\uparrow0}\frac{f'(t)}{t} \\ & =\left(2 b (1-s)+s^2-1\right) \ln b \\ &+\frac{(b-1) ((s-1) (b-s)+(b-1) b \ln s)}{b}<0. \end{aligned} \tag{40}\label{40} \end{equation*} Note that \begin{equation*} L_{20}(b,s):=\p_b^2 L(b,s)= 2 \ln s-\frac{(s-1) \left(2 b^2+b s+b-2 s\right)}{b^3}. \end{equation*} So, $\p_b L_{20}(b,b)=2 (1-b)^2/b^3>0$, so that $L_{20}(b,b)$ is increasing to $L_{20}(1,1)=0$, whence \begin{equation*} L_{20}(b,b)<0. \tag{50}\label{50} \end{equation*}

Next, \begin{equation*} L_{21}(b,s):=\frac{b^3 s}2\,\p_s L_{20}(b,s)= b^3 -(1+b^2) s + (2-b)s^2, \end{equation*} which is convex in $s$, with $L_{21}(b,b)=-(1-b)^2 b<0$ and $L_{21}(b,1)=(1-b)^2(1+b)>0$, so that $L_{21}(b,s)$ switches its sign just once, from $-$ to $+$, as $s$ increases from $b$ to $1$. So, for some $s_b\in(b,1)$ we have that $L_{20}(b,s)$ is decreasing in $s\in(b,s_b)$ and increasing in $s\in(s_b,1)$, to $L_{20}(b,1)=0$. Recalling now \eqref{50}, we conclude that $L_{20}(b,s)<0$, so that $L(b,s)$ is concave in $b$.

Moreover, $L(b,s)|_{b=s}=0=\p_b L(b,s)|_{b=s}$. Thus, $L(b,s)<0$ (if $0<b<s<1$, as was assumed in the very beginning); that is, the inequality in \eqref{40} is proved. $\quad\Box$

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  • $\begingroup$ Thank you so much again. I would never have guessed what it required to show this. $\endgroup$
    – MSquared
    Commented Oct 9 at 6:04

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