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Now comes the crucial point, which is choosing the value of $t$ as follows: \begin{equation*} t:=t(x):=k(x)g_0'(x)-\frac1A=\frac{ck(x)g_0(x)-1}A. \tag{150}\label{150} \end{equation*}

Now comes the crucial point, which is choosing the value of $t$ as follows: \begin{equation*} t:=t(x):=k(x)g_0'(x)-\frac1A=\frac{ck(x)g_0(x)-1}A. \tag{150}\label{150} \end{equation*}

$\newcommand{\R}{\mathbb R}$By Cram´er's large deviation principle (see e.g. Corollary 2.2.19), for all real $x$ \begin{equation*} s(x)=-\inf_{y\in[x,\infty)}L^*(y), \tag{1}\label{1} \end{equation*} where \begin{equation*} L^*(y):=\sup_{t\in\R}(ty-L(t)),\quad L(t):=\ln Ee^{tX}, \tag{2}\label{2} \end{equation*} $X:=X_1$.

If $L(t)=\infty$ for all real $t>0$, then $L^*(y)=0$ for all real $y\ge0$, and hence, by \eqref{1}, $s(x)=0$ for all real $x\ge0$, so that the desired relation \begin{equation*} s(x)\sim t(x)=\ln P(X>x) \tag{3}\label{3} \end{equation*} will not hold.

So, without loss of generality (wlog) \begin{equation} \text{$L(t_0)<\infty$ for some real $t_0>0$, } \tag{4}\label{4} \end{equation} and then, by \eqref{2}, $L^*(y)\ge t_0y-L(t_0)\to\infty$ as $y\to\infty$. Also, the function $L^*$ is convex, being the supremum of affine functions. So, by \eqref{1}, for all large enough real $x>0$ \begin{equation*} s(x)=-L^*(x)=\ln\inf_{t\in\R}Q(x,t), \tag{1a}\label{1a} \end{equation*} where \begin{equation*} Q(x,t):=e^{-tx}M(t),\quad M(t):=Ee^{tX}. \tag{5}\label{5} \end{equation*} The condition that $L(t_0)<\infty$ for some real $t_0>0$ implies that there exists $\mu:=EX\in[-\infty,\infty)$. So, by Jensen's inequality, for $x>\mu$ and $t<0$ we have $Q(x,t)\ge e^{t(\mu-x)}>1=Q(x,0)$. So, for all large enough real $x$ \begin{equation*} s(x)=-L^*(x)=T(x):=\ln\inf_{t\ge0}Q(x,t). \tag{1b}\label{1b} \end{equation*}

Let \begin{equation*} q(x):=P(X>x), \end{equation*} so that \begin{equation*} t(x)=\ln q(x). \tag{5.5}\label{5.5} \end{equation*} By Markov's inequality, $q(x)\le Q(x,t)$ for all real $t\ge0$. So, by \eqref{1b}, \begin{equation*} s(x)\ge t(x) \tag{5.75}\label{5.75} \end{equation*} eventually (that is, for all large enough real $x>0$).

It should be clear that, for \eqref{3} to hold, the tail function $q$ should vary regularly enough. (Indeed, the function $s$ is always concave; so, if, for instance, $t$ is a function decreasing to $-\infty$ in a neighborhood of $\infty$ so that $t$ is not asymptotically equivalent to any concave function, then, of course, \eqref{3} cannot hold.) On the other hand, it follows from \eqref{4} that $q$ cannot decrease near $\infty$ slower than all decreasing exponential functions. So, it seems reasonable to assume that the tail function $q$ is log concave in a neighborhood of $\infty$ (especially because its desirably "matching" function $e^s$ is log concave). It turns out that this is enough for \eqref{3}:

Proposition 1: If the tail function $q$ is log concave and nonzero in a neighborhood of $\infty$, then \eqref{3} holds.

Proof: By \eqref{5}, for real $t>0$,
\begin{equation*} \begin{aligned} M(t)=Ee^{tX}&=-\int_{-\infty}^\infty dq(u)e^{tu} \\ &=-\int_{-\infty}^\infty dq(u)\int_{-\infty}^u t\,dv\,e^{tv} \\ &=-\int_{-\infty}^\infty t\,dv\,e^{tv}\int_v^\infty dq(u) \\ &=t\int_{-\infty}^\infty \,dv\,e^{tv} q(v) \\ &=I_1+I_2, \end{aligned} \tag{6}\label{6} \end{equation*} where \begin{equation*} I_1:=t\int_{-\infty}^{x-A} \,dv\,e^{tv} q(v) \le t\int_{-\infty}^{x-A} \,dv\,e^{tv}=e^{t(x-A)}, \tag{7}\label{7} \end{equation*} \begin{equation*} I_2:=t\int_{x-A}^\infty \,dv\,e^{tv} e^{-g(v)}, \tag{8}\label{8} \end{equation*} $A$ is any real number, and $g:=-\ln q$, so that $g$ is (strictly) increasing, convex, $>0$, and $<\infty$ in a neighborhood of $\infty$ -- say on the interval \begin{equation*} [a,\infty) \end{equation*} for some real $a$. So, $g'>0$ on $[a,\infty)$, where $g'$ is (say) the right derivative of the convex function $g$. In what follows, by default $x\in[a,\infty)$.

Take now any $c\in(0,1)$ and let
\begin{equation*} A:=\frac{cg(x)}{g'(x)}, \quad t:=g'(x)-\frac1A=\frac{cg(x)-1}A, \tag{9}\label{9} \end{equation*} so that eventually $A>0$ and $t>0$. Note also that, by the convexity of $g$ on $[a,\infty)$, we have $(x-a)g'(x)\ge g(x)-g(a)$ (for $x\in[a,\infty)$), that is, $xg'(x)-g(x)\ge ag'(x)-g(a)$, so that
\begin{equation*} x-A=(1-c)x+c\frac{xg'(x)-g(x)}{g'(x)} \ge(1-c)x+c\frac{ag'(x)-g(a)}{g'(x)}. \tag{10}\label{10} \end{equation*} Also, again by the convexity of $g$ on $[a,\infty)$, there exists $g(\infty):=\lim_{x\to\infty}g'(x)\in(0,\infty]$. So, by \eqref{10}, $x-A\to\infty$ and hence wlog $x-A\ge a$, which will be assumed by default in the sequel. Using the convexity of $g$ on $[a,\infty)$ again, we have \begin{equation} g(v)\ge h_x(v):=g(x)+g'(x)(v-x) \end{equation} for real $v\ge x-A$. So, by \eqref{8} and \eqref{9}, \begin{equation*} e^{-tx}I_2\le t\int_{x-A}^\infty \,dv\,e^{tv} e^{-h_x(v)} \\ =e^{-g(x)+1}(cg(x)-1)=e^{-g(x)(1+o(1))}=q(x)^{1+o(1)} \tag{11}\label{11} \end{equation*} (as $x\to\infty$). Also, by \eqref{7} and \eqref{9}, \begin{equation*} e^{-tx}I_1\le e^{-tA}=e^{-cg(x)+1}=eq(x)^c. \end{equation*} Since $c\in(0,1)$ was arbitrary, we now get \begin{equation*} e^{-tx}I_1\le q(x)^{1+o(1)}. \tag{12}\label{12} \end{equation*} It follows from \eqref{1b}, \eqref{5}, \eqref{6}, \eqref{12}, \eqref{11}, and \eqref{5.5} that \begin{equation} s(x)\le\ln\big(q(x)^{1+o(1)}\big)\sim\ln q(x)=t(x). \tag{13}\label{13} \end{equation} Finally, \eqref{3} follows from \eqref{5.75} and \eqref{13}. $\quad\Box$

Remark: It seems that, following the lines of the above proof, one can show that it is necessary and sufficient for \eqref{3} that the function $t$ be asymptotically equivalent to some concave function. I am going to check this later, after getting some sleep.

$\newcommand{\R}{\mathbb R}$

Proposition 1: For \begin{equation*} s(x)\sim T(x)=\ln P(X>x) \tag{00}\label{00} \end{equation*} to hold (as $x\to\infty$), it is necessary and sufficient that \begin{equation*} T(x)\sim T_0(x) \tag{10}\label{10} \end{equation*} for some real-valued concave function $T_0$.

The relation $a\sim b$ is understood here as $a/b\to1$. In particular, $a\sim b$ implies that eventually the values of $|b|$ are not in the set $\{0,\infty\}$.

Proof of Proposition 1:

Part 1: Necessity: Assume that \eqref{00} holds; in particular, this implies that $T(x)\ne-\infty$ eventually (that is, for all large enough $x>0$); it is clear that $T(x)\ne0$ eventually. We have to show that then \eqref{10} holds as well.

By Cram´er's large deviation principle (see e.g. Corollary 2.2.19), for all real $x$ \begin{equation*} s(x)=-\inf_{y\in[x,\infty)}L^*(y), \tag{20}\label{20} \end{equation*} where \begin{equation*} L^*(y):=\sup_{t\in\R}(ty-L(t)),\quad L(t):=\ln Ee^{tX}, \tag{30}\label{30} \end{equation*} $X:=X_1$.

If $L(t)=\infty$ for all real $t>0$, then $L^*(y)=0$ for all real $y\ge0$, and hence, by \eqref{20}, $s(x)=0$ for all real $x\ge0$, so that \eqref{00} will not hold.

So, without loss of generality (wlog) \begin{equation} \text{$L(t_0)<\infty$ for some real $t_0>0$, } \tag{40}\label{40} \end{equation} and then, by \eqref{30}, $L^*(y)\ge t_0y-L(t_0)\to\infty$ as $y\to\infty$. Also, the function $L^*$ is convex, being the supremum of affine functions. So, by \eqref{20}, for all large enough real $x>0$ \begin{equation*} s(x)=-L^*(x)=\ln\inf_{t\in\R}Q(x,t), \tag{20a}\label{20a} \end{equation*} where \begin{equation*} Q(x,t):=e^{-tx}M(t),\quad M(t):=Ee^{tX}. \tag{50}\label{50} \end{equation*} Letting $T_0:=-L$ in a neighborhood of $\infty$ and looking at the first equality in \eqref{20a}, we see that \eqref{00} does imply \eqref{10}, for such a real-valued concave function $T_0$, which completes the necessity part of the proof.

Part 2: Sufficiency: Assume that \eqref{10} holds for some real-valued concave function $T_0$. We have to show that then \eqref{00} holds as well.

Since $T(x)\to-\infty$ (as $x\to\infty$), \eqref{10} and the concavity of $T_0$ imply that for some $\tau\in[-\infty,0)$ we have $T_0(x)/x\to\tau$ and $T(x)/x\to\tau$. So, \eqref{40} and \eqref{20a} hold.

It also follows that there exists $\mu:=EX\in[-\infty,\infty)$. So, by Jensen's inequality, for $x>\mu$ and $t<0$ we have $Q(x,t)\ge e^{t(\mu-x)}>1=Q(x,0)$. So, for all large enough real $x$ \begin{equation*} s(x)=-L^*(x)=\ln\inf_{t\ge0}Q(x,t). \tag{20b}\label{20b} \end{equation*}

Let \begin{equation*} q(x):=P(X>x), \end{equation*} so that \begin{equation*} T(x)=\ln q(x). \tag{60}\label{60} \end{equation*} By Markov's inequality, $q(x)\le Q(x,t)$ for all real $t\ge0$. So, by \eqref{20b}, \begin{equation*} s(x)\ge T(x) \tag{70}\label{70} \end{equation*} eventually (that is, for all large enough real $x>0$).

By \eqref{50}, for real $t>0$,
\begin{equation*} \begin{aligned} M(t)=Ee^{tX}&=-\int_{-\infty}^\infty dq(u)e^{tu} \\ &=-\int_{-\infty}^\infty dq(u)\int_{-\infty}^u t\,dv\,e^{tv} \\ &=-\int_{-\infty}^\infty t\,dv\,e^{tv}\int_v^\infty dq(u) \\ &=t\int_{-\infty}^\infty \,dv\,e^{tv} q(v) \\ &=I_1+I_2, \end{aligned} \tag{80}\label{80} \end{equation*} where \begin{equation*} I_1:=t\int_{-\infty}^{x-A} \,dv\,e^{tv} q(v) \le t\int_{-\infty}^{x-A} \,dv\,e^{tv}=e^{t(x-A)}, \tag{90}\label{90} \end{equation*} \begin{equation*} I_2:=t\int_{x-A}^\infty \,dv\,e^{tv} e^{-g(v)}, \tag{100}\label{100} \end{equation*} $A$ is a real number (to be specified later), and $$g(v):=-\ln q(v)=-T(v).$$ Let $$g_0(v):=-T_0(v).$$

The function $g_0$ is convex, (strictly) increasing, $>0$, and $<\infty$ in a neighborhood of $\infty$ -- say on the interval \begin{equation*} [a,\infty) \end{equation*} for some real $a$. So, $g_0'>0$ on $[a,\infty)$, where $g_0'$ is (say) the right derivative of the convex function $g_0$. In what follows, by default $x\in[a,\infty)$.

Take now any $c\in(0,1)$ and let
\begin{equation*} A:=A(x):=\frac{cg_0(x)}{g_0'(x)}, \tag{110}\label{110} \end{equation*} so that eventually $A>0$. Note also that, by the convexity of $g_0$ on $[a,\infty)$, we have $(x-a)g_0'(x)\ge g_0(x)-g_0(a)$ (for $x\in[a,\infty)$), that is, $xg_0'(x)-g_0(x)\ge ag_0'(x)-g_0(a)$, so that
\begin{equation*} x-A=(1-c)x+c\frac{xg_0'(x)-g_0(x)}{g_0'(x)} \\ \ge(1-c)x+c\frac{ag_0'(x)-g_0(a)}{g_0'(x)}. \tag{120}\label{120} \end{equation*} Also, again by the convexity of $g_0$ on $[a,\infty)$, there exists $g_0(\infty):=\lim_{x\to\infty}g_0'(x)\in(0,\infty]$. So, by \eqref{120}, $x-A\to\infty$ and hence wlog $x-A\ge a$, which will be assumed by default in the sequel.

Let now \begin{equation*} k(x):=\inf_{v\in[x-A,\infty)}\frac{g(v)}{g_0(v)}, \tag{130}\label{130} \end{equation*} so that, in view of \eqref{10} and because $x-A\to\infty$, we have \begin{equation*} k(x)\to1. \tag{140}\label{140} \end{equation*}

Now comes the crucial point, which is choosing the value of $t$ as follows: \begin{equation*} t:=t(x):=k(x)g_0'(x)-\frac1A=\frac{ck(x)g_0(x)-1}A. \tag{150}\label{150} \end{equation*}

Using the convexity of $g_0$ on $[a,\infty)$ again, we have \begin{equation*} g_0(v)\ge h_x(v):=g_0(x)+g_0'(x)(v-x) \end{equation*} for real $v\ge x-A$. So,
by \eqref{100}, \eqref{130}, \eqref{150}, \eqref{140}, and \eqref{10}, \begin{equation*} e^{-tx}I_2\le \int_{x-A}^\infty \,dv\,e^{tv} e^{-k(x)h_x(v)} \\ =e^{-k(x)g_0(x)+1}(ck(x)g_0(x)-1)=e^{-k(x)g_0(x)(1+o(1))}=q(x)^{1+o(1)} \tag{160}\label{160} \end{equation*} (as $x\to\infty$). Also, by \eqref{90} and \eqref{150}, \begin{equation*} e^{-tx}I_1\le e^{-tA}=e^{-ck(x)g_0(x)+1}=eq(x)^{ck(x)}. \end{equation*} In view of \eqref{140} and because $c\in(0,1)$ was arbitrary, we now get \begin{equation*} e^{-tx}I_1\le q(x)^{1+o(1)}. \tag{170}\label{170} \end{equation*} It follows from \eqref{20b}, \eqref{50}, \eqref{80}, \eqref{170}, \eqref{160}, and \eqref{60} that \begin{equation*} s(x)\le\ln\big(q(x)^{1+o(1)}\big)\sim\ln q(x)=T(x). \tag{180}\label{180} \end{equation*} Finally, \eqref{00} follows from \eqref{70} and \eqref{180}.

This completes the sufficiency part of the proof as well. $\quad\Box$

$\newcommand{\R}{\mathbb R}$BBy Cram´er's large deviation principle (see e.g. Corollary 2.2.19), for all real $x$ \begin{equation*} s(x)=-\inf_{y\in[x,\infty)}L^*(y), \tag{1}\label{1} \end{equation*} where \begin{equation*} L^*(y):=\sup_{t\in\R}(ty-L(t)),\quad L(t):=\ln Ee^{tX}, \tag{2}\label{2} \end{equation*} $X:=X_1$.

Take now any $c\in(0,1)$ and let
\begin{equation*} A:=\frac{cg(x)}{g'(x)}, \quad t:=g'(x)-\frac1A=\frac{cg(x)-1}A, \tag{9}\label{9} \end{equation*} so that eventually $A>0$ and $t>0$. Note also that, by the convexity of $g$ on $[a,\infty)$, we have $(x-a)g'(x)\ge g(x)-g(a)$ (for $x\in[a,\infty)$), that is, $xg'(x)-g(x)\ge ag'(x)-g(a)$, so that
\begin{equation*} x-A=(1-c)x+c\frac{xg'(x)-g(x)}{g'(x)} \ge(1-c)x+c\frac{ag'(x)-g(a)}{g'(x)}. \tag{10}\label{10} \end{equation*} Also, again by the convexity of $g$ on $[a,\infty)$, there exists $g(\infty):=\lim_{x\to\infty}g'(x)\in(0,\infty]$. So, by \eqref{10}, $x-A\to\infty$ and hence wlog $x-A\ge a$, which will be assumed by default in the sequel. Using the convexity of $g$ on $[a,\infty)$ again, we have \begin{equation} g(v)\ge h_x(v):=g(x)+g'(x)(v-x) \end{equation} for real $v\ge x-A$. So, by \eqref{8} and \eqref{9}, \begin{equation*} e^{-tx}I_2\le t\int_{x-A}^\infty \,dv\,e^{tv} e^{-h_x(v)} \\ =e^{-g(x)+1}(cg(x)-1)=e^{-g(x)(1+o(1))}=q(x)^{1+o(1)} \tag{11}\label{11} \end{equation*} (as $x\to\infty$). Also, by \eqref{7} and \eqref{9}, \begin{equation*} e^{-tx}I_1\le e^{-tA}=e^{-cg(x)+1}=eq(x)^c. \end{equation*} Since $c\in(0,1)$ was arbitrary, we now get \begin{equation*} e^{-tx}I_1\le q(x)^{1+o(1)}. \tag{12}\label{12} \end{equation*} It follows from \eqref{1b}, \eqref{5}, \eqref{6}, \eqref{12}, \eqref{11}, and \eqref{5.5} that \begin{equation} s(x)\le\ln\big(q(x)^{1+o(1)}\big)\sim\ln q(x)=t(x). \tag{13}\label{13} \end{equation} Finally, \eqref{3} follows from \eqref{5.75} and \eqref{13}. $\quad\Box$

$\newcommand{\R}{\mathbb R}$By Cram´er's large deviation principle (see e.g. Corollary 2.2.19), for all real $x$ \begin{equation*} s(x)=-\inf_{y\in[x,\infty)}L^*(y), \tag{1}\label{1} \end{equation*} where \begin{equation*} L^*(y):=\sup_{t\in\R}(ty-L(t)),\quad L(t):=\ln Ee^{tX}, \tag{2}\label{2} \end{equation*} $X:=X_1$.

Take now any $c\in(0,1)$ and let
\begin{equation*} A:=\frac{cg(x)}{g'(x)}, \quad t:=g'(x)-\frac1A=\frac{cg(x)-1}A, \tag{9}\label{9} \end{equation*} so that eventually $A>0$ and $t>0$. Note also that, by the convexity of $g$ on $[a,\infty)$, we have $(x-a)g'(x)\ge g(x)-g(a)$ (for $x\in[a,\infty)$), that is, $xg'(x)-g(x)\ge ag'(x)-g(a)$, so that
\begin{equation*} x-A=(1-c)x+c\frac{xg'(x)-g(x)}{g'(x)} \ge(1-c)x+c\frac{ag'(x)-g(a)}{g'(x)}. \tag{10}\label{10} \end{equation*} Also, again by the convexity of $g$ on $[a,\infty)$, there exists $g(\infty):=\lim_{x\to\infty}g'(x)\in(0,\infty]$. So, by \eqref{10}, $x-A\to\infty$ and hence wlog $x-A\ge a$, which will be assumed by default in the sequel. Using the convexity of $g$ on $[a,\infty)$ again, we have \begin{equation} g(v)\ge h_x(v):=g(x)+g'(x)(v-x) \end{equation} for real $v\ge x-A$. So, by \eqref{8} and \eqref{9}, \begin{equation*} e^{-tx}I_2\le t\int_{x-A}^\infty \,dv\,e^{tv} e^{-h_x(v)} \\ =e^{-g(x)+1}(cg(x)-1)=e^{-g(x)(1+o(1))}=q(x)^{1+o(1)} \tag{11}\label{11} \end{equation*} (as $x\to\infty$). Also, by \eqref{7} and \eqref{9}, \begin{equation*} e^{-tx}I_1\le e^{-tA}=e^{-cg(x)+1}=eq(x)^c. \end{equation*} Since $c\in(0,1)$ was arbitrary, we now get \begin{equation*} e^{-tx}I_1\le q(x)^{1+o(1)}. \tag{12}\label{12} \end{equation*} It follows from \eqref{1b}, \eqref{5}, \eqref{6}, \eqref{12}, \eqref{11}, and \eqref{5.5} that \begin{equation} s(x)\le\ln\big(q(x)^{1+o(1)}\big)\sim\ln q(x)=t(x). \tag{13}\label{13} \end{equation} Finally, \eqref{3} follows from \eqref{5.75} and \eqref{13}. $\quad\Box$

Remark: It seems that, following the lines of the above proof, one can show that it is necessary and sufficient for \eqref{3} that the function $t$ be asymptotically equivalent to some concave function. I am going to check this later, after getting some sleep.