2.1. Regularly varying and asymptotically smooth density functions

Regularly varying survival functions have long been used in probability theory. For the study of the properties of $m_{X}(\!\cdot\!)$ , it is preferable to consider regularly varying probability density functions because $m_{X}(\!\cdot\!)$ possesses a useful representation in terms of density functions as explained in Section 2.2. Note that variables with regularly varying density functions have regularly varying survival functions, but the reverse is not true. This is a direct consequence of Proposition 1.5.10 in Bingham et al. (Reference Bingham, Goldie and Teugels1987). Recall that a positive measurable function $L(\!\cdot\!)$ is said to be slowly varying if it is defined on some neighborhood $\left(x_0, \infty\right)$ of infinity, with $x_0\geq 0$ , and $\lim _{x \rightarrow \infty} \frac{L(t x)}{L(x)} =1$ for all $t\gt0$ . Intuitively speaking, a slowly varying function is a function that grows/decays asymptotically slower than any polynomial. For example, $\log(\!\cdot\!)$ or any function converging to a constant is a slowly varying function. We are now in a position to recall the definition of a regularly varying probability density function for positive random variables.

An important property of regularly varying densities of positive random variables is that they form a stable family under convolution. Precisely, the convolution of two regularly varying densities is still regularly varying with a tail index equal to the minimum of the tail indices of the convoluted density functions. More precisely, we have the following property:

(2.1) \begin{equation}f_X\text{ and }f_Y\text{ regularly varying}\Rightarrow\lim_{s\rightarrow \infty }\frac{f_{X+Y}\left( s\right) }{f_{X}\left(s\right) +f_{Y}\left( s\right) }=1,\end{equation}

where $f_{X+Y}(\!\cdot\!)$ is the convolution of $f_X(\!\cdot\!)$ and $f_Y(\!\cdot\!)$ . See, for example, Theorem 1.1 in Bingham et al. (Reference Bingham, Goldie and Omey2006) for a proof.

In Table 1, we summarize some distributions with regularly varying densities, along with the index and slowly varying function $L(\!\cdot\!)$ appearing in the representation of the density $f(\!\cdot\!)$ as stated in Definition 2.1. The notation $\Gamma (\!\cdot\!)$ stands for the Gamma function and $\zeta (\!\cdot\!)$ for the Riemann Zeta function. Note that the parameters of the distribution and density functions have been chosen to have index $\alpha$ in all cases. To this end, we may deviate from the standard parameter choices for these distributions. Type I and Type II Pareto, Log-Gamma, and Dagun distributions are often defined with a parameter $\beta=\alpha-1$ instead of the parameter $\alpha$ which appears in Table 1. Type III and Type VI Pareto distributions are often specified with parameter $\gamma=\frac{\lambda}{\alpha-1}$ ( $\lambda=1$ in the case of Type III) instead of $\alpha$ . Note that the support is $(\vartheta,\infty)$ for certain distributions listed in Table 1. In the remainder of this paper, we assume that the supports of X and Y (and hence of their sum S) are $(0,\infty)$ . If the support of the variables is of the form $(\vartheta,\infty)$ , we can either extend the density function $f\left(\cdot\right)$ by assuming $f(x)=0$ for $x\in \left(0,\vartheta\right]$ or shift the random variable under consideration by $\vartheta$ . Note that, even though assuming a regularly varying density is more restrictive than assuming regular variation of the survival function, it is not such a restrictive assumption in practice since, as seen in Table 1, it is verified by many well-known heavy-tailed distributions.

Table 1. Families of distributions with regularly varying densities with index $\alpha$ .

To conduct our study, we need to impose a condition known in the literature as asymptotic smoothness after Barbe and McCormick (Reference Barbe and McCormick2005) and recalled next.

Definition 2.2. A density function $f(\!\cdot\!)$ defined on $\left(0,\infty\right)$ is said to be asymptotically smooth with index $\alpha \gt1$ if

\begin{equation*}\lim_{\delta \rightarrow 0}\limsup_{t\rightarrow \infty }\sup_{0\lt|x|\leq\delta }\left\vert \frac{f(t(1-x))-f(t)}{xf(t)}-\alpha \right\vert =0.\end{equation*}

Asymptotically smooth functions are related to regularly varying ones with the same index. When $f(\!\cdot\!)$ is asymptotically smooth and differentiable, then it is also regularly varying with the same index. Conversely, if $f(\!\cdot\!)$ is regularly varying, differentiable and has an ultimately monotone derivative, then it is asymptotically smooth (see Proposition 2.1 in Barbe and McCormick, Reference Barbe and McCormick2005). It must be remarked that all regularly varying densities considered in Table 1 have ultimately monotone derivatives and therefore are asymptotically smooth. The proof of this statement is given in Appendix A.

2.2. Representation of the conditional expectation given the sum in terms of size biasing

Let us consider a nonnegative random variable X with distribution function $F_{X}$ and finite and strictly positive expected value. If X is a random variable having a regularly varying density with index $\alpha \gt1$ , it is well known that $k\lt\alpha -1$ implies $\mathrm{E}[X^{k}]\lt\infty $ . Hence, $\mathrm{E}[X]\lt\infty$ if $\alpha\gt2$ , and we retain this assumption for all variables considered throughout this paper. The distribution function of the size-biased version $\widetilde{X}$ of X is then given by

\begin{equation*}\mathrm{P}[\widetilde{X}\leq t]=\frac{1}{\mathrm{E}[X]}\int_{0}^{t}x \mathrm{d}F_{X}(x),\hspace{2mm}t\geq 0.\end{equation*}

We refer the reader to Arratia et al. (Reference Arratia, Goldstein and Kochman2019) for an introduction to the size-biased transform. Note that for any random variables X, Y with joint density f, the regression function can be expressed as

\begin{align*}m_X(s)=\frac{1}{f_{X+Y}(s)}\int_0^s x f(x,s-x)\mathrm{d}x.\end{align*}

Under our framework, that is, considering independent random variables, in Denuit (Reference Denuit2019) a representation for the regression function $m_{X}(\!\cdot\!)$ is established in terms of the size-biased transform of X. Assume that X and Y have finite means and respective probability density functions $f_X(\!\cdot\!)$ and $f_Y(\!\cdot\!)$ . Since X possesses a density function $f_{X}(\!\cdot\!)$ and its expected value is finite, its size-biased version $\widetilde{X}$ also possesses a density given by $f_{\widetilde{X}}(x)=xf_{X}(x)/\mathrm{E}[X]$ . If $\widetilde{X}$ is independent of Y, then the representation

(2.2) \begin{equation}m_{X}(s)=\mathrm{E}[X] \frac{f_{\widetilde{X}+Y}(s)}{f_{X+Y}(s)}\end{equation}

holds true for any $s\gt 0$ .

2.3. Minimum limit value of ${\textbf{\textit{m}}}_{\textbf{\textit{X}}}( \!\cdot\! ) $

Understanding the asymptotic behavior of $m_{X}\left( s\right) $ as s gets large is important in practice. For example, in the case of risk sharing, $m_{X}\left( s\right) $ represents the contribution to be paid by the participant bringing X in the event of a large pool loss. If the supports of X and Y are not bounded, we might expect this contribution to tend toward infinity when s tends to infinity. We shall see that this is not always the case. However, under some technical assumptions, there exists a minimum value of the limit below which the contribution cannot fall, which is $\mathrm{E}[X]$ as established below. This limit is reached in particular when X and Y have regularly varying densities with tail indices whose difference is larger than 1 as it will become clear in the next section.

Proposition 2.3. Assume that X and Y have densities $f_{X}( \!\cdot\! ) $ and $ f_{Y}( \!\cdot\! ) $ such that $f_{Y}( \!\cdot\! ) $ is bounded and ultimately decreasing, that is there exists $y_{0}$ such that $ f_{Y}( \!\cdot\! ) $ is decreasing over $(y_{0},\infty )$ . Moreover, assume that $\sup_{ x\in (s-y_{0},s)}f_{X}\left( x\right) =\mathrm{o} (f_{X+Y}(s))$ as $s\rightarrow \infty $ , then

\begin{equation*} \liminf_{s\rightarrow \infty } m_{X}\left( s\right) \geq \mathrm{E}[X]. \end{equation*}

Proof. Note that, for $s\gt y_0$ ,

(2.3) \begin{equation} m_{X}\left( s\right) =\frac{\int_{0}^{s}xf_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}{\int_{0}^{s}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x} \geq \frac{\int_{0}^{s-y_{0}}xf_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x }{\int_{0}^{s-y_{0}}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x+\int_{s-y_{0}}^{s}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}. \end{equation}

Since $f_{Y}\left( s-\cdot \right) $ is increasing on $\left( 0,s-y_{0}\right) $ , we have

\begin{equation*} \mathrm{E}[Xf_{Y}\left( s-X\right) |X\leq s-y_{0}]\geq \mathrm{E}[X|X\leq s-y_{0}]\mathrm{E}[f_{Y}\left( s-X\right) |X\leq s-y_{0}] \end{equation*}

or equivalently

\begin{equation*} \frac{\int_{0}^{s-y_{0}}xf_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}{ \int_{0}^{s-y_{0}}f_{X}\left( x\right) \mathrm{d}x}\geq \frac{ \int_{0}^{s-y_{0}}xf_{X}\left( x\right) \mathrm{d}x}{\int_{0}^{s-y_{0}}f_{X}\left( x\right) \mathrm{d}x}\frac{\int_{0}^{s-y_{0}}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}{\int_{0}^{s-y_{0}}f_{X}\left( x\right) \mathrm{d}x}, \end{equation*}

what implies

\begin{equation*} \frac{\int_{0}^{s-y_{0}}xf_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}{\int_{0}^{s-y_{0}}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}\geq \frac{ \int_{0}^{s-y_{0}}xf_{X}\left( x\right) \mathrm{d}x}{\int_{0}^{s-y_{0}}f_{X}\left( x\right) \mathrm{d}x}=\mathrm{E}[X|X\leq s-y_{0}]. \end{equation*}

Therefore, we deduce that

(2.4) \begin{eqnarray} &&\frac{\int_{0}^{s-y_{0}}xf_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}{ \int_{0}^{s-y_{0}}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x+\int_{s-y_{0}}^{s}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x} \\ &=&\frac{\int_{0}^{s-y_{0}}xf_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}{ \int_{0}^{s-y_{0}}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x}\frac{1}{ 1+\int_{s-y_{0}}^{s}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x/\int_{0}^{s-y_{0}}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x} \nonumber\\ &\geq &\mathrm{E}[X|X\leq s-y_{0}]\frac{1}{1+\int_{s-y_{0}}^{s}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x/\left(f_{X+Y}(s)-\int_{s-y_{0}}^{s}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x\right)} \nonumber. \end{eqnarray}

Moreover,

\begin{equation*}\frac{\int_{s-y_{0}}^{s}f_{X}\left( x\right)f_{Y}\left( s-x\right) \mathrm{d}x}{f_{X+Y}(s)}\leq F_Y(y_0)\frac{\left(\sup_{x\in (s-y_{0},s)}f_{X}\left( x\right) \right)}{f_{X+Y}(s)}\end{equation*}

and, since $\sup_{x\in (s-y_{0},s)}f_{X}\left( x\right) =\mathrm{o} (f_{X+Y}(s))$ as $s\rightarrow \infty $ , it follows that

\begin{equation*}\lim_{s\rightarrow \infty }\frac{\int_{s-y_{0}}^{s}f_{X}\left( x\right)f_{Y}\left( s-x\right) \mathrm{d}x}{f_{X+Y}(s)}=0\end{equation*}

And, therefore

\begin{equation*} \lim_{s\rightarrow \infty }\frac{1}{1+\int_{s-y_{0}}^{s}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x/ \left(f_{X+Y}(s)-\int_{s-y_{0}}^{s}f_{X}\left( x\right) f_{Y}\left( s-x\right) \mathrm{d}x\right)}=1 \end{equation*}

and, as $\lim_{s\rightarrow \infty }\mathrm{E}[X|X\leq s-y_{0}]=\mathrm{E}[X]$ , from (2.3) and (2.4), $\liminf_{s\rightarrow \infty } m_{X}\left( s\right) \geq \mathrm{E}[X]$ .

2.4. Representation of the first derivative of the conditional expectation given the sum

Let us assume that the first derivative $f_Y'(\!\cdot\!)$ of the probability density function $f_Y(\!\cdot\!)$ exists and is well defined. Let $(X_{1},Y_{1})$ and $(X_{2},Y_{2})$ be independent copies of (X, Y), with respective sums $S_{1}=X_{1}+Y_{1}$ and $S_{2}=X_{2}+Y_{2}$ . Then Denuit and Robert (Reference Denuit and Robert2021a) demonstrated that the first derivative $m_X'(\!\cdot\!)$ of $m_X(\!\cdot\!)$ exists and can be represented as

\begin{equation*}m_{X}^{\prime }(s)=\frac{1}{4}\mathrm{E}\left[ \left. \left(X_{1}-X_{2}\right) \left( \frac{f_{Y}^{\prime }(Y_{1})}{f_{Y}(Y_{1})}-\frac{f_{Y}^{\prime }(Y_{2})}{f_{Y}(Y_{2})}\right) \mathrm{I}\left[ X_{1} \gt X_{2}\right] \right\vert S_{1}=s,S_{2}=s\right]\end{equation*}

where $\mathrm{I}[\cdot]$ denotes the indicator function (equal to 1 when the condition appearing within brackets is fulfilled and to 0 otherwise). In the latter expression, the inequality $Y_{1}\lt Y_{2}$ must hold true because $X_{1}\gt X_{2}$ and the sums $X_{1}+Y_{1}$ and $X_{2}+Y_{2}$ are constrained to be equal to some fixed value s. We deduce that if $f_{Y}(\!\cdot\!)$ is assumed to be log-concave, that is $\log f_{Y}(\!\cdot\!)$ is concave, then the ratio $f_{Y}^{\prime }(\!\cdot\!)/f_{Y}(\!\cdot\!)$ is decreasing and the second factor in the last conditional expectation must thus be positive. Therefore $m_{X}^{\prime }(s)\geq 0$ for any $s\gt 0$ . Log-concavity of $f_Y(\!\cdot\!)$ thus guarantees that $m_X(\!\cdot\!)$ is non-decreasing.

This expression clarifies the role that log-concavity plays in the monotonicity of $m_X(\!\cdot\!)$ . However, it must be remarked that, under the assumption of log-concavity, not only $m_X(\!\cdot\!)$ is non-decreasing, but the variable $\lbrace X\mid S=s\rbrace$ is non-decreasing in s in the usual stochastic order. This means that the expected value of any increasing transformation of the variable is non-decreasing in the conditioning value s of S. This result is a consequence of Section 2 in Efron (Reference Efron1965), and a proof computing $m_{X}^{\prime }(s)$ is provided in Saumard and Wellner (Reference Saumard and Wellner2014) using symmetrization arguments for independent log-concave variables.

Consider a regularly varying density $f(\!\cdot\!)$ with tail index $\alpha$ such that $f^{\prime }(\!\cdot\!)$ is ultimately monotonic (from Property A.1, this is the case for all distributions listed in Table 1). Then, we can derive from Karamata’s theorem (Karamata, Reference Karamata1933) that

\begin{equation*}\frac{f^{\prime }(x)}{f(x)}\sim -\alpha \frac{1}{x}\end{equation*}

and therefore $f^{\prime }(\!\cdot\!)/f(\!\cdot\!)$ cannot be ultimately decreasing. Hence, it is natural to investigate whether $m_{X}(\!\cdot\!)$ can be decreasing for some tail indices of X and Y and for some values s of S. This is precisely one of the questions investigated in the present paper. Therefore, the following sections will delve into, for X, Y independent, nonnegative random variables with well-defined density functions, how a heavier-tail of Y affects the monotonicity of $m_{X}(\!\cdot\!)$ .

3.1. Difference in tail indices larger than 1 ( $\alpha_X-\alpha_Y\gt1$ )

The next result shows that when the difference between tail indices exceeds 1, the conditional expectation given the sum tends to the expected value for the term with a larger index, as the sum tends to infinity.

Proposition 3.2. If X and Y possess regularly varying densities $f_X(\!\cdot\!)$ and $f_Y(\!\cdot\!)$ with respective indices $\alpha _{X}$ and $\alpha _{Y}$ such that $\alpha_{X}\gt\alpha _{Y}+1$ , then

\begin{equation*}\lim_{s\rightarrow \infty }m_{X}(s)=\mathrm{E}\left[ X\right].\end{equation*}

Proof. Since $f_{X}(\!\cdot\!)$ is regularly varying with index $\alpha _{X}\gt2$ , we have that $f_{\widetilde{X}}(\!\cdot\!)$ is regularly varying with index $\alpha _{\widetilde{X}}=\alpha _{X}-1$ . As $\alpha _{X}\gt\alpha _{Y}+1$ , then $\alpha _{\widetilde{X}}\gt\alpha _{Y}$ and, therefore, $f_{\widetilde{X}}\left( s\right) =o\left(f_{Y}\left( s\right) \right) $ and $f_{X}\left( s\right) =o\left( f_{Y}\left( s\right) \right) $ , As a consequence of (2.1), then

\begin{equation*}\lim_{s\rightarrow \infty }m_{X}(s)=\mathrm{E}\left[ X\right]\lim_{s\rightarrow \infty }\frac{f_{\widetilde{X}}\left( s\right) +f_{Y}\left(s\right) }{f_{X}\left( s\right) +f_{Y}\left( s\right) } = \mathrm{E}[X].\end{equation*}

This ends the proof.

Note that a direct implication of Proposition 3.2 for absolutely continuous distributions is that $m_X(\!\cdot\!)$ is bounded. The following result shows that $m_X(\!\cdot\!)$ cannot be monotonic over $(0,\infty)$ in the case considered in Proposition 3.2.

The following example illustrates the behavior of $m_X(\!\cdot\!)$ in the case where the difference in the tail indices exceeds 1.

Example 3.4. Let us consider two independent random variables $X\sim LG(\alpha_{X},\lambda_X)$ and $Y\sim P(IV)(\theta,\alpha_{Y},\vartheta,\lambda_Y)$ such that $\alpha_{X}\gt\alpha_Y+1$ . Figure 1 shows a numerical representation considering the parameter values $\vartheta =\theta =1$ , $\lambda_X=\lambda_Y=2$ , $\alpha_{X}=5$ , and $\alpha_Y=2$ . We can observe that $m_{X}(\!\cdot\!)$ is not monotonic over $(0,\infty)$ , in accordance with Proposition 3.3. Here, $m_{X}(\!\cdot\!)$ reaches its maximum and starts decreasing beyond $F_S^{-1}(0.91)$ .

Figure 1. Conditional expectation $m_X(\!\cdot\!)$ (solid line) and horizontal line at $\mathrm{E}[X]$ (dashed line) when $X\sim LG(\alpha_{X},\lambda_X)$ and $Y\sim P(IV)(\theta,\alpha_{Y},\vartheta,\lambda_Y)$ with $\vartheta =\theta =1$ , $\lambda_X=\lambda_Y=2$ , $\alpha_{X}=5$ , and $\alpha_Y=2$ .

Let us remark that the preceding results serve to enhance the discussion provided in Section 4 of Major and Mildenhall (Reference Major and Mildenhall2020) and Section 14.3 of Mildenhall and Major (Reference Mildenhall and Major2022), in which several examples where, for combinations of heavy-tailed units, $m_X(\!\cdot\!)$ may not be monotone. The authors also discussed that a light-tailed unit combined with a heavy-tailed unit could lead to analogous behaviors, and this is, indeed, the case. Note that Propositions 3.2 and 3.3 can be extended to scenarios where X does not have a regularly varying density but has a density dominated by $f_Y(\!\cdot\!)$ in the tail. This follows from Theorem 2.1 in Bingham et al. (Reference Bingham, Goldie and Omey2006), which states that, if $f_X(\!\cdot\!)$ and $f_Y(\!\cdot\!)$ are probability densities on $\mathbb{R}$ , with $f_Y(\!\cdot\!)$ regularly varying and $f_X(x)=o(f_Y(x))$ as x tends to infinity, then $\lim_{s\rightarrow \infty}\frac{f_{X+Y}(s)}{f_Y(s)}=1$ .

Since $m_{X}(s)+m_{Y}(s)=s$ , both functions cannot decrease simultaneously. Therefore, if either $m_{X}(\!\cdot\!)$ and $m_{Y}(\!\cdot\!)$ have a decreasing interval, as the other increases in such interval, it implies that, locally, $m_{X}(S)$ and $m_{Y}(S)$ are negatively dependent. This means that, given that S belongs to such interval, $m_{X}(S)$ tends to take larger values as $m_{Y}(S)$ takes smaller values and vice-versa. Although in the scenario considered in this section, even if the conditional expectation given the sum must decrease over some values of the sum according to Proposition 3.3, it is interesting to point out that, globally, $m_{X}(S)$ and $m_{Y}(S)$ always remain positively correlated. This is precisely stated next.

While this paper considers random variables X and Y possessing regularly varying density functions, the next result holds more generally for any positive variables with finite second-order moments. Let us remark that, if X and Y possess regularly varying densities with tail indices $\alpha _{X}$ and $\alpha _{Y}$ , respectively, then the assumption of finite second-order moments is equivalent to $\min\{\alpha _{X},\alpha _{Y}\}\gt3$ .

Proposition 3.6. Considering random variables X and Y, if their second-order moments are finite, then

(3.2) \begin{equation} \mathrm{Cov}[m_{X}(S),m_{Y}(S)]\geq 0. \end{equation}

Proof. First notice that for any s in the support of $S=X+Y$ , we have

\begin{equation*}\mathrm{Var}\left[ X|S=s\right] =\mathrm{Var}\left[ X-s|S=s\right] =\mathrm{\ Var}\left[ -Y|S=s\right] =\mathrm{Var}\left[ Y|S=s\right] .\end{equation*}

This implies that

(3.3) \begin{equation} \mathrm{Var}[Y\mid S=s]=\mathrm{Var}[X\mid S=s].\end{equation}

Now, we can write

\begin{equation*}0=\mathrm{Var}[S|S=s]=\mathrm{Var}[X|S=s]+\mathrm{Var}[Y\mid S=s]+2\mathrm{Cov}[X, Y|S=s].\end{equation*}

It follows from (3.3) that

(3.4) \begin{equation} \mathrm{Cov}[X, Y|S=s]=-\mathrm{Var}[X|S=s].\end{equation}

Then,

\begin{equation*}\begin{split}\mathrm{Cov}[X, Y]&= \mathrm{E}\big[\mathrm{Cov}[X, Y \mid S]\big]+\mathrm{Cov}\big[\mathrm{E}[X \mid S], \mathrm{E}[Y \mid S]\big] \\&= -\mathrm{E}\big[\mathrm{Var}[Y \mid S]\big]+\mathrm{Cov}\big[\mathrm{E}[X\mid S], \mathrm{E}[Y \mid S]\big] \\&= \mathrm{Var}\big[\mathrm{E}[Y \mid S]\big]-\mathrm{Var}[Y]+\mathrm{Cov}\big[\mathrm{E}[X \mid S], \mathrm{E}[Y \mid S]\big].\end{split}\end{equation*}

so that we get

(3.5) \begin{equation} \mathrm{Cov}\big[\mathrm{E}[X \mid S], \mathrm{E}[Y \mid S]\big]=\mathrm{Cov}[X, Y]+\mathrm{Var}[Y]-\mathrm{Var}\big[\mathrm{E}[Y \mid S]\big].\end{equation}

Considering (3.5), Jensen’s inequality ensures that $\mathrm{Var}[Y]\geq\mathrm{Var}\big[\mathrm{E}[Y \mid S]\big]$ . The announced inequality (3.2) finally follows from $\mathrm{Cov}[X, Y]=0$ .

We can now briefly comment on the result stated in Proposition 3.5:

• • As a direct consequence of (3.3), $\mathrm{Var}[X\mid S]=\mathrm{Var}[Y\mid S]$ a.s.

• • We see from (3.4) that $\mathrm{Cov}[X, Y|S=s]\leq 0$ for all s, which could be expected since both variables are positive and the sum is fixed. Therefore, a greater value taken by one of the variables is negatively influencing the second variable. Again, (3.4) implies $\mathrm{Cov}[X, Y|S]=-\mathrm{Var}[X\mid S]$ a.s. Note that, considering variables with log-concave densities, X and Y would not only have a negative covariance given S but would also be negatively associated (see Theorem 2.8 in Joag-Dev and Proschan, Reference Joag-Dev and Proschan1983).

Regarding risk-sharing, positive dependence means that, overall, participants have common interests as their contributions are likely to be large or small together. This positive global relationship between $m_{X}(S)$ and $m_{Y}(S)$ holds true even if the functions $m_{X}(\!\cdot\!)$ and $m_{Y}(\!\cdot\!)$ are not everywhere increasing. When $m_{X}(\!\cdot\!)$ and $m_{Y}(\!\cdot\!)$ refer to the contributions to the pool, if one of them decreases in terms of the total loss, it will then be at the expense of the other contribution assuming a greater part of such total loss. Inequality (3.2) indicates that if there are values for which the monotonicity of $m_{X}(\!\cdot\!)$ and $m_{Y}(\!\cdot\!)$ differ, those must be values with a small probability of occurrence or with slight differences in the increasing/decreasing rate, as overall, the dependence remains positive.

3.2. Difference in tail indices less than 1 ( $0\leq\alpha_X-\alpha_Y \lt 1$ )

When the difference between tail indices is less than 1, the conditional expectation given the sum is not bounded as the sum tends to infinity. This is in contrast with the preceding case.

Proof. Since $f_X(\!\cdot\!)$ is regularly varying with index $\alpha_X\gt2$ , $f_{\widetilde{X}}(\!\cdot\!)$ is regularly varying with index $\alpha_{\widetilde{X}}=\alpha_X-1\lt\alpha_Y$ . Because $f_Y(\!\cdot\!)$ is a regularly varying density, it implies that $f_Y(s)=o\left(f_{\widetilde{X}}(s)\right)$ . Since $\alpha_{\widetilde{X}}\lt\alpha_X$ , it also follows that $f_X(s)=o\left(f_{\widetilde{X}}(s)\right)$ . Therefore, by (2.1):

\begin{align*}m_X(s)=\mathrm{E}[X] \frac{f_{\widetilde{X}+Y}(s)}{f_{X+Y}(s)}\sim \mathrm{E}[X] \frac{f_{\widetilde{X}}(s)+f_Y(s)}{f_X(s)+f_Y(s)}=\mathrm{E}[X] \frac{f_{\widetilde{X}}(s)+o\left(f_{\widetilde{X}}(s)\right)}{o\left(f_{\widetilde{X}}(s)\right)+o\left(f_{\widetilde{X}}(s)\right)}.\end{align*}

From this expression, we can see that, as $s\rightarrow \infty$ , $m_X(s)\rightarrow \infty$ and the assertion follows.

From the previous result, we know that $m_X\left(\cdot\right)$ diverges as $s\rightarrow \infty$ . However, this is not sufficient to ensure that it is an increasing function. A variety of different behaviors may occur when the difference in tail indices is less than 1. Example 3.8 shows an increasing behavior of $m_X(\!\cdot\!),$ while Example 3.9 shows that it may be decreasing over a range of central values. Note that we cannot ensure the monotonicity of $m_X(\!\cdot\!)$ , but we can check it numerically over an interval (0,b) for a large b (e.g. $b\gt \mathrm{E}[X+Y]+ 10^2\mathrm{Var}[X+Y]$ ). Finally, in Section 4, we will see under which cases it is asymptotically monotonic.

Example 3.8. Let us consider two independent random variables $X\sim P(II)(\theta, \alpha_{X},\vartheta)$ and $Y\sim P(II)(\theta, \alpha _{Y},\vartheta)$ such that $\alpha _{X}=4.5$ , $\alpha_Y=4$ , $\theta=1$ , and $\vartheta=0$ . Figure 2 shows that $m_X(\!\cdot\!)$ increases over (0, 120).

Figure 2. Conditional expectation $m_X(\!\cdot\!)$ when $X\sim P(II)(\theta, \alpha_{X},\vartheta)$ and $Y\sim P(II)(\theta, \alpha _{Y},\vartheta)$ with $\alpha _{X}=4.5$ , $\alpha_Y=4$ , $\theta=1$ , and $\vartheta=0$ .

Example 3.9. Let us consider two independent random variables $X\sim P(I)(\alpha _{X},\theta)$ and $Y\sim LG(\alpha _{Y},\lambda )$ with $\alpha_{Y}+1\gt\alpha _{X}\gt\alpha _{Y}$ . Figure 3 shows that the conditional expectation given the sum decreases over a range of central values.

Figure 3. Conditional expectation $m_X(\!\cdot\!)$ when $X\sim P(I)(\alpha_X, \theta )$ and $Y\sim LG(\alpha_Y, \lambda )$ with $\alpha_X=8$ , $\alpha_Y=7.8$ , $\lambda =2.5$ , and $\theta =1$ .

3.3. Boundary case: Difference in tail indices equal to 1 ( $\alpha_X-\alpha_Y=1$ )

The next result shows that when the difference between tail indices is exactly 1, the asymptotic behavior of the conditional expectation given the sum depends on the asymptotic behavior of the ratio between the slowly varying functions associated with the densities.

• • If $\frac{L_X(s)}{L_Y(s)}\rightarrow\infty $ as s tends to infinity, then $m_{X}(s) \rightarrow\infty$ .

• • If $\lim_{s\rightarrow \infty }\frac{L_X(s)}{L_Y(s)}=k$ with $k \geq 0$ , then $\lim_{s\rightarrow \infty }m_{X}(s)=\mathrm{E}[X]+k$ .

Proof. Since $f_X$ is regularly varying with index $\alpha_X\gt3$ , then $f_{\widetilde{X}}(\!\cdot\!)$ is regularly varying with index $\alpha_{\widetilde{X}}=\alpha_X-1=\alpha_Y$ . As a consequence of (2.1), taking into account that $f_X(\!\cdot\!)$ and $f_Y(\!\cdot\!)$ are regularly varying densities, $f_X(s)=o\left(f_{Y}(s)\right)$ and $f_{\widetilde{X}}(s)=\frac{s}{\mathrm{E}[X]}f_X(s)$ then

(3.6) \begin{equation} m_X(s)=\mathrm{E}[X] \frac{f_{\widetilde{X}+Y}(s)}{f_{X+Y}(s)}\sim \mathrm{E}[X]\frac{f_{\widetilde{X}}(s)+f_Y(s)}{f_Y(s)+o(f_Y(s))} =\frac{L_X(s)+\mathrm{E}[X] L_Y(s) }{L_Y(s)+o(L_Y(s))}\sim \frac{L_X(s) }{L_Y(s)}+\mathrm{E}[X]. \end{equation}

Hence, if $\frac{L_X(s)}{L_Y(s)}\rightarrow\infty$ as s tends to infinity, it is direct that $m_{X}(s)$ diverges. On the other hand, if $\lim_{s\rightarrow \infty }\frac{L_X(s)}{L_Y(s)}=k$ with $k \geq 0$ , from (3.6) then $\lim_{s\rightarrow \infty }m_{X}(s)=\mathrm{E}[X]+k$ .

The following example illustrates the different cases of Proposition 3.10.

Example 3.11 (Log-Gamma). Consider two independent random variables $X\sim LG(\alpha_X, \lambda_X)$ and $Y\sim LG(\alpha_Y, \lambda_Y)$ with $\alpha _{X}=\alpha_{Y}+1$ . Here,

\begin{equation*} f_X(x)=\frac{(\alpha_X-1)^{\lambda_X}}{\Gamma(\lambda_X)} (\log x)^{\lambda_X-1}x^{-\alpha_X},\hspace{2mm}x\geq 1. \end{equation*}

Then $L_X(x)=\frac{(\alpha_X-1)^{\lambda_X}}{\Gamma(\lambda_X)} (\log x)^{\lambda_X-1}$ and,

\begin{align*} \frac{L_X(x)}{L_Y(x)}=\frac{(\alpha_X-1)^{\lambda_X}}{(\alpha_X-2)^{\lambda_Y}} \frac{\Gamma(\lambda_Y)}{\Gamma(\lambda_X)} (\log x)^{\lambda_X-\lambda_Y}. \end{align*}

Hence, Figure 4 illustrates that in accordance with Proposition 3.10:

Figure 4. Conditional expectation $m_X(\!\cdot\!)$ when $X\sim LG(\alpha_X, \lambda_X)$ and $Y\sim LG(\alpha_Y, \lambda_Y)$ with $\alpha _{X}=\alpha_{Y}+1$ with $\alpha_X=5$ , $\alpha_Y=4$ , $\lambda_X=3$ and considering different values of $\lambda_Y$ in each case.

• • If $\lambda_X\lt\lambda_Y$ , the latter ratio converges to zero and $\lim _{s \rightarrow \infty} m_X(s)=\mathrm{E}[X]=\left(\frac{\alpha_X-1}{\alpha_X-2}\right)^{\lambda_X};$

• • If $\lambda_X=\lambda_Y$ , then $\frac{L_X(x)}{L_Y(x)}=\mathrm{E}[X]$ and, therefore, $\lim _{s \rightarrow \infty} m_X(s)=2 \mathrm{E}[X];$

• • If $\lambda_X\gt\lambda_Y$ , the ratio $\frac{L_X(x)}{L_Y(x)}$ diverges and $ m_X(s)\rightarrow\infty$ as $s\rightarrow\infty$ .

Note that in the case where the difference between the tail indices is exactly one and the constant considered in Proposition 3.10 is $k=0$ , similarly to Proposition 3.3, we can ensure that there exists a nonempty interval in $(0,\infty)$ where $m_X(\!\cdot\!)$ is decreasing. However, contrary to this case or to the case where the difference in tail indices exceeds 1, there is no guarantee that there exists a nonempty interval of $(0,\infty)$ where $m_X(\!\cdot\!)$ is decreasing when tail indices differ by one and $k\gt0$ . Nevertheless, when the ratio among the slowly varying functions associated with the densities is finite, since the limit is finite, $m_X(\!\cdot\!)$ is necessarily bounded. Under this framework, we can now encounter a variety of different behaviors for $m_X(\!\cdot\!)$ , as illustrated in Example 3.12. In the first situation considered there, the conditional expectation given the sum is monotonically increasing over an interval (0,b) for a large b. However, in the second situation, the conditional expectation given the sum peaks before decreasing to tend to its limit from above.

1. 1. Figure 5 shows that there are situations where $m_X(\!\cdot\!)$ increases over (0,50) and, numerically, we observe that it tends to its limit from below.

   

   Figure 5. Conditional expectation $m_X(\!\cdot\!)$ (blue solid line) and horizontal line at $2\mathrm{E}[X]$ (orange dashed line) when $X\sim P(I)(\theta, \alpha_X )$ and $Y\sim P(I)(\theta, \alpha_Y )$ with $\theta =1$ , $\alpha_X =6$ and $\alpha_Y =5$ .

2. 2. Figure 6 shows that there are situations where $m_X(\!\cdot\!)$ attains a maximum before decreasing to its limit. Specifically, we can see there that, considering the values $\theta =1$ , $\alpha_X =3.5$ and $\alpha_Y =2.5$ , $m_X(\!\cdot\!)$ increases over $(0,64.46)$ where $64.46\simeq F^{-1}(0.998)$ and then decreases over $(64.46,1000)$ . From the plot, we conjecture that it tends to its limit from above.

   

   Figure 6. Conditional expectation $m_X(\!\cdot\!)$ (blue solid line) and horizontal line at $2\mathrm{E}[X]$ (orange dashed line) when $X\sim P(I)(\theta, \alpha_X )$ and $Y\sim P(I)(\theta, \alpha_Y )$ with $\theta =1$ , $\alpha_X =3.5$ , and $\alpha_Y =2.5$ .

From this section, we can conclude that if the difference between the indices exceeds one, $m_X(\!\cdot\!)$ will decrease in an interval; if the difference is less than one, $m_X(\!\cdot\!)$ diverges; and in the boundary case (i.e. if the difference is exactly one) it can be either bounded or not, depending on the asymptotic behavior of the ratio between the associated slowly varying functions.

In the following section, in addition to the asymptotic levels, we aim to delve into the rate at which $m_X(\!\cdot\!)$ is either converging or diverging as the argument gets larger.

4.1. Asymptotic expansion of the conditional mean given the sum

Define the truncated mean function for a nonnegative random variable Z as

(4.1) \begin{equation}\mu_{Z}\left( t\right) =\int_{0}^{t}xf_{Z}\left( x\right) \mathrm{d} x.\end{equation}

If the expectation is finite (i.e. $\alpha _{Z}\gt2$ ), the truncated mean tends to the expected value, which is denoted by $\mu_Z$ :

\begin{equation*}\lim_{t\rightarrow \infty }\mu_{Z}\left( t\right) =\mu_{Z}=\mathrm{E}[Z]\lt\infty.\end{equation*}

The variance of Z will be denoted by $\sigma^2_Z$ . The following result provides an asymptotic expansion of the density function of the sum $X+Y$ when X and Y both possess asymptotically smooth densities. This is an extension of Theorem 1.1 in Bingham et al. (Reference Bingham, Goldie and Omey2006).

Proposition 4.1. Let X and Y be positive random variables with asymptotically smooth density functions $f_{X}(\!\cdot\!)$ and $f_{Y}(\!\cdot\!)$ with tail indices $\alpha _{X}$ and $\alpha _{Y}$ such that $\min\{\alpha _{X},\alpha _{Y}\}\geq2$ . Then, as s tends to infinity,

\begin{equation*}f_{X+Y}\left( s\right) =f_{X}\left( s\right) \left( 1+\alpha _{X}\frac{\mu_{Y}\left( s\right) }{s}\left( 1+o\left( 1\right) \right) \right)+f_{Y}\left( s\right) \left( 1+\alpha _{Y}\frac{\mu _{X}\left( s\right) }{s}\left( 1+o\left( 1\right) \right) \right) .\end{equation*}

In particular, if $\min\{\alpha _{X},\alpha _{Y}\}\gt2$ ,

\begin{equation*} f_{X+Y}\left( s\right) =f_{X}\left( s\right) \left( 1+\alpha _{X}\frac{\mu _{Y} }{s}\left( 1+o\left( 1\right) \right) \right) +f_{Y}\left( s\right) \left( 1+\alpha _{Y}\frac{\mu _{X} }{s} \left( 1+o\left( 1\right) \right) \right) .\end{equation*}

Proof. The density function of $X+Y$ can be written as

\begin{equation*}f_{X+Y}\left( t\right) =\int_{0}^{t}f_{X}\left( t-x\right) f_{Y}\left(x\right) \mathrm{d} x=T_{Y}f_{X}\left( t\right) +T_{X}f_{Y}\left( t\right)\end{equation*}

where

\begin{equation*}T_{Y}f_{X}\left( t\right) =\int_{0}^{t/2}f_{X}\left( t-x\right) f_{Y}\left(x\right) \mathrm{d} x.\end{equation*}

The announced result then follows from the expressions for $T_{Y}f_{X}\left( t\right)$ and $T_{X}f_{Y}\left( t\right)$ derived in Proposition B.1 in Appendix B.

An asymptotic expansion for the conditional mean given the sum is then deduced from this result.

Proposition 4.2. Let X and Y be positive random variables with asymptotically smooth density functions $f_{X}(\!\cdot\!)$ and $f_{Y}(\!\cdot\!)$ with tail indices $\alpha _{X}$ and $\alpha _{Y}$ such that $\min\{\alpha _{X}-1,\alpha _{Y}\}\gt2$ . Then, as s tends to infinity,

(4.2) \begin{equation}m_{X}(s)=\frac{f_{X}\left( s\right) \left( s+(\alpha _{X}-1)\mu _{Y}\big(1+o\left( 1\right) \big) \right)+\mu _{X}f_{Y}\left( s\right) \left(1+\frac{\alpha _{Y}\mu _{\widetilde{X}}}{s}\big( 1+o\left( 1\right) \big) \right)}{f_{X}\left( s\right) \left( 1+\frac{\alpha _{X}\mu _{Y}}{s}\big( 1+o\left(1\right) \big) \right)+f_{Y}\left( s\right) \left( 1+\frac{\alpha_{Y}\mu _{X}}{s}\big( 1+o\left( 1\right) \big) \right) }.\end{equation}

Proof. Representation (2.2) shows that $m_{X}(\!\cdot\!)$ can be expressed with the help of the ratio of the densities of $\widetilde{X}+Y$ of $X+Y$ . Applying Proposition 4.1 to these densities, and taking into account that $\alpha_{\widetilde{X}}=\alpha_X-1\gt2$ , we get

\begin{eqnarray*} f_{\widetilde{X}+Y}\left( s\right) &=&f_{\widetilde{X}}\left( s\right)+f_{Y}\left( s\right)+\left( \alpha_{\widetilde{X}} \frac{f_{\widetilde{X}}\left( s\right)}{s}\mu _{Y}+\alpha_Y \frac{f_{Y}\left( s\right) }{s}\mu _{\widetilde{X}}\right) \left( 1+o\left( 1\right) \right)\\&=&\frac{sf_{X}\left( s\right)}{\mu_X}+f_{Y}\left( s\right)+\left( (\alpha_X-1) \frac{f_{X}\left( s\right)}{\mu_X}\mu _{Y}+\alpha_Y \frac{f_{Y}\left( s\right) }{s}\mu _{\widetilde{X}}\right) \left( 1+o\left( 1\right) \right)\end{eqnarray*}

and

\begin{equation*} f_{X+Y}\left( s\right) =f_{X}\left( s\right)+f_{Y}\left( s\right)+\left( \alpha_X \frac{f_{X}\left( s\right)}{s}\mu _{Y}+\alpha_Y \frac{f_{Y}\left( s\right) }{s}\mu _{X}\right) \left( 1+o\left( 1\right) \right)\end{equation*}

as s tends to infinity. The announced result then follows from (2.2).

Proposition 4.2 allows us to study the asymptotic behavior of the conditional expectation given the sum according to the values of the tail indices. In the following sections, we will study it under different frameworks and we assume again that $\alpha_{X}\geq \alpha_Y$ . Note that, if $\alpha_Y\gt\alpha_X$ , the asymptotic expression of $m_X(\!\cdot\!)$ can be obtained by computing the asymptotic expression of $m_Y(\!\cdot\!)$ and noting that $m_{X}(s)=s-m_Y(s)$ . Although it must be noted that, in order to obtain the asymptotic expression of $m_X\left(\cdot\right)$ , it is required that $\alpha_X\gt3$ and $\alpha_Y\gt2$ . Hence, if computing the asymptotic expression of $m_Y\left(\cdot\right)$ , we need to assume $\alpha_Y\gt3$ and $\alpha_X\gt2$ .

Based on Proposition 4.2, we derive different types of asymptotic behaviors of the conditional mean of X given the sum, as detailed in the next sections. We consider two cases where $m_X\left(\cdot\right)$ converges to the unconditional expected value: when the difference of the indices is greater than two and when it is in between one and two. In addition, we compare the asymptotic behavior of $m_X\left(\cdot\right)$ as it diverges, by deriving the asymptotic expansion when indices are “near” (their difference is positive but smaller than one) and when both variables have the same index. Figure 7 illustrates the different cases considered in the next sections within the plane $( \alpha _{X},\alpha_{Y})$ . As is common to the different cases, throughout the next sections, we will assume $\alpha_X\gt3$ and $\alpha_Y\gt2$ . In the following sections, we will denote by $\widehat{m}_X\left(\cdot\right)$ an approximation of $m_X\left(\cdot\right)$ at infinity.

Figure 7. Discussion according to the position of $( \alpha _{X},\alpha_{Y})$ in $(3,\infty)\times(2,\infty)$ with $\alpha_{X}\geq \alpha_Y$ .

4.2. Difference in the tail indices larger than 2 ( $\alpha_X-\alpha_Y\gt2$ )

From Proposition 4.2, we can write $m_X\left(\cdot\right)$ as the ratio of two asymptotic expansions, each considering four terms. Note that, assuming $\alpha_X-\alpha_Y\gt1$ , the terms of the denominator are regularly varying with arranged indices $\lbrace \alpha_Y, \alpha_Y+1,\alpha_X,\alpha_X+1 \rbrace$ . Therefore, $\alpha_Y$ and $\alpha_Y+1$ are the two smallest indices (i.e., the two greatest orders). With respect to the numerator, the (non-arranged) indices of the terms are $\lbrace \alpha_X-1,\alpha_X, \alpha_Y, \alpha_Y+1 \rbrace$ . Under the assumption $\alpha_X-\alpha_Y\gt1$ , the smallest index is $\alpha_Y$ . However, if $\alpha_X-\alpha_Y\gt2$ , the second smallest index is $\alpha_Y+1$ and, if $\alpha_X-\alpha_Y\lt2$ , it is $\alpha_X-1$ . This motivates the study of the behavior of $m_X\left(\cdot\right)$ separately if the difference between indices trespasses two.

Proposition 4.3. Let X and Y be positive random variables possessing asymptotically smooth density functions $f_X(\!\cdot\!)$ and $f_Y(\!\cdot\!)$ with respective tail indices $\alpha _{X}$ and $\alpha _{Y}$ . If $\alpha _{X}\gt\alpha _{Y}+2$ , then

\begin{equation*}m_{X}(s)=\mu _{X}+\alpha _{Y}\frac{1}{s}\sigma^2_X\left( 1+o\left( 1\right) \right)\end{equation*}

as s tends to infinity.

Proof. Notice that $\sigma^2_X=\mu_{\widetilde{X}}\mu_{X}-\mu_{X}^2.$ Since $\min\{\alpha _{X}-1,\alpha _{Y}\}\gt2$ , (4.2) allows us to write

\begin{equation*}m_{X}(s)=\frac{sf_{X}\left( s\right) +\mu _{X}f_{Y}\left( s\right)+\left( (\alpha_X -1)f_{X}\left( s\right)\mu _{Y}+\alpha_Y \frac{f_{Y}\left( s\right) }{s}\left(\sigma^2_X+\mu _{X}^2\right)\right) \left( 1+o\left( 1\right) \right)}{ f_{X}\left( s\right)+f_{Y}\left( s\right)+\left( \alpha_X \frac{f_{X}\left( s\right)}{s}\mu _{Y}+\alpha_Y \frac{f_{Y}\left( s\right) }{s}\mu _{X}\right) \left( 1+o\left( 1\right) \right)}\end{equation*}

as s tends to infinity. As $\alpha _{Y}\gt2$ and $\alpha _{X}\gt\alpha _{Y}+2$ , we have that $f_{X}\left( s\right) =o(\frac{f_{Y}\left( s\right) }{s^2})$ , and we can write

\begin{eqnarray*}m_{X}(s) &=&\mu _{X}\frac{f_{Y}\left( s\right)+\alpha _{Y}\frac{f_{Y}\left( s\right) }{s}(\mu _{X}+\frac{\sigma^2_X}{ \mu _{X}})\left(1+o\left( 1\right) \right) }{f_{Y}\left( s\right) +\alpha _{Y}\frac{f_{Y}\left( s\right) }{s}\mu _{X}\left( 1+o\left( 1\right) \right) } \\&=&\mu _{X}\frac{1+\alpha _{Y}\frac{1}{s}(\mu _{X}+\frac{\sigma^2_X}{\mu _{X}})\left( 1+o\left( 1\right) \right) }{1+\alpha _{Y}\frac{1}{s}\mu_{X}\left( 1+o\left( 1\right) \right) } \\&& \\&=&\mu _{X}\left( 1+\alpha _{Y}\frac{1}{s}\frac{\sigma^2_X}{\mu _{X}}\left( 1+o\left( 1\right) \right) \right) .\end{eqnarray*}

This ends the proof.

The following example illustrates Proposition 4.3.

Example 4.4. Let us consider two independent random variables $X\sim P(II)(\theta, \alpha_{X},\vartheta)$ and $Y\sim P(II)(\theta, \alpha _{Y},\vartheta)$ such that $\alpha _{X}=10$ , $\alpha_Y=5$ , $\theta=1$ , and $\vartheta=0$ . Figure 8 shows the same asymptotic behavior of $m_X(\!\cdot\!)$ and $\widehat{m}_X(s)=\mu _{X}\left( 1+\alpha _{Y}\frac{1}{s}\frac{\sigma^2_X}{\mu _{X}}\right)$ .

Figure 8. Conditional expectation $m_X(\!\cdot\!)$ when $X\sim P(II)(\theta, \alpha_{X},\vartheta)$ and $Y\sim P(II)(\theta, \alpha _{Y},\vartheta)$ with $\alpha _{X}=10$ , $\alpha_Y=5$ , $\theta=1$ , and $\vartheta=0$ .

4.3. Difference in the tail indices between 1 and 2 ( $1 \lt \alpha_X-\alpha_Y \lt 2$ )

Proposition 4.5. Let X and Y be positive random variables possessing asymptotically smooth density functions $f_X(\!\cdot\!)$ and $f_Y(\!\cdot\!)$ with respective tail indices $\alpha _{X}$ and $\alpha _{Y}$ . If $\alpha_Y+2\gt\alpha _{X}\gt\alpha _{Y}+1$ , then

\begin{equation*}m_{X}(s)=\mu _{X}+s\frac{f_{X}\left( s\right) }{f_{Y}\left(s\right) }\left( 1+o\left( 1\right) \right)\end{equation*}

as s tends to infinity.

Proof. Since $\alpha _{Y}\gt2$ and $\alpha _{Y}+2\gt \alpha _{X}\gt\alpha_{Y}+1$ , we have that $f_{X}\left( s\right) =o\left(\frac{f_{Y}\left( s\right)}{s}\right)$ and $ \frac{f_{Y}\left( s\right)}{s} =o(sf_{X}\left( s\right) )$ . Hence, we can write

\begin{eqnarray*}m_{X}(s) &=&\frac{\mu _{X}f_{Y}\left( s\right) +sf_{X}\left( s\right) \left( 1+o\left( 1\right) \right) }{f_{Y}\left(s\right) +\alpha _{Y}\frac{f_{Y}\left( s\right) }{s}\mu _{X}\left( 1+o\left(1\right) \right) } \\&=&\mu _{X}+s\frac{f_{X}\left( s\right) }{f_{Y}\left(s\right) }\left( 1+o\left( 1\right) \right).\end{eqnarray*}

This ends the proof.

The following example illustrates Proposition 4.5

Example 4.6. Let us consider two independent random variables $X\sim Davis(\alpha_{X},b,\vartheta)$ and $Y\sim Davis(\alpha _{Y},b,\vartheta)$ such that $\alpha _{X}=6$ , $\alpha_Y=4.5$ , $b=2$ , and $\vartheta=0$ . Figure 9 shows the same asymptotic behavior of $m_X(\!\cdot\!)$ and $\widehat{m}_X(s)=\mu _{X}+s\frac{f_{X}\left( s\right) }{f_{Y}\left(s\right) }$ .

Figure 9. Conditional expectation $m_X(\!\cdot\!)$ when $X\sim Davis(\alpha_{X},b,\vartheta)$ and $Y\sim Davis(\alpha _{Y},b,\vartheta)$ with $\alpha _{X}=6$ , $\alpha_Y=4.5$ , $b=2$ , and $\vartheta=0$ .

From Proposition 4.3, if $\alpha_X-\alpha_Y\gt2$ , the asymptotic behavior of $m_X\left(\cdot\right)$ can be approximated by $\widehat{m}_{X}(s)=\mu _{X}+\alpha_{Y}\frac{1}{s}\sigma^2_X$ , where the second term behaves proportionally to $1/s$ . However, if $1\lt\alpha_X-\alpha_Y\lt2$ , by Proposition 4.5, asymptotically, $m_X\left(\cdot\right)$ can be approximated by $\widehat{m}_{X}(s)=\mu _{X}+s\frac{f_{X}\left( s\right) }{f_{Y}\left(s\right)}$ , whose second term is $s^{-\alpha_X+1+\alpha_Y}\frac{L_{X}\left( s\right) }{L_{Y}\left(s\right)}$ . Note that $-\alpha_X+1+\alpha_Y\gt-1$ . Hence, as we could expect, if $\alpha_X-\alpha_Y\gt2$ , asymptotically, $m_X\left(\cdot\right)$ decreases faster as it does if $1\lt\alpha_X-\alpha_Y\lt2$ .

4.4. Difference in the tail indices less than 1 ( $0 \lt \alpha_X-\alpha_Y \lt 1$ )

When the indices are similar, and by similar, we mean here that they do not differ by more than one unit, the following result provides the asymptotic expression of $m_X\left(\cdot\right)$ , where the second and third terms vary depending on how similar are the indices. Note that the expressions are arranged in terms of their order, so even if the terms considered are the same in the two first cases, they are given separately to emphasize that the second-order term differs in the two cases.

1. (a) if $1\gt\alpha_X-\alpha_Y\gt\frac{1}{2}$ , then:

   \begin{equation*} m_{X}(s)=s\frac{f_X(s) }{f_Y(s)}\left(1+\mu_{X}\frac{f_Y(s) }{s f_X(s)}-\frac{f_X(s) }{f_Y(s)}(1+o(1))\right);\end{equation*}

2. (b) if $\frac{1}{2}\gt\alpha_X-\alpha_Y\gt\frac{1}{3}$ , then:

   \begin{equation*} m_{X}(s)=s\frac{f_X(s) }{f_Y(s)}\left(1-\frac{f_X(s) }{f_Y(s)}+\mu_{X}\frac{f_Y(s) }{s f_X(s)}(1+o(1))\right);\end{equation*}

3. (c) if $\frac{1}{3}\gt\alpha_X-\alpha_Y\gt0$ , then:

   \begin{equation*} m_{X}(s)=s\frac{f_X(s) }{f_Y(s)}\left(1-\frac{f_X(s) }{f_Y(s)}+\left(\frac{f_X(s) }{f_Y(s)}\right)^2(1+o(1))\right)\end{equation*}

Proof. If $\alpha _{Y}+1\gt\alpha _{X}\gt\alpha _{Y}$ , since $\alpha_X\gt3$ and $\alpha_Y\gt2$ , then, as s tends to infinity:

\begin{eqnarray*} m_{X}(s)&=&\frac{f_{X}\left( s\right) \big( s+(\alpha _{X}-1)\mu _{Y}\left( 1+o\left( 1\right) \right) \big) +\mu _{X}f_{Y}\left( s\right) \big( 1+\alpha _{Y}\mu _{\widetilde{X}}s^{-1}\left( 1+o\left( 1\right) \right) \big) }{ f_{X}\left( s\right) \big( 1+\alpha _{X}\mu _{Y}s^{-1}\left( 1+o\left( 1\right) \right) \big) +f_{Y}\left( s\right) \big( 1+\alpha _{Y}\mu _{X}s^{-1}\left( 1+o\left( 1\right) \right)\big) }\\ &=&s\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\frac{ 1+\mu _{X}\frac{f_{Y}\left( s\right)}{s f_{X}\left( s\right)}+\frac{1}{s}(\alpha _{X}-1)\mu _{Y}\left( 1+o\left( 1\right) \right) +\alpha _{Y}\mu _{\widetilde{X}}\mu _{X}\frac{f_{Y}\left( s\right)}{s^2 f_{X}\left( s\right)} \left( 1+o\left( 1\right) \right) }{ 1+\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}+\alpha _{Y}\mu _{X}\frac{1}{s}\left( 1+o\left( 1\right) \right) +\alpha _{X}\mu _{Y} \frac{f_{X}\left( s\right)}{s f_{Y}\left( s\right)}\left( 1+o\left( 1\right) \right) }\\ &=&s\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\frac{ 1+\mu _{X}\frac{f_{Y}\left( s\right)}{s f_{X}\left( s\right)}+\frac{1}{s}(\alpha _{X}-1)\mu _{Y}\left( 1+o\left( 1\right) \right) }{ 1+\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}+\alpha _{Y}\mu _{X}\frac{1}{s}\left( 1+o\left( 1\right) \right)}.\end{eqnarray*}

where the last step holds since we can write $\alpha _{Y}\mu _{\widetilde{X}}\mu _{X}\frac{f_{Y}\left( s\right)}{s^2 f_{X}\left( s\right)}=\frac{1}{s} o(1)$ and $\alpha _{X}\mu _{Y}\frac{f_{X}\left( s\right)}{s f_{Y}\left( s\right)}=\frac{1}{s} o(1)$ . The term $\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}+\alpha_{Y}\mu _{X}\frac{1}{s}$ tends to zero as s tends to infinity and therefore we can write:

\begin{equation*}\frac{1}{ 1+\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}+\alpha_{Y}\mu _{X}\frac{1}{s}\left( 1+o\left( 1\right) \right)}=1-\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}-\alpha_{Y}\mu _{X}\frac{1}{s}\left( 1+o\left( 1\right) \right) +\left(\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\right)^2 \left( 1+o\left( 1\right) \right) .\end{equation*}

Therefore, as s tends to infinity, $ m_{X}(s)=s\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}g(s)$ with

\begin{equation*} g(s)=1-\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}+\mu _{X}\frac{f_{Y}\left( s\right)}{s f_{X}\left( s\right)}+\left(\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\right)^2 \left( 1+o\left( 1\right) \right)+\frac{(\alpha _{X}-1)\mu _{Y}-(\alpha _{Y}-1)\mu _{X}}{s}\left( 1+o\left( 1\right) \right) .\end{equation*}

We can now consider the different cases.

1. (a) If $\alpha_X-\alpha_Y\gt \frac{1}{2}$ , then $\frac{f_X(s) }{f_Y(s)}=o(\mu_{X}\frac{f_Y(s) }{s f_X(s)})$ and, by noticing that $\left(\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\right)^2=o(\frac{f_X(s) }{f_Y(s)})$ and $\frac{1}{s}=o(\frac{f_X(s) }{f_Y(s)})$ , as s tends to infinity:

   \begin{equation*} m_{X}(s)=s\frac{f_X(s) }{f_Y(s)}\left(1+\mu_{X}\frac{f_Y(s) }{s f_X(s)}-\frac{f_X(s) }{f_Y(s)}(1+o(1))\right).\end{equation*}

2. (b) If $\frac{1}{2}\gt\alpha_X-\alpha_Y\gt \frac{1}{3}$ , then $\mu_{X}\frac{f_Y(s) }{s f_X(s)}=o(\frac{f_X(s) }{f_Y(s)})$ and $\left(\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\right)^2=o(\frac{f_Y(s) }{s f_X(s)})$ . Therefore, because $\frac{1}{s}=o(\frac{f_Y(s) }{s f_X(s)})$ , as s tends to infinity:

   \begin{equation*} m_{X}(s)=s\frac{f_X(s) }{f_Y(s)}\left(1-\frac{f_X(s) }{f_Y(s)}+\mu_{X}\frac{f_Y(s) }{s f_X(s)}(1+o(1))\right).\end{equation*}

3. (c) If $\frac{1}{3}\gt\alpha_X-\alpha_Y\gt 0$ , then $\mu_{X}\frac{f_Y(s) }{s f_X(s)}=o(\frac{f_X(s) }{f_Y(s)})$ and $\frac{f_Y(s) }{s f_X(s)}=o(\left(\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\right)^2)$ . Therefore, because $\frac{1}{s}=o(\left(\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\right)^2)$ , as s tends to infinity:

   \begin{equation*} m_{X}(s)=s\frac{f_X(s) }{f_Y(s)}\left(1-\frac{f_X(s) }{f_Y(s)}+\left(\frac{f_{X}\left( s\right)}{f_{Y}\left( s\right)}\right)^2(1+o(1))\right).\end{equation*}

This ends the proof.

The following example illustrates Proposition 4.7.

Example 4.8. Let us consider two independent random variables $X\sim P(II)(\theta, \alpha_{X},\vartheta)$ and $Y\sim P(II)(\theta, \alpha _{Y},\vartheta)$ . Figure 10 shows the same asymptotic behavior of $m_X(\!\cdot\!)$ and:

Figure 10. Conditional expectation $m_X(\!\cdot\!)$ when $X\sim P(II)(\theta, \alpha_{X},\vartheta)$ and $Y\sim P(II)(\theta, \alpha _{Y},\vartheta)$ with $\alpha _{X}=5$ , $\theta=1$ , and $\vartheta=0$ and considering different values of $\alpha_Y$ in each case.

• • $\widehat{m}_X(s)=s\frac{f_X(s) }{f_Y(s)}\left(1+\mu_{X}\frac{f_Y(s) }{s f_X(s)}-\frac{f_X(s) }{f_Y(s)}\right)$ in cases (a) and (b).

• • $\widehat{m}_X(s)=s\frac{f_X(s) }{f_Y(s)}\left(1-\frac{f_X(s) }{f_Y(s)}+\left(\frac{f_X(s) }{f_Y(s)}\right)^2\right)$ in case (c).

4.5. Equal indices ( $\alpha_X=\alpha_Y$ )

The following results describe the asymptotic behavior of $m_X(\!\cdot\!)$ when both variables have the same index, and under the assumption of asymptotic proportionality between the densities. Let us note that if $X\overset{\mathrm{d}}{=} Y$ , then $m_X(s)=m_Y(s)$ for all s in the support of S. Hence, as $m_X(s)+m_Y(s)=s$ , it is direct to see that, for equally distributed variables, $m_X(s)=\frac{s}{2}$ . The following result provides the asymptotic expression of $m_X(\!\cdot\!)$ when both densities have the same index and are asymptotically proportional but not necessarily equal.

Proposition 4.9. Let X and Y be positive random variables possessing asymptotically smooth density functions $f_X(\!\cdot\!)$ and $f_Y(\!\cdot\!)$ with respective tail indices $\alpha _{X}$ and $\alpha _{Y}$ such that $\alpha _{Y}=\alpha _{X}$ and $f_{X}(s)=c f_{Y}(s)\left( 1+o\left( 1\right) \right)$ with $c\gt0$ . Then

(4.3) \begin{equation} m_{X}(s)=\frac{c}{1+c}s \left(1+ \frac{1}{s}\left(\frac{\alpha}{1+c}\left(\mu_Y-\mu_X\right) +\frac{\mu_X}{c}-\mu_Y\right)\left( 1+o\left( 1\right) \right) \right) \end{equation}

as s tends to infinity.

Proof. We consider $\alpha=\alpha _{Y}=\alpha _{X}$ , and we can write:

\begin{eqnarray*} m_{X}(s)&=&\frac{sf_{X}\left( s\right) +\mu _{X}f_{Y}\left( s\right)+\left( (\alpha_X -1)f_{X}\left( s\right)\mu_{Y}+\alpha_Y \frac{f_{Y}\left( s\right) }{s}\left(\sigma^2_X+\mu _{X}^2\right)\right) \left( 1+o\left( 1\right) \right)}{ f_{X}\left( s\right)+f_{Y}\left( s\right)+\left( \alpha_X \frac{f_{X}\left( s\right)}{s}\mu _{Y}+\alpha_Y \frac{f_{Y}\left( s\right) }{s}\mu _{X}\right) \left( 1+o\left( 1\right) \right)}\\ &=&\frac{c}{1+c}s \frac{f_{Y}\left( s\right) +\frac{f_{Y}\left( s\right) }{s}\left((\alpha-1)\mu _{Y} +\frac{\mu_X}{c}\right)\left( 1+o\left( 1\right) \right) }{f_{Y}\left( s\right) +\frac{f_{Y}\left( s\right) }{s}\left( \alpha\mu _{Y}\frac{c}{1+c}+\mu_X \alpha\frac{1}{1+c}\right)\left( 1+o\left( 1\right) \right) } \\ &=&\frac{c}{1+c}s \left(1+ \frac{1}{s}\frac{\left(\mu_{Y}\left(\frac{\alpha}{1+c}-1\right) +\mu_X\left(\frac{1}{c}-\frac{\alpha}{1+c}\right)\right)\left( 1+o\left( 1\right) \right) }{1 +\frac{1}{s}\left( \alpha \mu _{Y}\frac{c}{1+c}+\mu_X \alpha\frac{1}{1+c}\right)\left( 1+o\left( 1\right) \right) } \right) \\ &=&\frac{c}{1+c}s \left(1+ \frac{1}{s}\left(\frac{\alpha}{1+c}\left(\mu_Y-\mu_X\right) +\frac{\mu_X}{c}-\mu_Y\right)\left( 1+o\left( 1\right) \right) \right) .\end{eqnarray*}

This ends the proof.

If $X\overset{\mathrm{d}}{=} Y$ , since $c=1$ and $\mu_X=\mu_Y$ , the term $\frac{1}{s}\left(\frac{\alpha}{1+c}\left(\mu_Y-\mu_X\right) +\frac{\mu_X}{c}-\mu_Y\right)$ in (4.3) is zero and the expression provided in Proposition 4.9 corresponds to $m_X(s)=\frac{s}{2}$ as expected.

Example 4.10. Consider two independent random variables $X\sim P(IV)(\theta_X, \alpha,\vartheta,\lambda_X)$ and $Y\sim P(IV)(\theta_Y,\alpha,\vartheta,\lambda_Y)$ with $\alpha\gt3$ . Without loss of generality, we can assume $\vartheta=0$ . Here, the slowly varying function associated to the density $f_X(\!\cdot\!)$ is given by:

\begin{equation*} L_X(x)=(\alpha-1)\theta_X^{-\frac{\alpha-1}{\lambda_X}} \frac{x^{\alpha}(x)^{-1+\frac{\alpha-1}{\lambda_X}}}{\left(1+\left(\frac{x}{\theta_X}\right)^{\frac{\alpha-1}{\lambda_X}}\right)^{\lambda_X+1}}. \end{equation*}

Note that $\frac{f_X(s)}{f_Y(s)}=\frac{L_X(s)}{L_Y(s)}$ . Therefore, we can write $f_X(s)=c f_{Y}(s)\left( 1+g_{IV}\left(x\right) \right)$ , where

\begin{align*} g_{IV}(x)=x^{-\frac{(\alpha-1)(\lambda_X-\lambda_Y)}{\lambda_X \lambda_Y}} \theta_X^{\frac{1-\alpha}{\lambda_X}}\left(1+\left(\frac{x}{\theta_X}\right)^{\frac{\alpha-1}{\lambda_X}}\right)^{-(1+\lambda_X)}\left(\frac{\theta_X}{\theta_Y}\right)^{1-\alpha} \theta_Y^{\frac{\alpha-1}{\lambda_Y}}\left(1+\left(\frac{x}{\theta_Y}\right)^{\frac{\alpha-1}{\lambda Y}}\right)^{1+\lambda_Y}-1, \end{align*}

verifying $g_{IV}(x)=o(1)$ and $c=\left(\frac{\theta_X}{\theta_Y}\right)^{\alpha-1}$ . In particular, if $\lambda=\lambda_X=\lambda_Y$ , then

\begin{align*} g_{IV}(x)=\left(\frac{\theta_Y^{\frac{\alpha-1}{\lambda}}+x^{\frac{\alpha-1}{\lambda}}}{\theta_X^{\frac{\alpha-1}{\lambda}}+x^{\frac{\alpha-1}{\lambda}}}\right)^{(1+\lambda)}-1. \end{align*}

Let us consider two independent random variables $X\sim P(IV)(\theta_X, \alpha,\vartheta,\lambda)$ and $Y\sim P(IV)(\theta_Y,\alpha,\vartheta,\lambda)$ such that $\alpha=7$ , $\theta_X=1$ , $\theta_Y=2$ , $\lambda=2$ and $\vartheta=0$ . Figure 11 shows the same asymptotic behavior of $m_X(\!\cdot\!)$ and

\begin{align*}\widehat{m}_X(s)=\frac{c}{1+c}s \left(1+ \frac{1}{s}\left(\frac{\alpha}{1+c}\left(\mu_Y-\mu_X\right) +\frac{\mu_X}{c}-\mu_Y\right) \right), \; \; c=\frac{1}{64}.\end{align*}

Figure 11. Conditional expectation $m_X(\!\cdot\!)$ when $X\sim P(IV)(\theta_X, \alpha,\vartheta,\lambda)$ and $Y\sim P(IV)(\theta_Y,\alpha,\vartheta,\lambda)$ with $\alpha=7$ , $\theta_X=1$ , $\theta_Y=2$ , $\lambda=2$ and $\vartheta=0$ .

From Proposition 4.9, when the densities have equal tail indices and they are asymptotically proportional, $m_X\left(\cdot\right)$ can be approximated by a function where the term which dominates is $\frac{c}{1+c}s$ . However, from Proposition 4.7, if $1\gt\alpha_X-\alpha_Y\gt0$ , the asymptotic behavior of $m_X\left(\cdot\right)$ can be approximated by a function where the dominant term is $s^{\alpha_Y+1-\alpha_X}\frac{L_{X}\left( s\right) }{L_{Y}\left(s\right)}$ , with $1\gt\alpha_Y+1-\alpha_{X}\gt0$ . Hence, if $0 \lt \alpha_X-\alpha_Y \lt 1$ , asymptotically, $m_X\left(\cdot\right)$ increases faster than it does if the indices are equal. And, the nearer the indices, the faster that $m_X\left(\cdot\right)$ asymptotically increases.