Stochastic orders to approach investments in condor financial derivatives

Abstract

The comparison of investments is a key research topic in mathematical finance. Financial derivatives are popular tools for economic investments. A common financial derivative is the so-called condor derivative. A new mathematical framework for the comparison of investments in condor derivatives is introduced in this manuscript. That model is based on the theory of stochastic orders. Namely, a new family of stochastic orders to approach such comparison problems is introduced. That family is analyzed in detail providing characterizations of the new orders, properties and connections with other stochastic orderings. Results which permit to compare condor derivatives, when the prices of the underlying assets follow Brownian movements, or geometric Brownian movements, are developed. Moreover, an analysis with the DOWJONES and EUROSTOXX indexes shows how to use the new stochastic orders to compare investments in condor derivatives based on those indexes. On the other hand, it is shown how well-known stochastic orders can be applied to compare investments in other financial derivatives, like future derivatives, bull call spreads, call options or long straddle derivatives.

Appendix

Proofs of mathematical results are included in this appendix.

Proof of Lemma 1

Observe that $f_{p_2,p_2,p_3,p_4,k}= f_{p_2,p_2,p_3,p_4,0}+k$ for any $p_1,p_2,p_3,p_4,k \in \mathbb {R}$ with $p_1<p_2<p_3 <p_4$, which leads to the result. Note that $\, \mathcal{F}_0^{p_2,p_3} \,\subset \, \mathcal{F}^{p_2,p_3} \,$. $\square$

Proof of Lemma 2

Let us see (i). Consider $\{f_{p_2-1/n,p_2,p_3,p_4,k}\}_n \subset \, \mathcal{F}^{p_2,p_3} \,$. Condition $X \, \preceq _{con}^{p_2,p_3} \,Y$ implies $E(f_{p_2-1/n,p_2,p_3,p_4,k}(X)) \le E(f_{p_2-1/n,p_2,p_3,p_4,k}(Y))$ for all $n \in \mathbb {N}.$ Note that for any $x \in \mathbb {R}$ we have that $\lim _n f_{p_2-1/n,p_2,p_3,p_4,k}(x)=f_{p_2,p_2,p_3,p_4,k}(x)$, and the above sequence of mappings is uniformly bounded. The dominated convergence theorem leads to (i). In a similar way, it is possible to prove (ii) and (iii). $\square$

Proof of Proposition 1

Let us suppose that $X \, \preceq _{con}^{p_2,p_3} \,Y$. Let $f \in \mathcal{F}_0^{-\mu /2, \mu /2}.$ Note that this class of mapping is a generator of $\preceq _{con}^{-\mu /2, \mu /2}$ as Lemma 1 reads. Let $T:\mathbb {R}\rightarrow \mathbb {R}$ with $T(x)=x-(p_2+p_3)/2.$ Observe that $P_X \circ T^{-1}(B)=P_{X-(p_2+p_3)/2}(B)$ for any $B \in \mathcal{B}.$ By a change of variable (see, for instance, Halmos (1950)), we have that

$$\begin{aligned} \int _{\mathbb {R}} f(x)\, dP_{X-(p_2+p_3)/2}=\int _{\mathbb {R}} f\circ T(x)\, dP_X= \int _{\mathbb {R}} f(x-(p_2+p_3)/2)\, dP_X. \end{aligned}$$

It is not hard to prove that the map $x \rightarrow f(x-(p_2+p_3)/2)$ belongs to the class $\, \mathcal{F}_0^{p_2,p_3} \,$. In fact, if $f=f_{p_1,-\mu /2,\mu /2,p_4,0}$ for some $p_1 < -\mu /2$ and $p_4> \mu /2$, then $f(x-(p_2+p_2)/2)=f_{p_1+(p_2+p_3)/2,p_2,p_3,p_4+(p_2+p_3)/2 }(x).$ Since $X \, \preceq _{con}^{p_2,p_3} \,Y$, we obtain that

$$\begin{aligned} \int _{\mathbb {R}} f(x-(p_2+p_3)/2)\, dP_X \le \int _{\mathbb {R}} f(x-(p_2+p_3)/2)\, dP_Y=\int _{\mathbb {R}} f\, dP_{Y-(p_2+p_3)/2}, \end{aligned}$$

and so $X-(p_2+p_3)/2 \, \preceq _{con}^{-\mu /2, \mu /2} \, Y-(p_2+p_3)/2.$ The converse can be proved in a similar way. $\square$

Proof of Proposition 2

Assume that the condition $X \, \preceq _{con}^{-\delta ,\delta } \,Y$ is satisfied.

Let $t >0.$ Take the mapping $f_{-\delta , -\delta ,\delta , \delta +t,0}$, that is, $f_{-\delta , -\delta ,\delta , \delta +t,0}(x)=(x-\delta )I_{(\delta ,\delta +t]}(x)+tI_{(\delta +t,+\infty )}(x)$ for any $x \in \mathbb {R}$.

By Lemma 2, $E(f_{-\delta , -\delta ,\delta , \delta +t,0}(X))\le E(f_{-\delta , -\delta ,\delta , \delta +t,0}(Y))$ holds. Note that

$$\begin{aligned}&\int _{\mathbb {R}} f_{-\delta , -\delta ,\delta , \delta +t,0}(x)\, dP_X =\int _{\mathbb {R}} (x-\delta )I_{(\delta ,\delta +t]}(x)\, dP_X + \int _{\mathbb {R}} tI_{(\delta +t,+\infty )}(x)\, dP_X \nonumber \\&\quad =\int _{\mathbb {R}} xI_{(\delta ,\delta +t]}(x)\, dP_X -\delta P_X((\delta ,\delta +t])+tP_X((\delta +t,+\infty ))\nonumber \\&\quad = E(XI_{(\delta , \delta +t]}(X))-\delta (F_X(\delta +t)-F_X(\delta ))+t(1-F_X(\delta +t)), \end{aligned}$$

(2)

which implies (i).

Let $t<0.$ Take the mapping $f_{-\delta +t,-\delta ,\delta ,\delta ,0}$. That mapping is given by $f_{-\delta +t,-\delta ,\delta ,\delta ,0}(x)=\vert t \vert I_{(-\infty ,-\delta +t]}(x)+(-\delta -x)I_{(-\delta +t,-\delta ]}(x)$ for any $x \in \mathbb {R}$.

By Lemma 2, $E(f_{-\delta +t,-\delta ,\delta ,\delta ,0}(X)) \le E(f_{-\delta +t,-\delta ,\delta ,\delta ,0}(Y)).$ Now

$$\begin{aligned}&\int _{\mathbb {R}} f_{-\delta +t,-\delta ,\delta ,\delta ,0} (x) \, dP_X= \int _{\mathbb {R}} \vert t \vert I_{(-\infty ,-\delta +t]}(x) \, dP_X \nonumber \\&\qquad + \int _{\mathbb {R}} (-\delta -x)I_{(-\delta +t,-\delta ]}(x) \, dP_X \nonumber \\&\quad =\vert t \vert P_X((-\infty ,-\delta +t]) -\delta P_X((-\delta +t,-\delta ]) - \int _{\mathbb {R}} x I_{(-\delta +t,-\delta ]}(x) \, dP_X \nonumber \\&\quad =\vert t \vert F_X(-\delta +t)-\delta (F_X(-\delta )-F_X(-\delta +t))-E(XI_{(-\delta +t,-\delta ]}(X)), \nonumber \\&\quad =\vert t \vert F_X(-\delta +t)-\delta (F_X(-\delta )-F_X(-\delta +t))+E(\vert X\vert I_{(-\delta +t,-\delta ]}(X)), \end{aligned}$$

(3)

what leads to (ii).

Now suppose that (i) and (ii) hold. Let $f_{p_1,-\delta ,\delta ,p_4,0} \in \, \mathcal{F}_0^{-\delta ,\delta } \,$ $(p_1 < -\delta ,\, p_4> \delta )$. Note that $f_{p_1,-\delta ,\delta ,p_4,0}= f_{(p_1+\delta )-\delta ,-\delta ,\delta ,\delta ,0} +f_{-\delta ,-\delta ,\delta ,\delta +(p_4-\delta ),0}.$ Conditions (i) and (ii), jointly with formulas (2) and (3), imply that $E(f_{p_1,-\delta ,\delta ,p_4,0}(X)) \le E(f_{p_1,-\delta ,\delta ,p_4,0}(Y)),$ and so $X \, \preceq _{con}^{-\delta ,\delta } \,Y$. $\square$

Proof of Proposition 3

Let $t>0.$ We have that

$$\begin{aligned}&E(XI_{(\delta , \delta +t]}(X)) =\int _{(0,+\infty )} P(XI_{(\delta , \delta +t]}(X)>x)\, \hbox {d}x \\&\quad = \int _{(0,+\infty )} P(X\in (\delta , \delta +t],\, X>x)\, \hbox {d}x = \int _{(0,\delta +t]} P(X\in (\delta , \delta +t],\, X>x)\, \hbox {d}x\\&\quad = \int _{(0,\delta ]}P(X\in (\delta , \delta +t])\, \hbox {d}x + \int _{(\delta ,\delta +t]} P(x<X \le \delta +t)\, \hbox {d}x\\&\quad = \delta (F_X(\delta +t)-F_X(\delta )) +tF_X(\delta +t)- \int _{(\delta ,\delta +t]}F_X(x)\, \hbox {d}x \end{aligned}$$

Thus, formula (i) of Proposition 2 is equivalently to

$$\begin{aligned} \int _{\delta }^{\delta +t} F_X(x)\, \hbox {d}x \ge \int _{\delta }^{\delta +t} F_Y(x)\, \hbox {d}x \end{aligned}$$

for any $t>0$.

On the other hand, if $t<0$ then

$$\begin{aligned}&E(\vert X \vert I_{(-\delta +t, -\delta ]}(X))=\int _{(0,+\infty )} P(\vert X\vert I_{(-\delta +t, -\delta ]}(X) \ge x)\, \hbox {d}x \\&\quad = \int _{(0,+\infty )} P(X\in (-\delta +t, -\delta ],\, \vert X\vert \ge x)\, \hbox {d}x\\&\quad = \int _{(0,\delta -t]} P(X\in (-\delta +t, -\delta ],\, X \le -x)\, \hbox {d}x\\&\quad = \int _{(0,\delta ]}P(X\in (-\delta +t, -\delta ])\, \hbox {d}x + \int _{(\delta ,\delta -t]} P(X\in (-\delta +t, -x])\, \hbox {d}x\\&\quad =\delta (F_X(-\delta )-F_X(-\delta +t))+ \int _{(\delta ,\delta -t]} F_X(-x)\, \hbox {d}x+tF_X(-\delta +t). \end{aligned}$$

As a consequence, condition (ii) in Proposition 2 is the same as

$$\begin{aligned} \int _{-\delta +t}^{-\delta } F_X(x)\, \hbox {d}x \le \int _{-\delta +t}^{-\delta } F_Y(x)\, \hbox {d}x \end{aligned}$$

for any $t<0.$ $\square$

Proof of Proposition 4

Observe that for any random variable W, $F_{W_+}(x)=F_W(x)I_{[0,\infty )}(x)$ and $F_{W_-}(x)=(1-F_W(-x^-))I_{[0,\infty )}(x)$.

When $t>0$, we have that

$$\begin{aligned} \int _{\delta }^{\delta +t} F_X(x)\, \hbox {d}x=\int _{0}^{t} F_X(x+\delta )\, \hbox {d}x= \int _{0}^{t} F_{X-\delta }(x)\, \hbox {d}x= \int _{-\infty }^{t} F_{(X-\delta )_+}(x)\, \hbox {d}x, \end{aligned}$$

and then, condition (i) of Proposition 3 is the same as

$$\begin{aligned} \int _{-\infty }^{t} F_{(X-\delta )_+}(x)\, \hbox {d}x \ge \int _{-\infty }^{t} F_{(Y-\delta )_+}(x)\, \hbox {d}x \end{aligned}$$

for any $t \in \mathbb {R}$. By means of Theorem 4.A.2 in Shaked and Shanthikumar (2007), this is equivalent to $(X-\delta )_+ \preceq _{icv} (Y-\delta )_+$.

When $t<0$, we have that

$$\begin{aligned} \int _{-\delta +t}^{-\delta } F_X(x)\, \hbox {d}x= \int _{t}^{0} F_X(x-\delta )\, \hbox {d}x. \end{aligned}$$

Therefore, condition (ii) of Proposition 3 can be rewritten as

$$\begin{aligned}&t+\int _{t}^{0} F_X(x-\delta )\, \hbox {d}x \le t+\int _{t}^{0} F_Y(x-\delta )\, \hbox {d}x,\,\, \hbox { that is,}\\&\quad \int _t^0 (1-F_{X+\delta }(x))\, \hbox {d}x \ge \int _t^0 (1-F_{Y+\delta }(x))\, \hbox {d}x. \end{aligned}$$

The set of discontinuity points of an increasing map is at most countable; then, the above inequality is the same as

$$\begin{aligned}&\int _t^0 (1-F_{X+\delta }(x^-))\, \hbox {d}x \ge \int _t^0 (1-F_{Y+\delta }(x^-))\, \hbox {d}x, \,\, \hbox { equivalently,} \\&\quad \int _0^{-t} (1-F_{X+\delta }(-x^-))\, \hbox {d}x \ge \int _0^{-t} (1-F_{Y+\delta }(-x^-))\, \hbox {d}x, \,\, \hbox { that is,} \\&\quad \int _0^{-t} F_{(X+\delta )_-}(x)\, \hbox {d}x \ge \int _0^{-t} F_{(Y+\delta )_-}(x)\, \hbox {d}x \end{aligned}$$

for any $t<0$, or

$$\begin{aligned} \int _0^{t} F_{(X+\delta )_-}(x)\, \hbox {d}x \ge \int _0^{t} F_{(Y+\delta )_-}(x)\, \hbox {d}x \end{aligned}$$

for any $t>0$. It is clear that this is the same as

$$\begin{aligned} \int _{-\infty }^{t} F_{(X+\delta )_-}(x)\, \hbox {d}x \ge \int _{-\infty }^{t} F_{(Y+\delta )_-}(x)\, \hbox {d}x \end{aligned}$$

for any $t \in \mathbb {R}$. By Theorem 4.A.2 in Shaked and Shanthikumar (2007), this is $(X+\delta )_- \preceq _{icv} (Y+\delta )_-$. $\square$

Proof of Proposition 5

Assume that $X \, \preceq _{con}^{p_2,p_3} \,Y$. Let $f \in \mathcal{F}^{-p_3,-p_2}$. Consider $T:\mathbb {R}\rightarrow \mathbb {R}$ with $T(x)=-x$ for any $x \in \mathbb {R}$. Note that $P_X \circ T^{-1}=P_{-X}.$ By a change of variable

$$\begin{aligned} \int _{\mathbb {R}} f(x)\, dP_{-X}=\int _{\mathbb {R}} f(-x)\, dP_X. \end{aligned}$$

It is not hard to prove that if $f \in \mathcal{F}^{-p_3,-p_2}$, then the mapping $x \rightarrow f(-x)$ belongs to $\, \mathcal{F}^{p_2,p_3} \,,$ and so

$$\begin{aligned} \int _{\mathbb {R}} f(-x)\, dP_X \le \int _{\mathbb {R}} f(-x)\, dP_Y=\int _{\mathbb {R}} f(x)\, dP_{-Y}, \end{aligned}$$

which proves that $-X \, \preceq _{con}^{-p_3,-p_2} \, -Y$. The converse is implied by the proved part. $\square$

Proof of Proposition 6

Let $T:\mathbb {R}\rightarrow \mathbb {R}$ with $T(x)=\vert x\vert .$ Let $f \in \, \mathcal{F}_0^{-\delta ,\delta } \,$. A change of variable, the symmetry of the distribution and $f(0)=0$, imply that

$$\begin{aligned}&\int _{\mathbb {R}} f(x)\, dP_{\vert X\vert }=\int _{\mathbb {R}} f(\vert x\vert )\, dP_X=\int _{(-\infty ,0)} f(-x )\, dP_X+\int _{(0,+\infty )} f(x)\, dP_X\\&\quad =2\int _{(0,+\infty )} f(x)\, dP_X. \end{aligned}$$

Assume that $X \, \preceq _{con}^{-\delta ,\delta } \,Y$. Let $f \in \, \mathcal{F}_0^{-\delta ,\delta } \,$, thus $f=f_{p_1,-\delta ,\delta ,p_4,0}$ for some $p_1$ and $p_4$ with $p_1 < -\delta$ and $p_4 > \delta$. Note that $fI_{(0,\infty )}=f_{-\delta ,-\delta ,\delta ,p_4,0}$. Then

$$\begin{aligned}&\int _{\mathbb {R}} f(x)\, dP_{\vert X\vert }=2\int _{(0,+\infty )} f(x)\, dP_X=2\int _{(0,+\infty )} f_{-\delta ,-\delta ,\delta ,p_4,0}(x)\, dP_X \\&\quad =2\int _{\mathbb {R}} f_{-\delta ,-\delta ,\delta ,p_4,0}(x)\, dP_X \le 2\int _{\mathbb {R}} f_{-\delta ,-\delta ,\delta ,p_4,0}(x)\, dP_Y=\int _{\mathbb {R}} f(x)\, dP_{\vert Y\vert }. \end{aligned}$$

Therefore, we obtain that $\vert X \vert \, \preceq _{con}^{-\delta ,\delta } \,\vert Y \vert .$

Now let us suppose that $\vert X \vert \, \preceq _{con}^{-\delta ,\delta } \,\vert Y \vert .$ Let $f \in \, \mathcal{F}_0^{-\delta ,\delta } \,$. Note that the mapping $x \mapsto f(-x)$ belongs to $\, \mathcal{F}_0^{-\delta ,\delta } \,$. Thus

$$\begin{aligned}&\int _{\mathbb {R}} f\, dP_{X}=\int _{(0,+\infty )} f(x)\, dP_X + \int _{(0,+\infty )} f(-x)\, dP_X={1 \over 2} \int _{\mathbb {R}} f(x)\, dP_{\vert X\vert }\\&\quad + {1 \over 2} \int _{\mathbb {R}} f(-x)\, dP_{\vert X\vert }\le {1 \over 2} \int _{\mathbb {R}} f(x)\, dP_{\vert Y\vert } + {1 \over 2} \int _{\mathbb {R}} f(-x)\, dP_{\vert Y\vert }=\int _{\mathbb {R}} f\, dP_{Y}, \end{aligned}$$

which concludes the result. $\square$

Proof of Corollary 3

It follows from Propositions 6 and 4. $\square$

Proof of Proposition 7

By Proposition 3, we have that

• (i) $\int _{\delta }^{\delta +t} F_X(x)\, \hbox {d}x = \int _{\delta }^{\delta +t} F_Y(x)\, \hbox {d}x$ for any $t>0$, and

• (ii) $\int _{-\delta +t}^{-\delta } F_X(x)\, \hbox {d}x = \int _{-\delta +t}^{-\delta } F_Y(x)\, \hbox {d}x$ for any $t<0$.

By the First Fundamental Theorem of Calculus, we obtain that $F_X(\delta +t)=F_Y(\delta +t)$ for all $t>0$ such that $\delta +t$ is a continuity point of $F_X$ and $F_Y$. Using the right continuity of distributions functions and the density of the set of continuity points of both functions, we obtain that $F_X(x)=F_Y(x)$ for any $x \ge \delta .$ The same reasoning applied to condition (ii) provides that $F_X(x)=F_Y(x)$ for any $x \le -\delta .$ $\square$

Proof of Proposition 8

The condition $S^-(f_Y-f_X)=2$ with sign sequence $+,-,+$, implies that $S^-(F_Y-F_X)=1$ with sign sequence $+,-$ (see the proof of Theorem 3.A.44 in Shaked and Shanthikumar (2007)). The symmetry of the distributions implies that the crossing point is 0. Moreover, $F_{\vert X \vert }(x)=(2F_X-2)I_{[0,\infty )}(x)$. Therefore, $S^-(F_{\vert Y\vert }-F_{\vert X\vert })=0$ and $F_{\vert Y\vert }\le F_{\vert X\vert }$, which is the same as $\vert X\vert \preceq _{st}\vert Y\vert$. Since the mapping $g_{\delta ,t}:\mathbb {R}\rightarrow \mathbb {R}$ with $g_{\delta ,t}(x)=(x-\delta )_+$ is increasing, we conclude that $(\vert X\vert -\delta )_+ \preceq _{st}(\vert Y\vert -\delta )_+$. That implies $(\vert X\vert -\delta )_+ \preceq _{icv}(\vert Y\vert -\delta )_+$. Now Proposition 4 ensures that $\vert X\vert \, \preceq _{con}^{-\delta ,\delta } \,\vert Y\vert$, and Proposition 6 proves the result. $\square$

Proof of Proposition 9

Let us see (i). We have $F_{(X-\delta )_+}(x)=F_{X-\delta }(x)I_{[0,\infty )}(x)=F_X(x+\delta )I_{[0,+\infty )}(x)$ for any $x \in \mathbb {R}$, and the same formula holds for the random variable Y. Thus if $S^-(F_Y-F_X) \le 1$, we obtain that $S^-(F_{(Y-\delta )_+}-F_{(X-\delta )_+}) \le 1$, and if the second number of sign changes is 1, so is the first number of sign changes, the sequence of signs being equal. Applying Theorem 4.A.22 (b) in Shaked and Shanthikumar (2007), we conclude that $(X-\delta )_+ \preceq _{icv} (Y-\delta )_+$.

On the other hand $F_{(X+\delta )_-}(x)=(1-F_{(X+\delta )}(-x^-))I_{[0,\infty )}(x)=(1-F_X((-\delta -x)^-))I_{[0,\infty )}(x).$ As a consequence, we obtain that $S^-(F_{(Y+\delta )_-}-F_{(X+\delta )_-}) \le S^-(F_Y-F_X)$, and if $S^-(F_{(Y+\delta )_-}-F_{(X+\delta )_-})=1$ so is $S^-(F_Y-F_X)=1$, sharing the same sequence of signs. By Theorem 4.A.22 (b) in Shaked and Shanthikumar (2007), we obtain that $(X+\delta )_-\preceq _{icv} (Y+\delta )_-$. Now Proposition 4 implies that $X \, \preceq _{con}^{-\delta ,\delta } \,Y.$

In relation to (ii), observe that $F_{X_{(n)}}=F_X^n$. Therefore, $S^-(F_{Y_{(n)}}-F_{X_{(n)}})=S^-(F_Y-F_X)$. Moreover, $F_{Y_{(n)}}-F_{X_{(n)}}$ and $F_Y-F_X$ share the same sign changes. Hence (ii) is obtained by (i).

Taking into account $F_{X_{(1)}}=1-(1-F_X)^n$, statement (iii) can be obtained in a similar way. $\square$

Proof of Proposition 10

It holds that $P_X \circ h_i^{-1}$ is equal to $P_{h_i(X)}$ with $i=1,2.$ Let $f \in \, \mathcal{F}_0^{-\delta ,\delta } \,$. We have that

$$\begin{aligned} \int _{\mathbb {R}} f(x) \,dP_{h_1(X)}= & {} \int _{\mathbb {R}} f(h_1(x))\, dP_X\\= & {} \int _{[0,+\infty )} f(h_1(x))\, dP_X + \int _{(-\infty ,0)} f(h_1(x))\, dP_X. \end{aligned}$$

If $x\in [0,+\infty )$, we have that $h_1(x) \le h_2(x)$ and f is increasing on that set. On the other hand, if $x \in (-\infty ,0)$, then $h_1(x) \ge h_2(x)$ and on that subset the mapping f is decreasing. As a consequence

$$\begin{aligned}&\int _{[0,+\infty )} f(h_1(x))\, dP_X + \int _{(-\infty ,0)} f(h_1(x))\, dP_X\\&\quad \le \int _{[0,+\infty )} f(h_2(x))\, dP_X + \int _{(-\infty ,0)} f(h_2(x))\, dP_X=\int _{\mathbb {R}} f(x)\, dP_{h_2(X)}, \end{aligned}$$

which proves the result. $\square$

Proof of Proposition 11

The case $\alpha =0$ is trivial. Let $\alpha >0.$ Consider $T:\mathbb {R}\rightarrow \mathbb {R}$ with $T(x)= \alpha x$. It holds that $P_X \circ T^{-1}=P_{\alpha X}.$ Let $f \in \mathcal{F}_0^{\alpha p_2, \alpha p_3}.$ By a change of variable,

$$\begin{aligned} \int _{\mathbb {R}} f \, dP_{\alpha X}= \int _{\mathbb {R}} f(x)\, dP_X \circ T^{-1}= \int _{\mathbb {R}} f(\alpha x)\, dP_X. \end{aligned}$$

If $f \in \mathcal{F}_0^{\alpha p_2, \alpha p_3},$ then $f=f_{p_1,\alpha p_2, \alpha p_3,p_4,0}$ for some $p_1 < \alpha p_2$ and $p_4 > \alpha p_3.$ It is not hard to prove that the mapping $x \rightarrow f(\alpha x)$ is the function $\alpha f_{p_1/\alpha , p_2, p_3,p_4/\alpha ,0}$, which belongs to the class $\mathcal{F}_0^{p_2,p_3}$. As a consequence

$$\begin{aligned} \int _{\mathbb {R}} f(\alpha x)\, dP_X \le \int _{\mathbb {R}} f(\alpha x)\, dP_Y=\int _{\mathbb {R}} f\, dP_{\alpha Y}. \end{aligned}$$

Therefore, $\alpha X \, \preceq _{con}^{\alpha p_2, \alpha p_3} \,\alpha Y$.

Now let $\alpha <0$. By the proven part we conclude that $-\alpha X \, \preceq _{con}^{-\alpha p_2, -\alpha p_3} \,-\alpha Y.$ Applying Proposition 5, we deduce that $\alpha X \, \preceq _{con}^{\alpha p_3, \alpha p_2} \,\alpha Y$. $\square$

Proof of Proposition 12

Note that the mappings of $\, \mathcal{F}_0^{p_2,p_3} \,$ are continuous and bounded. $\square$

Proof of Proposition 13

The stochastic order is integral which implies the result (see Theorem 2.4.2 in Müller and Stoyan (2002)). $\square$

Proof of Proposition 14

Clearly $X-\delta \preceq _{st} X-\delta +a$. Then, $X-\delta \preceq _{icv} X-\delta +a$. Observe that $(X-\delta )_+=X-\delta \,\, a.s.$ and $(X+a-\delta )_+=X+a-\delta \,\, a.s.$ Moreover, $(X+\delta )_-=0=(X+a+\delta )_- \,\,a.s.$ Proposition 4 proves the result. $\square$

Proof of Proposition 15

In accordance with Proposition 5, $X \, \preceq _{con}^{-\delta ,\delta } \,X+b$ is equivalent to $-X \, \preceq _{con}^{-\delta ,\delta } \,-X-b.$ Now the result is a consequence of Proposition 14. $\square$

Proof of Proposition 16

A generator of the bidirectional order is the set $\mathcal{F}=\{\,f:\mathbb {R}\rightarrow \mathbb {R}\mid f \hbox { is bounded, increasing in } (0,\infty ), \hbox { decreasing in }(-\infty ,0)$ and with minimum at the point $0 \,\}$ (Proposition 5 in López-Díaz (2010)). Observe that $\mathcal{F}_0^{-\delta ,\delta } \subset \mathcal{F},$ which leads to the result. $\square$

Proof of Proposition 17

By Proposition 16, $X \preceq _{bd} Y$ implies $X \, \preceq _{con}^{-\delta ,\delta } \,Y$ for any $\delta >0$.

Now suppose that $X \, \preceq _{con}^{-\delta ,\delta } \,Y$ for any $\delta >0$. The condition $X \preceq _{bd} Y$ is the same as $F_X-F_Y$ pivots at 0, that is, $F_X(t)-F_Y(t) \ge 0$ for any $t \ge 0$ and $F_X(t)-F_Y(t) \le 0$ for any $t < 0$. Assume that $X \preceq _{bd} Y$ is false. Therefore, there exists $t_0 \ge 0$ with $F_X(t_0) < F_Y(t_0)$, or there is $t_0 <0$ satisfying that $F_X(t_0) > F_Y(t_0)$. Consider now the first possibility. By the right continuity of distribution functions, we can assume that $t_0 >0.$ For the same reason, there exists $\varepsilon >0$ such that $F_X < F_Y$ on $(t_0,t_0+\varepsilon ),$ which is a contradiction with Proposition 3 (i). The case of the existence of $t_0<0$ with $F_X(t_0) > F_Y(t_0)$, can be analyzed in the same way, using (ii) in Proposition 3. Thus, we conclude that $X \preceq _{bd} Y$. $\square$

Proof of Proposition 18

We have that $X \preceq _{peak} Y$ implies that $X-EX \preceq _{bd} Y-EY$ (Corollary 9 in López-Díaz (2010)). The result follows from Proposition 16. $\square$

Proof of Proposition 19

The relation $X \preceq _{w} Y$ is equivalent to $X-X' \preceq _{bd} Y-Y'$ (Corollary 10 in López-Díaz (2010)), and so we have the result. $\square$

Proof of Proposition 20

The sequences of mappings $\{(x-1/m)_+\}_m$ and $\{(x+1/m)_-\}_m$ are increasing. By the monotone convergence theorem, $\lim _m E((X-1/m)_+)= E(\lim _m (X-1/m)_+)=EX_+$ and $\lim _m E((X+1/m)_-)= E(\lim _m (X+1/m)_-)=EX_-$. On the other hand, $\lim _m (X-1/m)_+=X_+$ and $\lim _m (X+1/m)_-=X_-$ in the weak convergence since we have the pointwise convergence. The same results are satisfied by random variable Y.

By Proposition 4 in this manuscript, Theorem 1.5.9 in Müller and Stoyan (2002) and the relation between $\preceq _{icx}$ and $\preceq _{icv}$, we conclude that $X_+ \preceq _{icv} Y_+$ and $X_- \preceq _{icv} Y_-$.

If X and Y are negative a.s., then $Y_+ =0=X_+ \,\,a.s.$, $X_-=-X$ and $Y_-=-Y\,\,a.s.$, and so we conclude that $Y \preceq _{icx} X$. When X and Y are positive a.s., $Y_- =0=X_-\,\,a.s.$, $X_+=X$ and $Y_+=Y\,\,a.s.$, which implies that $X \preceq _{icv} Y.$ $\square$

Proof of Corollary 5

It is a consequence of Proposition 20 and Theorem 1.5.3 in Müller and Stoyan (2002). $\square$

Proof of Example 5

Proposition 19 in López-Díaz (2010) reads that $\sigma _X \le \sigma _Y$ implies that $X-\mu _X \preceq _{bd} Y-\mu _Y$, and as a consequence of Proposition 16, we deduce that $X-\mu _X \, \preceq _{con}^{-\delta ,\delta } \,Y-\mu _Y$.

Let us see the converse. We have that $X-\mu _X \, \preceq _{con}^{-\delta ,\delta } \,Y-\mu _Y$. Suppose that $\sigma _Y <\sigma _X$. By the proven part, we obtain that $Y-\mu _Y \, \preceq _{con}^{-\delta ,\delta } \,X-\mu _X.$ Applying Proposition 7, we deduce that $F_{X-\mu _X}(t)=F_{Y-\mu _Y}(t)$ for any $t \in (-\infty ,-\delta ] \cup [\delta ,+\infty )$. By Proposition 3 and the First Fundamental Theorem of Calculus, we deduce that the density mappings of $X-\mu _X$ and $Y-\mu _Y$ are the same on $(-\infty ,-\delta ] \cup [\delta ,+\infty )$, and so $\sigma _X=\sigma _Y,$ which contradicts the assumption $\sigma _Y <\sigma _X.$ Therefore, $\sigma _X \le \sigma _Y$. $\square$

Proof of Example 6

Lemma 1 in Finner et al. (2007) reads that the density mappings of X and Y satisfy that $S^-(f_Y-f_X)=2$ and the sign sequence is $+,-,+$. Now the result follows from Proposition 8. $\square$

Proof of Example 7

Lemma 1 and Theorem 3 in Finner et al. (2007) imply that $S^-(f_Y-f_X)=2$ and the sign sequence is $+,-,+$. Proposition 8 proves the result. $\square$

Proof of Proposition 21

The result follows from Example 5, note that $X_t \sim _{st} N(r_Xt,\sigma _X \sqrt{t})$ and $Y_t \sim _{st} N(r_Yt,\sigma _Y \sqrt{t})$. $\square$

Proof of Proposition 22

Proposition 21 reads that $\sigma _Y \ge \sigma _X$ if and only if $X_t-r_Xt \, \preceq _{con}^{-\delta ,\delta } \,Y_t-r_Yt$ for any $t \in [0,T]$ and any $\delta >0.$ Now take $p_2=k-\delta$ and $p_3=k+\delta$ with $k \in \mathbb {R}$ and $\delta >0$ in Proposition 1, which proves the result. $\square$

Proof of Proposition 23

We will prove the result by means of Proposition 3.

Let $z \le -\delta$. We have that $F_{X'_t}(z)=P(X'_t \le z)=P(X_t \le c e^{rt}+z)$. Assume that z satisfies that $c e^{rt}+z>0$, that is, $-c e^{rt}<z$, otherwise the above probability is equal to 0. Take $m_z=\ln (1+z/ce^{rt}),$ thus $c e^{rt}+z=ce^{rt+m_z}.$ The above probability satisfies that

$$\begin{aligned}&P(X_t \le c e^{rt}+z)=P(X_t \le c e^{rt+m_z})=P \left( -{1\over 2}\sigma ^2_X t+\sigma _X B_t \le m_z \right) \\&\quad =P \left( B_t \le {1 \over 2} \sigma _X t+ {m_z \over \sigma _X} \right) =F_{B_t} \left( {1 \over 2} \sigma _X t+ {m_z \over \sigma _X}\right) . \end{aligned}$$

The same result holds for the process $(Y'_t)_{t \in [0,T]}$.

Suppose that condition (ii) in Proposition 3 holds.

Recall that $B_t \sim _{st} \widetilde{B}_t \sim _{st} N(0,\sqrt{t})$. By the continuity of $F_{B_t}$ and $F_{\widetilde{B}_t}$, we obtain that

$$\begin{aligned} {1 \over 2} \sigma _X t+ {m_{-\delta } \over \sigma _X} \le {1 \over 2} \sigma _Y t+ {m_{-\delta } \over \sigma _Y}, \hbox { that is, } 0 \le \left( \sigma _Y - \sigma _X \right) \left( {t \over 2}-m_{-\delta }{1 \over \sigma _X \sigma _Y}\right) . \end{aligned}$$

Observe that $m_{-\delta }=\ln (1+(-\delta )/ce^{rt})$ is negative, and so $\sigma _Y \ge \sigma _X$. Conversely, if $\sigma _Y \ge \sigma _X$, then

$$\begin{aligned} F_{B_t} \left( {1 \over 2} \sigma _X t+ {m_z \over \sigma _X} \right) \le F_{\widetilde{B}_t} \left( {1 \over 2} \sigma _Y t+ {m_z \over \sigma _Y} \right) \end{aligned}$$

for any $z \le -\delta$ since $m_z$ is negative, and so condition (ii) in Proposition 3 is satisfied. Therefore, statement (ii) in Proposition 3 is equivalent to $\sigma _Y \ge \sigma _X$. Now take $z \ge \delta$. We have that $F_{X'_t}(z)=P(X'_t \le z)=P(X_t \le c e^{rt}+z)=P(X_t \le c e^{rt+m_z})$, where $m_z=\ln (1+z/ce^{rt}),$ note that in this case $m_z>0.$ Thus,

$$\begin{aligned}&P(X_t \le c e^{rt+m_z})=P \left( -{1\over 2}\sigma ^2_X t+\sigma _X B_t \le m_z \right) \\&\quad =P \left( B_t \le {1 \over 2} \sigma _X t+ {m_z \over \sigma _X}\right) =F_{B_t} \left( {1 \over 2} \sigma _X t+ {m_z \over \sigma _X}\right) . \end{aligned}$$

Assume that condition (i) in Proposition 3 is satisfied. By the right continuity of distribution functions, we obtain that

$$\begin{aligned} {1 \over 2} \sigma _X t+{m_\delta \over \sigma _X} \ge {1 \over 2} \sigma _Y t+{m_\delta \over \sigma _Y}, \hbox { that is, } 0 \ge (\sigma _Y-\sigma _X) \left( {t \over 2}-{m_\delta \over \sigma _x \sigma _Y}\right) . \end{aligned}$$

Since $\sigma _Y \ge \sigma _X$ by the first part of the proof, we conclude that $m_\delta \ge {t \over 2} \sigma _X \sigma _Y$, that is, $\delta \ge ce^{rt}(e^{{t \over 2} \sigma _X \sigma _Y} -1)$. Conversely, if $\delta \ge ce^{rt}(e^{{t \over 2} \sigma _X \sigma _Y} -1)$, that is, $m_\delta \ge {t \over 2} \sigma _X \sigma _Y$, then $m_z \ge m_\delta \ge {t \over 2} \sigma _X \sigma _Y$ for any $z \ge \delta$, and so $F_{B_t} ( {1 \over 2} \sigma _X t+ {m_z \over \sigma _X}) \ge F_{\widetilde{B}_t} ( {1 \over 2} \sigma _Y t+ {m_z \over \sigma _Y})$, which concludes (i) in Proposition 3, and so the proof of the result. $\square$

Proof of Proposition 24

It follows applying Propositions 23 and 1. $\square$

Proof of Proposition 25

Statement (i) is clear.

In relation to (ii), the real mapping on $\mathbb {R}$ given by $x \rightarrow S^x_{p_1,p_2,k_1,k_2}$ is increasing for any $p_1,p_2,k_1,k_2 \in \mathbb {R}$ with $p_1 \le p_2$. Therefore, $X \preceq _{st} Y$ implies that $E S^X_{p_1,p_2,k_1,k_2} \le E S^Y_{p_1,p_2,k_1,k_2}$ for all $p_1,p_2,k_1,k_2 \in \mathbb {R}$ with $p_1 \le p_2$.

Conversely, note that $E S^X_{p_1,p_2,k_1,k_2}=\pi _X(p_1)-\pi _X(p_2)+k_2-k_1$. Thus, $E S^X_{p_1,p_2,k_1,k_2} \le E S^Y_{p_1,p_2,k_1,k_2}$ for all $p_1,p_2,k_1,k_2 \in \mathbb {R}$ with $p_1 \le p_2$, implies that $\pi _X(p_1)-\pi _X(p_2) \le \pi _Y(p_1)-\pi _Y(p_2)$. That is $\pi _Y - \pi _X$ is decreasing. By Theorem 1.5.13 in Müller and Stoyan (2002), we obtain that $X \preceq _{st} Y.$

Statment (iii) follows from Theorem 1.5.7 in Müller and Stoyan (2002).

For the last statement, note that the real mapping on $\mathbb {R}$ defined by $x \rightarrow ST^x_{p,k_1,k_2}$ is convex for any $p,\,k_1,\, k_2 \in \mathbb {R}.$ Thus, $X \preceq _{cx} Y$ leads to $EST^X_{p,k_1,k_2} \le EST^Y_{p,k_1,k_2}$ for all $p,k_1,k_2 \in \mathbb {R}$.

For the converse, take $k_1=k_2=0.$ Now note that $\lim _{p' \rightarrow - \infty } {1 \over 2}(ST^X_{p,0,0} + ST^X_{p',0,0}-(p-p'))=C^X_{p,0}.$ Since $E(ST^X_{p,0,0}+ST^X_{p',0,0}) \le E(ST^Y_{p,0,0}+ST^Y_{p',0,0})$ for any $p,p' \in \mathbb {R}$, the dominated convergence theorem implies that $E(C^X_{p,0}) \le E(C^Y_{p,0})$. By Theorem 1.5.7 in Müller and Stoyan (2002), $X \preceq _{icx} Y$ holds.

On the other hand, since $ST^X_{p,0,0}=(X-p)_+ + (p-X)_+$ we obtain that $\lim _{p \rightarrow -\infty } ST^X_{p,0,0}+p=X \hbox { and } \lim _{p \rightarrow +\infty } ST^X_{p,0,0}-p=-X.$ By the dominated convergence theorem, we have that $\lim _{p \rightarrow - \infty } E(ST^X_{p,0,0}) + p=EX \hbox { and } \lim _{p \rightarrow + \infty } E(ST^X_{p,0,0}) - p=-EX.$ Therefore, we deduce that $EX=EY,$ which in conjunction with $X \preceq _{icx} Y$ implies that $X \preceq _{cx} Y.$ $\square$