Bivariate Risk Measures and Stochastic Orders

Abstract

Bivariate value-at-risks and bivariate average value-at-risks are defined with copula functions, and stochastic orders are induced from these bivariate value-at-risks and bivariate average value-at-risks. Value-at-risk order is almost equivalent to the first-order stochastic dominance, however it is made clear that average value-at-risk order is weaker than the second-order stochastic dominance and it has similar properties. We find that average value-at-risk order is an important criterion as a stochastic order in multi-object case-based reasoning with risk aversity.

Appendix

Proof of Lemma 2.1

(i) For $x_1 = {\mathrm{VaR}}_{q_1}(X_1)$ and $x_2 = {\mathrm{VaR}}_{q_2}(X_2)$, from (2.3) and (2.8) we have $c(q_1,q_2) = \frac{\partial ^2}{\partial q_1 \partial q_2} P(X_1 \le x_1, X_2 \le x_2) = w(x_1, x_2) \frac{dx_1}{d q_1} \frac{dx_2}{d q_2}.$ Thus we get (i). (ii) Since $P(X_1 \le x_1) = q_1$ and $P(X_2 \le x_2) = q_2$, we get $w_1(x_1) \, dx_1 = dq_1 \ \text{ and } \ w_2(x_2) \, dx_2 = dq_2.$ (iii) is trivial from (i) and (ii).    $\square$

Proof of Lemma 2.2

Let a random $(X_1,X_2) \in \mathcal{X}^2$ and a probability $p \in (0,1]$. (i) From (2.4), (2.5) and (2.6), we have

$$\begin{aligned}&{{\mathrm{VaR}}_{p}(X_1,X_2)} \\&= \{ (x_1,x_2) \in {\mathbb {R}}^2 \mid P(X_1 \le x_1)=p_1, P(X_2 \le x_2 )=p_2, C(p_1, p_2) =p \} \\&= \{ ({\mathrm{VaR}}_{p_1}(X_1), {\mathrm{VaR}}_{p_2}(X_2)) \mid (p_1, p_2) \in \varGamma (p) \}. \end{aligned}$$

Thus we get (i). (ii) From (i), (2.7) and Lemma 2.1(i), we also have

$$\begin{aligned}&{{\mathrm{AVaR}}_{p}(X_1,X_2)} \\&= \{ E( (X_1,X_2) \mid (X_1,X_2) \preceq {\mathrm{VaR}}_{p}(X_1,X_2) ) \} \\&= \{ E( (X_1,X_2) \mid (X_1,X_2) \preceq ({\mathrm{VaR}}_{p_1}(X_1), {\mathrm{VaR}}_{p_2}(X_2))) \mid (p_1, p_2) \in \varGamma (p) \} \\&= \{ E((X_1,X_2) \mid X_1 \le {\mathrm{VaR}}_{p_1}(X_1), X_2 \le {\mathrm{VaR}}_{p_2}(X_2) ) \mid (p_1, p_2) \in \varGamma (p) \} \\&= \left\{ \frac{1}{p} \iiint _{(-\infty ,{{\mathrm{VaR}}_{p_1}(X_1)}] \times (-\infty ,{{\mathrm{VaR}}_{p_2}(X_2)}]} \! \! \! \! \! \! \! \! \! \! \! \! (x_1,x_2) \, w(x_1,x_2) \, dx_1 dx_2 \mid (p_1, p_2) \in \varGamma (p) \right\} \\&= \left\{ \frac{1}{p} \iint _{[0,p_1] \times [0,p_2]} \! \! \! \! \! \! \! \! \left( {\mathrm{VaR}}_{q_1}(X_1), {\mathrm{VaR}}_{q_2}(X_2) \right) c(q_1, q_2) \, dq_1 dq_2 \mid (p_1, p_2) \in \varGamma (p) \right\} . \end{aligned}$$

Therefore this lemma holds.    $\square$

Proof of Lemma 2.3

Value-at-risks (2.1) has similar properties (Artzner  [2]). Therefore from Lemma 2.2(i) we can easily check (i)–(iii).    $\square$

Proof of Lemma 2.4

Value-at-risks (2.2) has similar properties to (i)–(iii) [2]. From Lemma 2.2(ii), we can easily check (i)–(iii). (iv) It is well-known in one-dimensional case (2.2) that average value-at-risks has sub-additivity:

$$\begin{aligned} {\mathrm{AVaR}}_p(X+Y) \ge {\mathrm{AVaR}}_p(X) + {\mathrm{AVaR}}_p(Y) \end{aligned}$$

(A.1)

for all $X,Y \in \mathcal{X}$ (Artzner  [2] and Yoshida  [9]). Let $(X_1,X_2), (Y_1,Y_2) \in \mathcal{X}^2$ and let $p \in (0,1]$ and $(p_1, p_2) \in \varGamma (p)$. From (2.9) and (2.14), maps $q_1 \mapsto \frac{\partial }{\partial q_1} C(q_1,p_2) = \int _{0}^{p_2} \! \! c(q_1, q_2) \, dq_2 \ \text{ and } \ q_2 \mapsto \frac{\partial }{\partial q_2} C(p_1, q_2) = \int _{0}^{p_1} \! \! c(q_1, q_2) \, dq_1$ are non-increasing. When we approximate these nonnegative-valued functions by sums of non-increasing step functions on (0, 1], $q \mapsto \gamma 1_{[0,\delta ]}(q) \, (\gamma >0, \, \delta \in (0,1] )$, from (A.1) we can check

$$\begin{aligned}&{ \frac{1}{p} \iint _{[0,p_1] \times [0,p_2]} \! \! \! \! \! \! \! \! \left( {\mathrm{VaR}}_{q_1}(X_1 + Y_1), {\mathrm{VaR}}_{q_2}(X_2 + Y_2) \right) c(q_1, q_2) \, dq_1 dq_2} \\&\;\; \succeq \frac{1}{p}\iint _{[0,p_1] \times [0,p_2]} \! \! \! \! \! \! \! \! \left( {\mathrm{VaR}}_{q_1}(X_1), {\mathrm{VaR}}_{q_2}(X_2) \right) c(q_1, q_2) \, dq_1 dq_2 \\&\quad \;\; +\, \frac{1}{p} \iint _{[0,p_1] \times [0,p_2]} \! \! \! \! \! \! \! \! \left( {\mathrm{VaR}}_{q_1}(Y_1), {\mathrm{VaR}}_{q_2}(Y_2) \right) c(q_1, q_2) \, dq_1 dq_2. \end{aligned}$$

Therefore we obtain (i) from Lemma 2.2(ii), and this lemma holds.    $\square$

Proof of Theorem 3.1

(a) $\Longrightarrow$ (b) is trivial when we let $x_1 \rightarrow \infty$ or $x_2 \rightarrow \infty$ in Definition 3.1. (b) $\Longrightarrow$ (a): From (2.9), maps $p_1 \mapsto C(p_1,p_2)$ and $p_2 \mapsto C(p_1,p_2)$ are non-decreasing. And we get (a) from (b) with (2.4). (a) $\Longrightarrow$ (c): Let $f \in \mathcal{L}^2$. We can approximate decreasing function $- f$ by sums of step functions $\gamma 1_{(-\infty , x_1] \times (-\infty ,x_2]}$ ($\gamma >0$) and we get (c) from (3.3). (c) $\Longrightarrow$ (a): We can easily check (a) when we approximate step functions $1_{(-\infty , x_1] \times (-\infty ,x_2]}$ by decreasing functions $- f$ with $f \in \mathcal{L}^2$.    $\square$

Proof of Theorem 3.2

(i) Letting $R \uparrow D$ in $M_w^f(R) \preceq M_v^f(R)$, we get $E(f(X_1,X_2)) \le E(f(Y_1,Y_2)).$ (ii) For simplicity, we use small order notation $o(\cdot )$ with Landau’s symbol, i.e. $\phi (t) \in o(\psi (t)) \iff \limsup _{t \downarrow 0} \left| \frac{\varphi (t)}{\psi (t)} \right| =0$ for functions $\varphi , \psi : (0,\infty ) \mapsto {\mathbb {R}}$. Let $(a,b) \in D$ and take an increasing utility function $f(x_1,x_2)=(x_1-a)+(x_2-b)$. Let $R=[a,a+h] \times [b,b+k] \in \mathcal{R}(D)$ with $h>0$ and $k>0$, and let $R(t)=[a,a+th] \times [b,b+tk] \subset R$ for $t \in (0,1)$. From $M^f_w(R(t)) \preceq M^f_v(R(t))$, we have

$$\begin{aligned} \frac{\iint _{R(t)} f(x_1,x_2) w(x_1,x_2) \, dx_1 dx_2}{\iint _{R(t)} w(x_1,x_2) \, dx_1 dx_2} \le \frac{\iint _{R(t)} f(x_1,x_2) v(x_1,x_2) \, dx_1 dx_2}{\iint _{R(t)} v(x_1,x_2) \, dx_1 dx_2}. \end{aligned}$$

(A.2)

Hence, by Taylor’s theorem, there exists $\theta \in [0,t]$ such that $w(x_1,x_2) = w(a,b) + w_{x_1}(a+ \theta h, b+ \theta k) (x_1-a) + w_{x_2}(a+ \theta h, b+ \theta k) (x_2-b).$ Then regarding $M^f_w(R(t))$ we have

$$\begin{aligned}&{\iint _{R(t)} f(x_1,x_2) w(x_1,x_2) \, dx_1 dx_2 - \frac{(h+k)t}{2} \iint _{R(t)} w(x_1,x_2) \, dx_1 dx_2 } \\&= \frac{h^2k^2t^4}{4} (w_{x_1}(a,b) + w_{x_2}(a,b) ) + \frac{hkt^4}{3} (h^2w_{x_1}(a,b) +k^2w_{x_2}(a,b)) \\&\quad - \frac{(h+k)hkt^4}{4} (h w_{x_1}(a,b) +k w_{x_2}(a,b)) + o(t^4). \end{aligned}$$

We also have similar equation holds for $M^f_v(R(t))$. Applying the equations to (A.1) and letting $t \downarrow 0$, we get $\frac{h^2w_{x_1}(a,b) +k^2w_{x_2}(a,b)}{w(a,b)} \le \frac{h^2v_{x_1}(a,b) +k^2v_{x_2}(a,b)}{v(a,b)}$ for all $(a,b) \in D$ and $h>$ and $k>0$. Thus we obtain $\frac{w_{x_1}}{w} \le \frac{v_{x_1}}{v}$ and $\frac{w_{x_2}}{w} \le \frac{v_{x_2}}{v}$ on D. (iii) In case of $c(\cdot ,\cdot ) = 1$, $w(x_1,x_2) = w_1(x_1) w_2(x_2)$ and $v(x_1,x_2) = v_1(x_1) v_2(x_2)$ for $(x_1,x_2) \in D$. Therefore (b) and (c) are equivalent. In case of $c(\cdot ,\cdot ) = 1$, (a) and (b) are equivalent from (ii) and the proof in Yoshida  [11, Theorem 2.3].    $\square$

Proof of Theorem 3.3

(a) $\Longrightarrow$ (b): From Theorem 3.1 we have $X_1 \preceq _{\text{ FSD }}X_2$ and $Y_1 \preceq _{\text{ FSD }}Y_2$, and then $P(X_i \le x_i) \ge P(Y_i \le x_i)$ for $x_i \in {\mathbb {R}}$ and $i=1,2$. They imply ${\mathrm{VaR}}_p(X_i) \le {\mathrm{VaR}}_p(Y_i)$ for all $p \in (0,1]$ and $i=1,2$. Thus we get (b) by Lemma 2.2(i) since maps $p_1 \mapsto C(p_1,p_2)$ and $p_2 \mapsto C(p_1,p_2)$ are non-decreasing from (2.9). (b) $\Longrightarrow$ (a): From (b) and (2.6), for any $(x_1,x_2) \in {\mathbb {R}}^2$ there exists $(y_1,y_2) \in {\mathbb {R}}^2$ such that $(x_1,x_2) \preceq (y_1,y_2)$ and $P((X_1,X_2) \preceq (x_1,x_2) ) = p = P((Y_1,Y_2) \preceq (y_1,y_2) )$. Then $P((X_1,X_2) \preceq (x_1,x_2) ) = P((Y_1,Y_2) \preceq (y_1,y_2) )$ $\ge P((Y_1,Y_2) \preceq (x_1,x_2) )$ and we get (a). The equivalence of (b) and (c) is trivial from (2.10).    $\square$

Proof of Lemma 4.1

(i) is trivial from the definitions (2.1), (2.2), (3.1) and (3.2). (ii) can be checked easily from the definition (2.1). (iii) is also trivial from the definitions (3.1) and (3.2). (iv) From $X \preceq _{\text{ SSD }}Y$ we have

$$\begin{aligned} \int _{-\infty }^x P(X \le z) \, dz \ge \int _{-\infty }^x P(Y \le z) \, dz \end{aligned}$$

(A.3)

for any $x \in {\mathbb {R}}$. Let $p \in (0,1]$. Let $Z \in \mathcal{X}$ have a density function u. By the integration of parts, we have $\frac{1}{p} \int _{- \infty }^{{{\mathrm{VaR}}}_p(Z)} P(Z \le z) \, dz = \frac{1}{p} \times {\mathrm{VaR}}_p(Z) P(Z \le {\mathrm{VaR}}_p(Z)) - \frac{1}{p} \int _{- \infty }^{{{\mathrm{VaR}}}_p(Z)} z u(z) \, dz = {\mathrm{VaR}}_p(Z) - {\mathrm{AVaR}}_p(Z).$ Thus we get

$$\begin{aligned} {\mathrm{AVaR}}_p(Z) = {\mathrm{VaR}}_p(Z) - \frac{1}{p} \int _{- \infty }^{{{\mathrm{VaR}}}_p(Z)} P(Z \le z) \, dz \quad \text{ for } \text{ all } \ Z \in \mathcal{X}. \end{aligned}$$

(A.4)

Put a function $g(x) = x - \frac{1}{p} \int _{- \infty }^x P(Y \le z) \, dz$ for $x \in {\mathbb {R}}$. Then $g'(x) = 1 - \frac{ P(Y \le x)}{p} \gtreqless 0 \iff x \lesseqgtr {\mathrm{VaR}}_p(Y),$ and g(x) has a maximum at $x ={\mathrm{VaR}}_p(Y)$. Together with (A.3) and (A.4) we get ${\mathrm{AVaR}}_p(Y) = g({\mathrm{VaR}}_p(Y)) \ge g({\mathrm{VaR}}_p(X)) = {\mathrm{VaR}}_p(X) - \frac{1}{p} \int _{- \infty }^{{{\mathrm{VaR}}}_p(X)} P(Y \le z) \, dz \ge {\mathrm{VaR}}_p(X) - \frac{1}{p} \int _{- \infty }^{{{\mathrm{VaR}}}_p(X)} P(X \le z) \, dz = {\mathrm{AVaR}}_p(X).$ Therefore we have $X \preceq _{\text{ AVaR }}Y$.    $\square$

Proof of Theorem 4.1

(i) is trivial from the definitions (3.1) and (4.1). (ii) Let $f_1 :{\mathbb {R}}\mapsto {\mathbb {R}}$ be a $C^2$-class strictly increasing concave function. Approximating a function $(x_1,x_2) (\in {\mathbb {R}}^2) \mapsto f_1(x_1)$ by concave functions $f \in \mathcal{L}^2_c$, we get $E(f_1(X_1)) \le E(f_1(Y_1))$ for all concave functions $f_1:{\mathbb {R}}\mapsto {\mathbb {R}}$. In the same way, we also have $E(f_2(X_2)) \le E(f_2(Y_2))$ for all concave functions $f_2:{\mathbb {R}}\mapsto {\mathbb {R}}$. It is well-known that these inequalities are equivalent to $X_1 \preceq _{\text{ SSD }}Y_1$ and $X_2 \preceq _{\text{ SSD }}Y_2$ (Levi  [4]). (iii) From $X_1 \preceq _{\text{ SSD }}Y_1$ and $X_2 \preceq _{\text{ SSD }}Y_2$, we have $\int _{-\infty }^{x_1} P(X_1 \le z_1) \, dz_1 \ge \int _{-\infty }^{x_1} P((Y_1 \le z_1) \, dz_1 \ge 0$ and $\int _{-\infty }^{x_2} P(X_2 \le z_2) \, dz_2 \ge \int _{-\infty }^{x_2} P((Y_2 \le z_2) \, dz_2 \ge 0$ for $(x_1,x_2) \in {\mathbb {R}}^2$. Since $c(\cdot ,\cdot ) = 1$, we get (2.4) and $(X_1,X_2) \preceq _{\text{ SSD }}(Y_1,Y_2)$ holds.    $\square$

Proof of Theorem 4.2

(i)–(iii) are trivial from Lemma 2.2 and (2.13). (iv) By Lemma 4.1(iv), from $X_1 \preceq _{\text{ SSD }}Y_1$ and $X_2 \preceq _{\text{ SSD }}Y_2$ we have $X_1 \preceq _{\text{ AVaR }}Y_1$ and $X_2 \preceq _{\text{ AVaR }}Y_2$. Thus we get (iii) together with (2.13).    $\square$