数理解析研究所講究録 1215 巻 2001 年 158-168

158

On Law Invariant Coherent Risk Measures
Shigeo KUSUOKA
Graduate School of Mathematical Sciences
The University of Tokyo

1

Introduction

The idea of coherent risk measures has been introduced by Artzner, Delbaen, Eber andHeath [1]. We think of special class of coherent risk measures and give acharacterization. Following [1],of it. Let
be aprobabilty space. We denoteby
we give the following definition.
$(\Omega,\mathcal{F},P)$$L^{\infty}$$L^{\infty}(\Omega,\mathcal{F}, P)$

Definition 1We say that a map$\rho:L^{\infty}arrow \mathrm{R}$are satisfied.

is a coherent risk measure(1) If $X\geq 0$ , then $\rho(X)\leq 0$ .
(2) SubaddUwity : $\rho(X_{1}+X_{2})\leq\rho(X_{1})+\rho(X_{2})$ .
(3) Positive homogeneity:for
we have(4) For every constant we have $\rho(X+c)=\rho(X)-c$ .
$\lambda>0$$\rho(\lambda X)=\lambda\rho(X)$$\dot{\iota}f$

the folloing.

$c$Then Delbaen [2] proved the following.
Theorem 2Let be a coherent risk measure. Then the following conditions are equivalent.
(1) There is (closed convex )set of probability measures
such that any $Q\in Q$ isabsolutely continuous with respect to and for$\rho$$Q$$a$$X\in L^{\infty}$$P$$\rho(X)=\sup\{E^{Q}[-X];Q\in Q\}$ .

(2)$\rho$

satisfies the Fatou property,

converging to$X$in probability, then$i.e.$, if$\{X_{n}\}_{n=1}^{\infty}\subset L^{\infty}$$\rho(X)\leq\lim_{1arrow}.\inf_{\infty}\rho(X_{n})$(3) If$X_{n}$

are unifo rmly bounded and.

is a uniformly bounded sequence that decreases to $X$ , then$\rho(X_{*}.)$

tends to $\rho(X)$ .

Now we introduce the following notion.

say that a map
Definition 3
X, Yhave the same probability laett.
$We^{\mathfrak{l}}$$\rho:L^{\infty}arrow \mathrm{R}$is law invariant, if$\rho(X)=\rho(\mathrm{Y})$whenever$\in L^{\infty}$

Our purpose is to characterize law invariant coherent risk measures with the Fatou
property.
Let I) be the set of probability distribution functions of bounded random variables,such that there arei.e., I) is the set of non-decreasing right-continuous functions $F$ on. Let us define $Z$ : $[0, 1)\cross D$,
for which $F(z)=0$ , $<z_{0}$ and $F(z)=1$ ,by$x\in[0,1)$ , $F\in D$ .
$Z(x, F)= \inf\{z;F(z)>x\}$ ,$\mathrm{R}$$z_{0}$$\mathrm{R}$$z_{1}\in \mathrm{R}$$z$$z\geq z_{1}$$arrow$

159
Then X(., F) [0,1)q R is non-decreasing and right continuous. We denote by F. theprobability distribution function of arandom variable X.
For each a E (0,1], let p.
be given by$\ovalbox{\tt\small REJECT}$$\ovalbox{\tt\small REJECT} L"-+\mathrm{R}$,$\rho_{\alpha}(X)=\alpha^{-1}\int_{1-\alpha}^{1}Z(x, F_{-X})dx$

Also, we define$\rho_{0}$:$L^{\infty}arrow \mathrm{R}$$X\in L^{\infty}$.

by$\rho_{0}(X)=ess.\sup(-X)$$X\in L^{\infty}$.

Then it is easy to see that $\rho.(X)$ :is anon-increasing continuous function forany.
Weshow later that , $\alpha\in[0,1]$ , is alaw invariant coherent risk measure with
the Fatou property. Actuallyis the same asin [1].
From now on, we assume the following.
(Assumption)is astandard probability space and $P$ is non-atomic.
Our main results are the following.
$[0, 1]arrow \mathrm{R}$$X\in L^{\infty}$$\mathrm{w}\mathrm{i}\mathrm{U}$$\mathrm{p}\mathrm{a}$$WCM_{\alpha}$$\rho_{\alpha}$$(\Omega,\mathcal{F}, P)$

Theorem 4Let
. Then the folloing conditions are equivalent.
(1) There is (compact convex) set
of probability measures on $[0, 1]$ such that$\rho:L^{\infty}arrow \mathrm{R}$$a$$\mathcal{M}_{0}$$\rho(X)=\sup\{\int_{0}^{1}\rho_{\alpha}(X)m(d\alpha);m\in \mathcal{M}_{0}\}$(2)$\rho$,$X\in L^{\infty}$.

is a law invariant coherent risk measure with the Fatou property.

Theorem 5If$m_{1}$

andare probability measures on $[0, 1]$ , and if$m_{2}$$\int_{0}^{1}\rho_{\alpha}(X)m_{1}(d\alpha)=\int_{0}^{1}\rho_{\alpha}(X)m_{2}(d\alpha)$

for all,$X\in L^{\infty}$,

then $m_{1}=m_{2}$ .
Definition 6(1) We say that a pair $X$ and$\mathrm{Y}$of random variables is comonotone, if$P(\mathrm{A}_{J})\otimes P(M’)-a.s$$(X(\omega)-X(\omega’))(\mathrm{Y}(\{v)-\mathrm{Y}(\omega’))\geq 0$(2) We sa that a map$y$$\rho:L^{\infty}arrow \mathrm{R}$is comonotone,.

if$\rho(X+\mathrm{Y})=\rho(X)+\rho(\mathrm{Y})$for any comonotone pair$X$,$\mathrm{Y}\in L^{\infty}$.

Theorem 7Let :. Then the following conditions are equivalent.
(1) There is a probability measure $m$ on $[0, 1]$ such that for$L^{\infty}arrow \mathrm{R}$$\rho$$X\in L^{\infty}$$\rho(X)=\int_{0}^{1}\rho_{\alpha}(X)m(d\alpha)$(2)$\rho$,$X\in L^{\infty}$.

is a larn invariant and comonotone coherent risk measure with the Fatou property.

160
Definition 8We define$VaR_{\alpha}$:$L^{\infty}arrow \mathrm{R}$,$\alpha\in(0,$1), by$VaR_{\alpha}(X)= \sup\{z\in \mathrm{R};F_{-X}(z)<1-\alpha\}$Theorem 9Let $\alpha\in(0,1)$ . If$\rho$.

is law invariant coherent risk measure such that$\rho(X)\geq VaR_{\alpha}(X)$then we have$\rho(X)\geq\rho_{\alpha}(X)$,,$X\in L^{\infty}$$X\in L^{\infty}$,.

The author thanks Prof. Delbaen for useful discussions. In particular, Theorems 7
and 9are suggested by him.

2Key Lemmais standard probabilty space and be non-atomic, we may
Since we assume thatis aLebesgue space, i.e., $\Omega=[0,1)$ ,assume that our basic probabilty spaceis the Borel algebra over $[0, 1)$ , and is the Lebesgue measure on $[0, 1)$ . Therefore weassume so throughout this paper.
Let be the set of non-decreasing right-continuous probabilty density functions on$[0, 1)$ . In this section, we will prove the following.
$P$$(\Omega,\mathcal{F})$$\mathcal{F}$$(\Omega,\mathcal{F}, P)$

$P$

$\mu$

$\mathcal{G}$Lemma 10 Let(1) There is a subset$\rho:L^{\infty}arrow \mathrm{R}$$\mathcal{G}_{0}$. Then the following conditions are equivalentof such that$\mathcal{G}$$\mathrm{p}(\mathrm{X})=\sup\{\int_{0}^{1}Z(x,F_{-X})g(x)dx;g\in \mathcal{G}_{0}\}$(2)$\rho$,

X$\in L^{\infty}$.

is a law invariant coherent risk measure with the Fatou property.

absolutely continuous withdenote the set of probability measures on,denotes the Radon-Nykodim density $dQ/dP$. Let
respect to $P$. For any $Q\in P$ ,be the, . Let$n\geq 1$ , be a
algebra of generated by $1_{[2^{-n}(k-1),2^{-n}k)},k=1$ ,
$X$
$X$
.
some
for
set of all bounded random variables such that is -measurable
Then we have the following.

Let

$P$

$(\Omega,\mathcal{F})$$\mathcal{F}_{n}$$\mathrm{Y}_{Q}$$\mathrm{s}\mathrm{u}\mathrm{b}-\sigma$$2^{n}$$\mathcal{F}$$n$$\mathcal{F}_{n}$Lemma 11 Let $Q\in P$ and$\mathcal{X}$$\ldots$$X\in \mathcal{X}$. Then we have$\int_{0}^{1}Z(x,F_{X})Z(x, F_{\mathrm{Y}_{Q}})dx=\sup\{E^{Q}[\tilde{X}];\tilde{X}\in \mathcal{X}, F_{\overline{X}}=F_{X}\}$$= \sup\{E^{\tilde{Q}}[X];\tilde{Q}\in P, F_{\mathrm{Y}_{Q}}=F_{\mathrm{Y}_{Q}}\}$

We make some preparations before proving Lemma 11.
We easily see the following.

.

161

.

.

Proposition 12 Let
kn\rangle be a sequence of numbers, and let$ ,n, be a sequence of non-negative numbers. If $1,2,{l2y\rangle
andS$12>\rangle$$\ovalbox{\tt\small REJECT}$$x_{k_{\rangle}}$$\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}$\rangle\ldots ,$\{\ovalbox{\tt\small REJECT} \mathrm{j}\mathrm{b}\mathrm{j}2\mathrm{t}^{\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}}\rangle$$\ovalbox{\tt\small REJECT}_{n}’ \mathit{1}^{\ovalbox{\tt\small REJECT}}\ovalbox{\tt\small REJECT}$$\{\mathrm{j}_{1\mathrm{t}\ovalbox{\tt\small REJECT}}\ovalbox{\tt\small REJECT}_{2\mathrm{t}^{\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}}}>:\ovalbox{\tt\small REJECT}_{\mathrm{n}}’\}\ovalbox{\tt\small REJECT}$$y_{in^{\rangle}}$$\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}$$\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}$$\sum_{k=1}^{n}x_{k}y_{k}\leq\sum_{k=1}^{n}x:_{k}y_{j_{k}}$$\ovalbox{\tt\small REJECT}$$\ovalbox{\tt\small REJECT}$$\mathrm{x}_{\ovalbox{\tt\small REJECT}_{2}}\ovalbox{\tt\small REJECT}$\cdots\ranglek$y_{h_{\rangle}}$$\ovalbox{\tt\small REJECT}$$\ovalbox{\tt\small REJECT}$$y_{\ovalbox{\tt\small REJECT}_{\mathit{2}}}\ovalbox{\tt\small REJECT}$$y_{\ovalbox{\tt\small REJECT}}$$\ovalbox{\tt\small REJECT}$$n’\}_{>}jpj_{p}^{l}en$.

Also we have the following (see Willams [3] Chapters 3and 17).
, the probability distribution function of the law of
Proposition 13 (1) For any$F$$Z(x, F)$ under $\mu(dx)$ is .
(2) If $F_{n}\in D$ converges to $F$ weakly, then $Z(x, F_{n})$ converges to $Z(x, F)$ for -a.s.x.
$F\in \mathrm{V}$$\mu$. Then $X$ is -measurable for some $n\geq 1$ .
Now let us prove Lemma 11. Let, $m\geq n$ . Then for any $m\geq n$ , we haveLet
$X\in \mathcal{X}$$\mathcal{F}_{n}$$\mathrm{Y}_{m}=E[\mathrm{Y}_{Q}|\mathcal{F}_{m}]$$X( \omega)=\sum_{k=1}^{2^{m}}x_{m,k}1_{[(k-1)2^{-m},k2^{-m})}(\omega)$,$\mathrm{Y}_{m}(\omega)=\sum_{k=1}^{2^{m}}y_{m,k}1_{[(k-1)2^{-m},k/2^{-m})}(\omega)$

where $x_{m,k}=2^{m}E^{P}[X,$ $[(k-1)2^{-m}, k2^{-m})]$ and$k=1$ , 2,
, . Letandbe apermutation on 1, 2,$y_{m,k}=2^{m}E^{P}[\mathrm{Y}_{Q},$$2^{m}$$\ldots$$\sigma_{m}$$\{$$\tau_{m}$$x_{m,\sigma_{m}(1)}\leq x_{m,\sigma_{m}(2)}\leq\ldots\leq x_{m,\sigma_{m}(2^{n})}$and$\ldots$,$2^{m}\}$$P-a.s.$ ,,$[(k-1)2^{-m}, k2^{-m})]$ ,such that$y_{m,\tau_{m}(1)}\leq y_{m,\tau_{m}(2)}\leq\ldots\leq ym,\tau_{m}(2^{n})$.

Then one can easily obtain that
$Z(x, F_{X})= \sum_{k=1}^{2^{m}}x_{\sigma_{m}(k)}1_{[(k-1)2^{-m},k2^{-m})}(x)$,$Z(x, F_{\gamma_{m}})= \sum_{k=1}^{2^{m}}y\tau_{m}(k)1[(k-1)2^{-m},k2^{-m})(x)$,and so$E^{Q}[X]=E[X \mathrm{Y}_{Q}]=E[X\mathrm{Y}_{m}]=2^{-m}\sum_{k=1}^{2^{m}}x_{m,k}y_{m,k}\leq 2^{-m}\sum_{k=1}^{2^{m}}x_{m,\sigma_{m}(k)}y_{m.,\tau_{m}}(k)$$= \int_{0}^{1}Z(x, F_{X})Z(x, FYq)dx$

canconverges to P-a.s., we see by Proposition 13 thatverges toare uniformly integrable,
for a.s.x. Since
are also uniformly integrable by Proposition 13 (1). Therefore letting $marrow\infty$ , we haveSince$Z(x,F_{\mathrm{Y}_{Q}})$$Z(x, F_{\mathrm{Y}_{m}})$$\mathrm{Y}$$\mathrm{Y}_{m}=E[\mathrm{Y}_{Q}|\mathcal{F}_{m}]|$$\mu$$\{Z(x,F_{\mathrm{Y}_{m}})\}_{m=n}^{\infty}$$\{\mathrm{Y}_{m}\}_{m=n}^{\infty}$(1)$E^{Q}[X] \leq\int_{0}^{1}Z(x, F_{X})Z(x, F_{\mathrm{Y}q})dx$

for any X$\in \mathcal{X}$. Let$\tilde{X}_{m}(\omega)=\sum_{k=1}^{2^{m}}x_{m,\sigma_{m}(\tau_{m}^{-1}(k))}1_{[(k-1)2^{-m},k2^{-m})}(\omega)$.

162
Then one can easily see that the probabilty distributions of X and X. under P are thesame andE . Also, we have$X_{m}$$\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}$$E^{Q}[ \tilde{X}_{m}]=2^{-m}\sum_{k=1}^{2^{m}}x_{m,\sigma_{m}(\tau_{\overline{m}^{1}}(k))}y_{m,k}$$= \int_{0}^{1}Z(x,F_{X})Z(x,F_{\mathrm{Y}_{m}})dx$.

So letting $marrow\infty$ , we have$\sup${$E^{Q}[\tilde{X}];\tilde{X}\in \mathcal{X}$,Fk $=F_{X}$ } $\geq\int_{0}^{1}Z(x,F_{X})Z(x,F_{\mathrm{Y}_{Q}})dx$ .

(2)

Let

.

$\tilde{\mathrm{Y}}_{m}(\omega)=\sum_{k=1}^{2^{m}}1_{[((k-1)2^{-m},k2^{-m})}(\omega)\mathrm{Y}_{Q}(\omega-k2^{-m}+\tau_{m}(\sigma_{m}^{-1}(k))2^{-m})$

Then one can easily see that the probability distributions of.
$P$. Thensame. Let
we have$\mathrm{Y}_{Q}$and$\tilde{\mathrm{Y}}_{m}$

under

$P$

are the$\tilde{Q}=\tilde{\mathrm{Y}}_{m}$$E^{\tilde{Q}}[X]=2^{-m} \sum_{k=1}^{2^{m}}x_{m,k}y_{m,\tau_{m}(\sigma_{\overline{m}^{1}}(k))}=\int_{0}^{1}Z(x,F_{X})Z(x,F_{\mathrm{Y}_{m}})dx$.

So letting $marrow\infty$ , we have$\sup\{E^{\tilde{Q}}[X];\tilde{Q}\in P, F_{\mathrm{Y}_{Q}}=F_{\mathrm{Y}_{Q}}\}\geq\int_{0}^{1}Z(x, F_{X})Z(x, F_{\mathrm{Y}_{Q}})dx$.

(3)

We have Lemma 11 from Equations (1), (2) and (3).
This completes the proof of Lemma 11.
Proposition 14 Let $Q\in P$ . Then for any$X\in L^{\infty}$, we have$\int_{0}^{1}Z(x,F_{X})Z(x, F_{\mathrm{Y}_{Q}})dx=\sup\{E^{\tilde{Q}}[X];\tilde{Q}\in P, F_{\mathrm{Y}_{\Phi}}=F_{\mathrm{Y}_{Q}}\}$

Proof. Let$\tilde{\mathrm{Y}}$$X_{n}=E[X|\mathcal{F}_{n}]$.

be arandom variable whose distribution is the same as that of,. Then for any $m\geq 1$ , we have$n\geq_{\iota}1$$E[|X-X_{n}|\tilde{\mathrm{Y}}]\leq E[|X-X_{n}|(\tilde{\mathrm{Y}}\wedge m)]+||X-X_{n}||_{\infty}E[\tilde{\mathrm{Y}},\tilde{\mathrm{Y}}>m]$$\leq mE[|X-X_{n}|]+2||X||_{\infty}E[\tilde{\mathrm{Y}},\tilde{\mathrm{Y}}>m]$.

So we have$\sup\{E^{\tilde{Q}}[|X-X_{n}|];\tilde{Q}\in P, F_{\mathrm{Y}_{\circ}}=F_{\mathrm{Y}_{Q}}\}arrow 0$,$narrow\infty$.

By Proposition 13, we have$\mathit{1}^{1}Z(x, F_{X_{n}})Z(x, F_{\mathrm{Y}_{Q}})dxarrow\int_{0}^{1}Z(x, F_{X})Z(x, F_{\mathrm{Y}_{Q}})dx$,$narrow\infty$.

$\mathrm{Y}_{Q}$

. Let

163
Therefore we have our assertion from Lemma 11. This completes the proof.
Now let us prove Lemma 10.
be given by
be asubset of , andLet
Proof of Lemma 10. (1)$\Rightarrow(2)$$\rho:L^{\infty}arrow \mathrm{R}$$\mathcal{G}$$\mathcal{G}_{0}$,$X\in L^{\infty}$$\mathrm{p}(\mathrm{X})=\sup\{\int_{0}^{1}Z(x,F_{-X})g(x)dx;g\in \mathcal{G}_{0}\}$.

Then it is obvious that is law invariant. So it is sufficient to prove that is acoherent
be the set of $Q\in P$ such thatrisk measure with the Fatou property. Let. Then by Proposition 14, we have$\rho$$\rho$$Z(\cdot,\mathrm{Y}Q)$$Q_{0}$$\in \mathcal{G}_{0}$$\rho(X)=\sup\{E^{Q}[-X];Q\in Q_{0}\}$ ,$X\in L^{\infty}$.

So by Theorem 2, we see that is acoherent risk measure with the Fatou property. Thisimplies our assertion.
Let be alaw invariant coherent risk measure with the Fatou property. Let(2). Then by Theorem 2
be the set of $Q\in P$ such that $E^{Q}[-X]\leq\rho(X)$ for all
we have$\rho$$\Rightarrow(1)$$\rho$$X\in L^{\infty}$$P_{0}$$\rho(X)=\sup\{E^{Q}[-X];Q\in P_{0}\}$ ,$X\in L^{\infty}$.

, and fix them for awhile. Let $\tilde{X}(\omega)=Z(\omega;Fx)$ ,Take a $Q\in P_{0}$ and, $n\geq 1$ , be random variables defined by$=[0,1)$ . Then we have $\rho(\tilde{X})=\rho(X)$ . Let$\omega\in\Omega$$X\in L^{\infty}$$U_{n}$$\tilde{X}(\omega+2^{-n})$$U_{n}=\{$$||X||_{\infty}$

Then we see that
$P-a.s$ . and that$\omega\in[0,1-2^{-n})$,$\omega\in[1-2^{-n}, 1)$. Then we see that $V_{n}\leq U_{n}$ ,, $P-a.s$ . Let, $P-a.s$ . So by Theorem 2we have$V_{n}=E^{P}[\overline{X}|\mathcal{F}_{n}]$$U_{n}\downarrow\tilde{X}$$V_{n}arrow\tilde{X}$,,$\lim_{narrow}\inf_{\infty}\rho(V_{n})\leq\lim_{narrow\infty}\rho(U_{n})=\rho(\tilde{X})$.

On the other hand, by Lemma 11 and Proposition 14 we have
$E^{Q}[-X]$$\leq$$\int_{0}^{1}Z(x, F_{-\overline{X}})Z(x, F_{\mathrm{Y}_{Q}})dx$$= \lim_{narrow\infty}\int_{0}^{1}Z(x, F_{-V_{n}})Z(x, F_{\mathrm{Y}_{Q}})dx$$= \lim_{narrow\infty}\sup\{E^{Q}[-\tilde{V}];\tilde{V}\in \mathcal{X}, F_{\overline{V}}=F_{V_{n}}\}$$\leq$

Thus letting$\lim_{narrow}\inf_{\infty}\rho(V_{n})\leq\rho(X)$$\mathcal{G}_{0}=\{Z(\cdot, F_{\mathrm{Y}_{\mathrm{Q}}});Q\in P_{0}\}$.

, we see that$\rho(X)=\sup\{\int_{0}^{1}Z(x, F_{-X})g(x)dx;g\in \mathcal{G}_{0}\}$.

This implies our assertion. This completes the proof of Lemma 10.

164
3Proof of Theorem 4
be given by:
In this section, we prove Theorem 4. Let $g\in(i$ , and let$t\geq1$$\tilde{g}(t)=g(1$$\tilde{g}(t)=0,t<0,\tilde{g}(t)=g(t),t\in[0,1)$ , and. Then we have for any,$\mathrm{R}arrow \mathrm{R}$$\tilde{g}$- $)$$X\in L^{\infty}$$\int_{0}^{1}Z(x, F_{-X})g(x)dx=\int_{[0,1)}(\int_{\mathrm{g}}^{1}Z(y;F_{-X})dy)d\tilde{g}(x)$$= \int_{[0,1)}\rho_{1-x}(X)(1-x)d\tilde{g}(x)$.

Letting $X=-1$ , we have
$1= \int_{0}^{1}g(x)dx=\int_{[0,1)}(1-x)d\tilde{g}(x)$.

From this observation and Lemma 10, we have the following.
Proposition 15 Let(1) There is a set. Then the folloing conditions are equivalent.
of probability measures on (0, 1] such that for X$\rho:L^{\infty}arrow \mathrm{R}$$\mathcal{M}_{0}$$\in L^{\infty}$$\rho(X)=\sup\{\int_{(0,1]}\rho_{\alpha}(X)m(d\alpha);m\in \mathcal{M}_{0}\}$(2)$\rho$.

is a law invariant coherent risk measure with the Fatou property.

Now we prove Theorem 4. For each probability measure $m$ on $[0, 1]$ , letbe aprobability measure on $(0, 1]$ given by$\nu_{||}(m)(A)=m(A\cap(0,1])+m(\{0\})\delta_{1/n}(A)$Then we see that for any$\nu_{n}(m)$, $n\geq 1$ ,for aBorel set in $[0, 1]$ .

,$X\in L^{\infty}$$\int_{[0,1]}\rho_{\alpha}(X)m(d\alpha)=\sup_{r\iota}\int_{(0,1]}.\rho_{\alpha}(X)\nu_{n}(m)(d\alpha)$.

This and Proposition 15 imply Theorem 4. This completes the proof of Theorem 4.

4Proof of Theorem 5
We give some computation on
Proposition 16 Let(1) We have$c\in$$(0, 1]$$\rho_{\alpha}$in this section.

and$X_{\mathrm{c}}(\omega)=1[1-\mathrm{C},1)(\omega)$$\rho_{\alpha}(-X_{\mathrm{c}})=1\wedge\frac{c}{\alpha}$, $\omega\in\Omega=[0,1)$ .

$\alpha\in(0,1]$(2) Let $m$ be a probability measure on $[0, 1]$ and let $f(s)= \int_{[0,1]}\rho_{\alpha}(-X_{s})m(d\alpha)$ , $s\in(0,1]$
Then $f(c)$ is differentiate at $s=c\in(0,1)$ such that $m(\{c\})=0$ , and$\frac{df}{ds}(c)=\int_{(\mathrm{c},1]}\frac{1}{\alpha}m(d\alpha)$.

165
Proof. Noting that

Then we have for$\mathrm{f}\mathrm{f}_{7}.(\mathrm{z})\ovalbox{\tt\small REJECT}$$0<\mathit{8}$$1_{(1-\cdot,1)}(\mathrm{z})$, xE [0,1), we easily have the assertion (1).

$<t<l$.

$\frac{f(t)-f(s)}{t-s}=\int_{(l,1]}\frac{1}{\alpha}m(d\alpha)+\frac{1}{t-s}\int_{(e,t]}\frac{\alpha-s}{\alpha}m(d\alpha)$This proves the assertion (2).
Theorem 5is an easy consequence of Proposition 16 (2).

5Supporting measures and Proof of Theorem 9
is acompact metricLetdenote the set of all probabilty measures on $[0, 1]$ . Then
space with the Prohorov metric. Let be alaw invariant coherent risk measure with the
Fatou property. Let$\mathcal{M}$$\mathcal{M}$$\rho$$\mathcal{M}(\rho)=${$m \in \mathcal{M};\int_{[0,1]}\rho_{\alpha}(X)m(d\alpha)\leq\rho(X)$is continuous in $\alpha\in[0,1]$ ,Since
Theorem 4we have$\rho_{\alpha}(X)$

for

au

,

$X\in L^{\infty}$}.

is aclosed convex subset of U. Then from$\mathcal{M}(\rho)$$X\in L^{\infty}$$\rho(X)=\sup\{\int_{|0,1]}\rho_{\alpha}(X)m(d\alpha);m\in \mathcal{M}(\rho)\}$For each$X\in L^{\infty}$.

let$\tilde{\mathcal{M}}(X;\rho)=\{m\in \mathcal{M}(\rho);\int_{[0,1]}\rho_{\alpha}(X)m(d\alpha)=\rho(X)\}$.

. It is obvious that
From the compactness of
we see that.
depends only on the distributionof $X$, and so we denote it by
Now we prove Theorem 9. Let be alaw invariant coherent risk measure such that$\tilde{\mathcal{M}}(X;\rho)$$\tilde{\mathcal{M}}(X;\rho)\neq\emptyset$$\mathcal{M}(\rho)$$\mathcal{M}(F_{X};\rho)$$F_{X}$$\rho$$\rho(X)\geq \mathrm{V}\mathrm{a}\mathrm{R}_{\alpha}(X)$, $\in\Omega=[0,1)$ ,Let
by Proposition 16 we see that
$X_{\epsilon}(\omega)=1_{[1-\alpha-\epsilon,1)}(\omega)$$\omega$$\epsilon$,$X\in L^{\infty}$$\in(0,1-\alpha)$.

, and let$\rho(-X_{e})=\int_{[0,1]}(1\Lambda\frac{\alpha+\epsilon}{s})m_{e}(ds)$On the other hand, we haveis compact, we see that there is an,
we see that.
This completes the proof of Theorem 9.

$\mathrm{V}\mathrm{a}\mathrm{R}_{\alpha}(-X_{\epsilon})=1$$\rho_{\alpha}(X)\leq\rho(X)$$X\in L^{\infty}$. Then.

. So we see that$m\in \mathcal{M}(\rho)$$\mathcal{M}(\rho)$$m_{\epsilon}\in\tilde{\mathcal{M}}(X_{e};\rho)$. Since. Therefore$m_{\epsilon}([0, \alpha+\epsilon])=1$

such that$m([0, \alpha])=1$

166
6Proof of Theorem 7
Proposition 17 Let$X,\mathrm{Y}$$a\}\subset\{\mathrm{Y}\geq b\}P-a.s$

be comonotone random variables and ,$a$$b\in \mathrm{R}$. Then $\{X\geq$. or $\{\mathrm{Y}\geq b\}\subset\{X\geq a\}P-a.s$ .

Proof. Let $C=\{X\geq a\}\cap\{\mathrm{Y}<b\}$ and $D=\{X<a\}\cap\{\mathrm{Y}\geq b\}$ . Then for$\in C\cross D$,$(X(\omega)-X(\omega’))(\mathrm{Y}(\omega)-\mathrm{Y}(\omega’))<0$$(\omega,\omega’)$.

This implies that $P(C)=0$ or $P(D)=0$ . So we have our assertion.
As an immediate consequence, we have the following.
Corollary 18 Let $X$,$\mathrm{Y}$be comonotone random variables. Then we have$P(X+\mathrm{Y}\geq a+b)\geq P(X\geq a)\Lambda P(\mathrm{Y}\geq b)$

Proposition 19 Let $X$ ,1$\mathrm{Y}\in L^{\infty}$

be comonotone and ,$a$$Z(x,F_{X+\mathrm{Y}})=Z(x,F_{X})+Z(x,F_{\mathrm{Y}})$,$a$$b\in \mathrm{R}$,,$b\in \mathrm{R}$.

. Then$x\in[0,1)$

Proof. By the definition of $Z(x, F_{X})$ we have $F_{X}(Z(x, F_{X})-)\leq x$ . So we have $P(X\geq$$Z(x, F_{X}))\geq 1-x$

we have. Similarly we have$P(\mathrm{Y}\geq Z(x,F_{\mathrm{Y}}))\geq 1-x$

$P(X+\mathrm{Y}\geq Z(x, F_{X})+Z(x,F_{\mathrm{Y}}))\geq 1-x$

. Therefore by Corollary 18,$x\in[0,1)$.

Note that Let $Z(x, F_{X+\mathrm{Y}})= \sup\{z\in \mathrm{R};F_{X+\mathrm{Y}}(z)\leq x\}$, $x\in(0,1)$ . So we see that
, $x\in(0,1)$ . On the other hand, we have$Z(x, F_{X})+Z(x, F_{\mathrm{Y}})\leq Z(x, F_{X+\mathrm{Y}})$$\int_{[0,1)}(Z(x, F_{X})+Z(x, F_{\mathrm{Y}}))\mu(dx)=E[X]+E[\mathrm{Y}]$ $= \int_{[0,1)}Z(x, F_{X+\mathrm{Y}}X)\mu(dx)$

So we see that
$Z(x,F_{X})+Z(x, F_{\mathrm{Y}})=Z(x,F_{X+\mathrm{Y}})$.

p-a.e.x.

,

Since both sides are right continuous, we have our assertion.
Proposition 20

Proof. For each$\mathrm{p}\mathrm{a}$, $\alpha\in[0,1]$ , are comonotone.

$\alpha\in(0,1]$, we see thatis comonotone from the definition of, we see that is also comonotone.
$\rho_{\alpha}$$\rho_{\alpha}$

andProposition 19. Letting$\alpha\downarrow 0$

Proposition 21 $Lei$
Fatou properry. Thenbe a comonotone law invariant coherent risk measure with thefor any $n\geq 1$ and , , , $F_{n}\in D$ .

Proof. Let$\rho$$\rho_{0}$$F_{1}$$\bigcap_{=1}^{n}.\cdot \mathcal{M}(F.\cdot;\rho)\neq\emptyset$$X\dot{.}(\omega)=Z(\omega, F_{})$

comonotone for each $k=1$ ,,$\omega$$\ldots$$\in\Omega=[0,1)$, $n-1$ . Let, $i=1$ ,$\ldots$$X= \sum_{=1}^{n}.\cdot$$\rho(X)=\dot{.}\sum_{=1}^{n}\rho(X_{\dot{1}})$.

$F_{2}$, . Then$n$$\ldots$$\sum_{=1}^{k}.\cdot X_{}$

X. . Then we have$\cdot$1

and $X_{k+1}$ are167
Let mC$\ovalbox{\tt\small REJECT} \mathrm{A}4(X;_{7}0)$. Then we have$. \cdot\sum_{=1}^{n}\int_{[0,1]}\rho_{\alpha}(X_{})m(d\alpha)=\rho(X)=\sum_{=1}^{n}\rho(X_{})$.

Also, we have$\int_{[0,1]}\rho_{\alpha}(X_{})m(d\alpha)\leq\rho(X.\cdot)$,$i=1$ , $\ldots,n$ .

$\int_{[0,1]}\rho_{\alpha}(X_{i})m(d\alpha)=\rho(X.\cdot)$,$i=1$ , $\ldots,n$ .

So we have

, . So we have our assertion.
, $i=1$ ,This implies
Now let us prove Theorem 7. Suppose that
and$m\in\tilde{\mathcal{M}}(X_{};\rho)$$\ldots$$n$$m\in \mathcal{M}$$\rho(X)=\int_{[0,1]}\rho_{\alpha}(X)m(d\alpha)$,$\rho:L^{\infty}arrow \mathrm{R}$$X\in L^{\infty}$

is given by.

Then by Theorem 4and Proposition 20, we see that is comonotone and law invariant.
On the other hand, suppose that is comonotone law invariant coherent risk measure
is compact, wewith the Fatou property. Then by Proposition 21 and the fact that$\cap\{\mathcal{M}(F;\rho);F\inD\}\neq\emptyset$
$m$
. Let
be an element of this set. Then we see that
see that$\rho$$\rho$$\mathcal{M}$$\rho(X)=\int_{[0,1]}\rho_{\alpha}(X)m(d\alpha)$,$X\in L^{\infty}$.

This completes the proof of Theorem 7.

7ARemark
For each $\alpha\in(0,1]$ let$\varphi_{\alpha}$: $[0, 1]arrow[0,1]$ be given by$\varphi_{\alpha}(t)=\frac{t}{\alpha}\Lambda 1$,$t\in[0,1]$ .

Then we have the following.
Proposition 22 For any $Q( \in(0,1]$ andfollowing.

$X\in L^{\infty}$

satisfying $X\leq 0$ , $P-a.s.$ , we have the$\rho_{\alpha}(X)=\int_{0}^{\infty}\varphi_{\alpha}(P(-X>y))dy$.

such that $X\leq 0$ and $X$ has acontinuous strictly
andincreasing distribution on (ess.inf $X$ , $ess. \sup X$ ). Then we see that $Z(x, F_{-X})=F_{-X}^{-1}(x)$ ,$x\in(0,1)$ . Letbe such that $F_{-X}(q_{\alpha})=1-\alpha$ . Then we have

Proof. Let$\alpha\in(0,1)$$X\in L^{\infty}$$q_{\alpha}\in(0, \infty)$$\rho_{\alpha}(X)=-\frac{1}{\alpha}q_{\alpha}\infty yd(1-F_{-X}(y))$$=- \frac{1}{\alpha}[y(1-F_{-X}(y))]_{q_{\alpha}}^{\infty}+\frac{1}{\alpha}9\alpha\infty(1-F_{-X}(y))dy$168$= \int_{0}^{\infty}\varphi_{\alpha}(P(-X>y))dy$.

Since any nonpositive random variables is approximated by such random variables in, we also have our assertion
probabilty, we have our assertion for $\alpha\in(0,1)$ . Letting
for a $=1$ . This completes the proof.
1:$t\in[0,1]$. Then we see that, and let,Let$[0, 1]arrow[0,1]$ is acontinuous increasing concave function with $\varphi(0)=0$ , and $\varphi(1)=$$1-m(\{0\})$ . We also see that$\alpha\uparrow 1$$m\in \mathcal{M}$$\varphi(\cdot,m)$$\varphi(t;m)=\int_{(0,1]}\varphi_{\alpha}(t)m(d\alpha)$$\frac{d}{dt}\varphi(t;m)=\int_{t}^{1}\frac{1}{\alpha}m(d\alpha)$

for any continuous point $t\in(0,1)$ of the measure $m$ . So
For any nonpositive
we have,$\varphi(\cdot, m)$determines $m$ .

$X\in L^{\infty}$$\int_{0}^{\infty}\rho_{\alpha}(X)m(\ )$$=m( \{0\})ess.\sup(-X)+\int_{0}^{\infty}\varphi(P(-X>y);m)dy$.

These observations imply the following.
Theorem 23 Let. Then the following are equivalent(1) is a law invariant and comonotone coherent risk measure with the Fatou property.
(2) There is a continuous nondecreasing concave function $\varphi:[0,1]arrow[0,1]$ such that$\rho:L^{\infty}arrow \mathrm{R}$$\rho$$\rho(X)=(1-\varphi(1))ess.\sup(-X)+\int_{0}^{\infty}\varphi(P(-X>y))dy$

for any nonpositive$X\in L^{\infty}$.

References[1] Artzner, Ph., F. Delbaen, J.-M. Eber, and D. Heath, Coherent Measures of Risk,
Math. Finance $9(1999)$ , 203228.

[2] Delbaen, F., Coherent Risk Measures on General Probability Spaces, Preprint 1999.
[3] Wilh.ms, D., Probability with Martingales Cambridge University Press 1991, Cambridge.