On the dynamic representation of some time-inconsistent risk measures in a Brownian filtration

Abstract

It is well-known from the work of Kupper and Schachermayer that most law-invariant risk measures are not time-consistent, and thus do not admit dynamic representations as backward stochastic differential equations. In this work we show that in a Brownian filtration the “Optimized Certainty Equivalent” risk measures of Ben-Tal and Teboulle can be computed through PDE techniques, i.e. dynamically. This can be seen as a substitute of sorts whenever they lack time consistency, and covers the cases of conditional value-at-risk and monotone mean-variance. Our method consists of focusing on the convex dual representation, which suggests an expression of the risk measure as the value of a stochastic control problem on an extended the state space. With this we can obtain a dynamic programming principle and use stochastic control techniques, along with the theory of viscosity solutions, which we must adapt to cover the present singular situation.

Appendix

Proof of Proposition 2.4

In our language, we easily obtain $\rho ^{l_z}(zX)=z\log \left( \frac{ E[e^{X}]}{z}\right) +z - 1$, and so for the value function, see Proposition 3.3, we have

$$\begin{aligned} V(s,y,z)&=-z\log z +z - 1+ z\log E \left[ \exp \left\{ f(Y_T^{s,y}) + \int _s^T g(t,Y_t^{s,y})dt \right\} \right] \\&=: -z\log z+z - 1 + z {\tilde{V}}(s,y). \end{aligned}$$

That $\exp {({\tilde{V}})}$ is a viscosity solution of the backward Kolmogorov PDE associated to the diffusion Y follows e.g. from Fleming and Soner [23, Chap. V.9]. That $\exp {({{\tilde{V}}})}$ is the uniqueness classical solution follows by e.g. Friedman [25, Theorems 1.7.12 and 2.4.10]. It is then clear that V is the unique classical solution of our HJB equation. $\square$

Proof of Lemma 3.1

A derivation of the dual representation

$$\begin{aligned} \rho (X)\,=\, \sup \left\{ E[XZ] -E[l^*( Z)]:Z\in L^{1}_{+} ,Z\in \text {dom}(l^*),\,E[ Z]=1 \right\} ,\quad X \in L^\infty \end{aligned}$$

(4.4)

can be obtained for instance from Ben-Tal and Teboulle [6], and elementary considerations. Deriving (3.1) from (4.4) is done by classical arguments which we give for completeness. We clearly have “$\ge$” in (3.1). Conversely, given $\varepsilon >0$ and a feasible Z for (4.4) such that $\rho (X)\le E[XZ] - E[l^*(Z)]-\varepsilon$, we must have $l^*(Z)< \infty$. For every $c\in (0,1)$, define $Z^{c}:= c +(1-c)Z$, which is likewise feasible for (4.4), satisfies $l^*(Z^c)<\infty$ and is such that $(Z^c)_c$ is uniformly integrable. Assume for the moment that $l^*(Z^{c})\rightarrow l^*(Z)$P-a.s. Then, by convexity, $0\le l^*(Z^c)\le (1-c)l^*(Z^c)\le l^*(Z)$. Thus, we conclude by dominated convergence that $\rho (X) \le \lim _{c\rightarrow 0}E[XZ^c]-E[l^*(Z^c)]-\varepsilon$, which yields the reverse inequality. The proof is finished after noticing that $l^*$ must be continuous throughout its domain. In fact, the domain of $l^*$ is an interval with end points denoted by $a\in {\mathbb {R}}_+$ and $b\in {\mathbb {R}}_+\cup \{+\infty \}$, and $l^*$ is continuous on (a, b). If $+\infty >b \in \text {dom}(l^*)$, let $x^n\rightarrow b$ and $\lambda ^n\in (0,1)$ such that $\lambda ^n\rightarrow 1$ and $x^n= \lambda ^nb + (1-\lambda ^n)a$. By convexity, we have $l^*(x^n) \le \lambda ^nl^*(b)+(1-\lambda ^n)l^*(a)$. This shows $\limsup _{n\rightarrow \infty } l^*(x^n)\le l^*(b)$. We conclude by lower semicontinuity that $l^*$ is continuous at b and the proof is similar if $a\in \text {dom}(l^*)$.

Proof of Proposition 3.2

By the dual representation of $\rho (X)$ in the completed Brownian filtration, we directly see that the r.h.s. of (3.2) is a lower bound for $\rho (X)$. We shall establish the opposite inequality by repeated approximation arguments. By Lemma 3.1 we need only consider $Z\in {{{\mathcal {Z}}}}$.

STEP 1: We may assume w.l.o.g. that $E[l^*(Z)]<\infty$, as otherwise this Z is irrelevant for the problem. Letting $Z_t:=E[Z|{\mathcal {F}}_t]$ and $\tau ^n:=\inf \{0<t\le T:Z_t=n\}\wedge T$, we have by optional sampling and Jensen’s inequality that $-E[l^*(Z)]\le -E[l^*(Z^{n})]$ and $E[l^*(Z^{n})]<\infty$, with $Z^n:=Z_{\tau ^n}\le n$. On the other hand $E[Z^nX]\rightarrow E[ZX]$ by the martingale convergence theorem. Thus

$$\begin{aligned} E[ZX]-E[l^*(Z)]\le \liminf _{n\rightarrow \infty }(E[Z^nX]-E[l^*(Z^n)]). \end{aligned}$$

So we assume w.l.o.g. that Z is essentially bounded from above and essentially bounded away from 0.

STEP 2: Define $Z^n:=E[Z|{\mathcal {G}}_n]$, where ${\mathcal {G}}_n=\sigma \{W_{kT2^{-n}}:k=0,\dots ,2^n\}$. It holds

$$\begin{aligned} Z^n = u_n(W_{T/n},\dots ,W_{kT/n} ,\dots ,W_{T} ), \end{aligned}$$

for some bounded positive Borel function $u_n$ which is bounded away from 0 and with range in $dom(l^*)$. Moreover, $Z^n\rightarrow Z$P-a.s. and by continuity of $l^*$ in its domain (see the proof of Lemma 3.1), $l^*(Z^n)$ is essentially bounded uniformly in n. Using martingale convergence again and dominated convergence we have $E[ZX]-E[l^*(Z)] = \lim _{n\rightarrow \infty }(E[Z^nX]-E[l^*(Z^n)])$. Thus, we may further assume w.l.o.g. that Z is of the form

$$\begin{aligned} Z = u(W_{T/n},\dots ,W_{kT/n} ,\dots ,W_{T} )\quad \text {for some } n\in {\mathbb {N}}, \end{aligned}$$

(4.5)

with u as above.

STEP 3: Let $\Phi \in C^\infty _c({\mathbb {R}}^n)$ be the mollifier $\textstyle \Phi (x)={\mathbb {I}}_{\{\Vert x\Vert <1\}}\lambda \exp \left\{ \frac{1}{\Vert x\Vert ^2-1} \right\}$, with $\lambda \ge 0$ such that $\int _{{\mathbb {R}}^n}\Phi (x)\,dx=1$. Define by convolution $u^{\delta }(x):= u*\delta ^{-n}\Phi (x/\delta )$, $\delta >0$. Then $u^\delta \in C^\infty ({\mathbb {R}}^n)$ is bounded from above, bounded away from 0 and the derivative $\nabla u^\delta = u*\nabla (\delta ^{-n}\Phi (\cdot /\delta ))$ is bounded. We can also choose a sequence $\delta _m\rightarrow 0$ so that $u^{\delta _m} \rightarrow u$, Lebesgue a.e. in ${\mathbb {R}}^n$; this follows from the convergence over compacts of $u^\delta$ to u in $L^1(dx)$ and a diagonalization argument. This shows that $u^{\delta _m}(W_{T/n},\dots ,W_{kT/n} ,\dots ,W_{T} )$ converges to Z almost surely as $m\rightarrow \infty$, since the law of the Gaussian vector $(W_{T/n},\dots ,W_{kT/n} ,\dots ,W_{T} )$ is equivalent to Lebesgue in ${\mathbb {R}}^n$. Arguing as in the previous step, we conclude that we may further assume w.l.o.g. that Z is given by (4.5) where u is smooth and with bounded derivatives.

STEP 4: From the previous steps, it remains to show that for every n, the random variable $Z = u(W_{T/n},\dots ,W_{kT/n} ,\dots ,W_{T} )$ can be written as ${{{\mathcal {E}}}}(\int \beta \,dW)_T$, with $\beta \in {{{\mathcal {L}}}}_b$. For $t\in \left[ \frac{n-1}{n}T,T\right]$ we have

$$\begin{aligned} E[Z|{\mathcal {F}}_t]&= \int u(W_{T/n},\dots ,W_{T(n-1)/n},W_{t}+x)] N^{T-t}(dx) \\&=: U^n\left( t,W_t\,;\,W_{T/n},\dots ,W_{\frac{n-1}{n}T}\right) , \end{aligned}$$

with $N^{T-t}(dx)$ the law of a centred Gaussian with variance $(T-t)\times I_d$, with $I_d$ the identity of ${\mathbb {R}}^{d\times d}$. By the mean value theorem and dominated convergence, the function $U^n$ is differentiable in the spacial arguments and the derivatives are bounded, uniformly in the time argument. Smoothness in the time argument is apparent from the density of $N^{T-t}(dx)$. In addition, $U^n$ is bounded away from 0.

We now proceed by reverse induction. Assume that we have constructed a function $U^{k+1}$ such that $E[Z|\mathcal{F}_t]=U^{k+1}(t,W_t\,;\,W_{T/n},\dots ,W_{\frac{k}{n}T})$ on $[kT/n,(k+1)T/n]$ with $U^{k+1}$ smooth, bounded from above and away from zero, as well as having bounded derivatives in $(x,w_1,\dots ,w_{k})$ uniformly in time. By the tower property, for $t\in [(k-1)T/n,kT/n]$ we have:

$$\begin{aligned} E[Z|{\mathcal {F}}_t]&= E\left[ U^{k+1}(kT/n,W_{kT/n}\,;\,W_{T/n},\dots ,W_{T(k-1)/n},W_{kT/n}) |{\mathcal {F}}_t\right] \\&= \int U^{k+1}(kT/n,W_{t}+x\,;\,W_{T/n},\dots ,W_{T(k-1)/n},W_{t}+x)] N^{(kT/n)-t}(dx) \\&=: U^k\left( t,W_t\,;\,W_{T/n},\dots ,W_{\frac{k-1}{n}T}\right) . \end{aligned}$$

By essentially the same argument as above, $U^k$ is smooth in time and space arguments, bounded from above and away from zero, and has bounded derivatives in $(x,w_1,\dots ,w_{k-1})$ uniformly in time.

Using Itô’s formula and by uniqueness in the martingale representation, we obtain that

$$\begin{aligned} \beta _t=\left[ { U^k\left( t,W_t\,;\,W_{T/n},\dots ,W_{\frac{k-1}{n}T}\right) }\right] ^{-1}{\nabla _x U^k\left( t,W_t\,;\,W_{T/n},\dots ,W_{\frac{k-1}{n}T}\right) }, \end{aligned}$$

on $t\in [(k-1)T/n,kT/n]$. Since there are only finitely many such intervals for fixed n, it follows that $\beta$ is essentially bounded. This concludes the proof. $\square$

We close the appendix including a technical lemma which was crucial in the proof of our main theorem. We assume ${{{\mathcal {O}}}}=(0,+\infty )$ from here on. The argument is inspired by [43].

Lemma 4.4

Let $w:[0,T]\times {\mathbb {R}}^m\times [0,+\infty )$ be a continuous viscosity supersolution of¹³

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu + H^n(t, y,z, Du, D^2u) = 0,\quad (t,y,z)\in [0,T)\times {\mathbb {R}}^m\times (0,+\infty )\\ u(T,y,z) = \psi (y,z),\quad (y,z)\in {\mathbb {R}}^m\times (0,+\infty ). \end{array}\right. } \end{aligned}$$

(4.6)

Then ${{\tilde{w}}}(s,y,z):=w(s,y,|z|)$ is a continuous viscosity supersolution of

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu + H^n(t, y,z, Du, D^2u) = 0,\quad (t,y,z)\in [0,T)\times {\mathbb {R}}^m\times {\mathbb {R}}\\ u(T,y,z) = \psi (y,|z|),\quad (y,z)\in {\mathbb {R}}^m\times {\mathbb {R}}. \end{array}\right. } \end{aligned}$$

(4.7)

The analogue statement holds for continuous viscosity (sub)solutions.

Proof

Let $x_0=(s_0,y_0,z_0)\in [0,T]\times {\mathbb {R}}^m\times {\mathbb {R}}$ and $\varphi \in C^2([0,T]\times {\mathbb {R}}^m\times {\mathbb {R}})$ such that $x_0$ is a local minimizer of ${{\tilde{w}}}-\varphi$ and $\varphi (x_0) = {{\tilde{w}}}(x_0)$. If $s_0=T$, and otherwise if $z_0 > 0$, there is nothing to prove, owing to the supersolution property of w. Otherwise, if $z_0<0$, we observe that the terms $\partial _{zz}\varphi$ and $z\partial _{yz}\varphi$ are of second-order in the z-variable, meaning that we can leverage again the supersolution property of w to easily obtain $\partial _t\varphi (x_0) + H^n(x_0, D\varphi (x_0), D^2\varphi (x_0)) \le 0$.

It remains to check the case $s_0<T$ and $z_0=0$. As conventional in this theory, we may further assume that $x_0$ was a strict local minimizer of ${{\tilde{w}}}-\varphi$. By definition, $x_0$ is also a strict local minimizer of $w-\varphi$ restricted to $[0,T]\times {\mathbb {R}}^m\times [0,+\infty )$. By Lemma 4.5 below, we have $\partial _t\varphi (x_0) + H^n(x_0, D\varphi (x_0), D^2\varphi (x_0)) \le 0$, finishing the proof for supersolution. Of course the same arguments apply to subsolution, and then to solutions too. $\square$

The above proof rests on the following lemmas, which are reminiscent of [7, Lemma 4.1]. We only provide the proof of Lemma 4.5, as the one for Lemma 4.6 is straightforward and lengthy.

Lemma 4.5

Take w as in Lemma 4.4 and let $\varphi$ be of class $C^{2}$ on $[0,T]\times {\mathbb {R}}^m\times [0,+\infty )$. Suppose that $w-\varphi$ has a strict local minimum at $x_0:=(s_0,y_0,0)$. Then $\partial _t\varphi (x_0) + H^n(x_0, D\varphi (x_0), D^2\varphi (x_0)) \le 0$.

Proof

Define $v:=w-\varphi$. We apply Lemma 4.6, obtaining the existence of $x_\epsilon :=(s_\epsilon ,y_\epsilon ,z_\epsilon )\rightarrow x_0:=(s_0,y_,0)$ with the properties listed therein. In particular, $x_\epsilon$ is a local minimum of $w-\varphi _\epsilon$ with $z_\epsilon >0$, where $\varphi _\epsilon (s,y,z):=\varphi (s,y,z)-\epsilon /z$. By definition of w, we have

$$\begin{aligned} \begin{aligned}&\partial _t\varphi _\epsilon (x_\epsilon )+b(s_\epsilon ,y_\epsilon )\partial _y\varphi _\epsilon (x_\epsilon )+\frac{1}{2}tr\left( \sigma (s_\epsilon ,y_\epsilon )\sigma (s_\epsilon ,y_\epsilon )'\partial ^2_{yy}\varphi _\epsilon (x_\epsilon )\right) \\&\quad +\textstyle \sup _{|\beta |\le n}\left[ \frac{1}{2} z_\epsilon ^2|\beta |^2\partial ^2_{zz}\varphi _\epsilon (x_\epsilon )+ z_\epsilon \,\partial ^2_{yz}\varphi _\epsilon (x_\epsilon ) \sigma (s_\epsilon ,y_\epsilon )\beta -\frac{\epsilon |\beta |^2}{z_\epsilon }\right] + z_\epsilon g(s_\epsilon ,y_\epsilon ) \le 0. \end{aligned} \end{aligned}$$

We conclude by sending $\epsilon \rightarrow 0$, by continuity and since the supremum defining the Hamiltonian $H^n$ is taken over a compact set. $\square$

Lemma 4.6

Let $v:[0,T]\times {\mathbb {R}}^m\times [0,\infty )\rightarrow {\mathbb {R}}$ be continuous and such that $(s_0,y_,0)$ is a strict local minimizer of v. Then there exists $(s_\epsilon ,y_\epsilon ,z_\epsilon )\rightarrow (s_0,y_,0)$ such that

$$\begin{aligned}&(1)\, z_\epsilon>0\, (\forall \epsilon >0),\quad (2)\, \frac{\epsilon }{z_\epsilon }\rightarrow 0 \text { as } \epsilon \searrow 0, \quad (3)\, (s_\epsilon ,y_\epsilon ,z_\epsilon ) \text { locally minimizes }v_\epsilon (s,y,z)\\&\quad := v(s,y,z)+\frac{\epsilon }{z}. \end{aligned}$$