Spectral Risk Measures are not Weak Acceptance Time Consistent

notes

mathematics

risk

pmir

insurance

pricing

Author

Stephen J. Mildenhall

Published

2026-02-17

Modified

2026-02-17

This note provides a short and motivated proof of Theorem 4.6 from Bielecki et al. (2024), showing that the risk measure associated with a strictly concave distortion on an atomless probability space is not weak acceptance time consistent.

Let $(\Omega,\mathcal{F},\mathsf{P})$ be an atomless probability space. Then for every $p\in(0,1)$ there exists $B\in\mathcal{F}$ with $\mathsf{P}(B)=p$, hence $(\Omega,\mathcal{F},\mathsf{P})$ supports Bernoulli$(p)$ variables (Föllmer and Schied 2016, Appendix A27). Let $\mathcal{F}_0\subset\mathcal{F}_1\subset\mathcal{F}_2=\mathcal{F}$ be a filtration at times $t\in\set{0,1,2}$. Let $g:[0,1]\to[0,1]$ be a concave distortion with $g(0)=0$ and $g(1)=1$.

For a bounded profit and loss variable $X$, the dynamic distortion risk measure of Bielecki et al. (2024) is the conditional Choquet integral \[ \rho_t^g(X) =\int_{0}^{\infty} g(\mathsf{P}(-X>y\mid\mathcal{F}_t))\,dy + \int_{-\infty}^{0}\bigl(g(\mathsf{P}(-X>y\mid\mathcal{F}_t))-1\bigr)\,dy. \]

Weak acceptance time consistency (WATC) for $\rho$ means: for $s>t$, \[ \rho_s^g(X)\le 0 \ \Rightarrow\ \rho_t^g(X)\le 0. \]

For a nonnegative bounded loss $L\ge 0$, define the induced price functional \[ \pi_t^g(L):=\rho_t^g(-L)=\int_{0}^{\infty} g(\mathsf{P}(L>y\mid\mathcal{F}_t))\,dy, \] since the second integral vanishes for $X=-L$.

By cash additivity, $\rho_t^g(m-L)=\pi_t^g(L)-m$. Hence WATC is equivalent to: for all bounded $L\ge0$ and all constants $m$, if $\pi_s^g(L)\le m$ a.s., then $\pi_t^g(L)\le m$ a.s.

Intuitively, WATC can fail even in a two-step tree. Consider a mixture of two conditional loss distributions, one low-spread and one high-spread, with the mixture component revealed at $t=1$. Choose them so that the time-1 conditional distortion price is the same on both branches, so $\rho_1(m-L)=0$ statewise. At $t=0$, however, $\rho_0$ applies the distortion to the unconditional mixture distribution, which exposes more extreme tail behavior than either conditional law alone. The theorem below formalizes this mechanism and shows that any non-trivial concave distortion yields such a counterexample, and hence fails WATC.

Theorem 1 (Nontrivial concave distortions fail WATC (Bielecki et al. 2024 Theorem 4.6)) Let $g$ be a concave distortion with $g(0)=0$, $g(1)=1$. Assume $g$ is neither the identity ($g(s)=s$) not the max distortion ($g_{\max}(s)=\set{{s>0}}$ indicator function). Then there exists a two-step filtered two-branch tree embedded in $(\Omega,\mathcal{F},\mathsf{P})$, and a bounded $X$ such that \[ \rho_1^g(X)\le 0 \ \text{a.s.}\quad\text{but}\quad \rho_0^g(X)>0, \] so $\rho^g$ is not weakly acceptance time consistent.

Proof. By assumption, there exist $r,s\in(0,1)$ such that $g(sr)>s g(r)$. (Otherwise $g(sr)=s\,g(r)$ for all $r,s$, forcing $g(u)=u$. Here we also rely on $g\ne g_{\max}$ for strict inequality.)

Fix such $(r,s)$ and set $w:=g(r)\in(0,1)$.

Next, we build a tree and match time-1 prices. Let $S$ be a Bernoulli random variable with $\mathsf{P}\set{S=1}=s$, and let $\mathcal{F}_1=\sigma(S)$. On each state, let the loss $L$ take two values with the same conditional probabilities: \[ \begin{aligned} \begin{cases} \mathsf{P}\set{L=b_1\mid S=0}&=r \\ \mathsf{P}\set{L=a_1\mid S=0}&=1-r \end{cases} \qquad\mathrm{and}\qquad \begin{cases} \mathsf{P}\set{L=b_2\mid S=1}&=r \\ \mathsf{P}\set{L=a_2\mid S=1}&=1-r \end{cases} \end{aligned} \] with $a_2<a_1<b_1<b_2$.

For a two-point loss $Y\in\set{a,b}$ with $\mathsf{P}\set{Y=b}=r$, one has \[ \pi^g(Y)=a+(b-a)w. \] Hence, for any $m>0$ and any choice of $a\in(0,m)$, setting \[ b=\frac{m-(1-w)a}{w} \] gives $\pi^g(Y)=m$. Choose $0<a_2<a_1<m$ and define $b_1,b_2$ by this formula; then $a_2<a_1<b_1<b_2$ and \[ \pi_1^g(L)=m \] in each state $S=0,1$.

We now show the time-0 price strictly exceeds $m$. Unconditionally, $L$ has four atoms with probabilities \[ s(1-r), \ (1-s)(1-r),\ (1-s)r,\ \ sr \] on the ordered values $a_2<a_1<b_1<b_2$. Writing out $\pi_0^g(L)$ as the integral of survival probabilities on the three intervals $(a_2,a_1)$, $(a_1,b_1)$, $(b_1,b_2)$ yields \[ \pi_0^g(L)=m+(a_1-a_2)D, \] where \[ D:=g(1-s(1-r))-1+\frac{1-w}{w}\,g(sr). \] Concavity gives the chord bound $g(1-s(1-r))\ge 1-s+s w$ (mixing $1$ and $r$), and the choice of $(r, s)$ gives $g(sr)>s w$. Substituting, \[ D>(1-s+s w)-1+\frac{1-w}{w}\,s w=0, \] so $\pi_0^g(L)>m$.

Finally, let $X:=m-L$. Then \[ \begin{aligned} \rho_1^g(X)&=\pi_1^g(L)-m=0\le 0 \\ \rho_0^g(X)&=\pi_0^g(L)-m>0, \\ \end{aligned} \] showing that WATC fails.

Remark 1.

Bielecki et al. (2024) parameterize distortions via a mixing measure $\mu$ (Kusuoka/spectral representation). This is an equivalent encoding of the same spectral functional.
The counterexample requires $g(r)\in(0,1)$ and a strict chord inequality at some tail level. It fails only for the two degenerate endpoint cases: the identity and the max distortion $g_{\max}$ (which yields essential supremum for nonnegative losses).
The argument uses only concavity, so it covers discontinuous distortions (e.g. with $g(0+)>0$) and does not require absolute continuity.

References.

References

Bielecki, T. R., Cialenco, I., & Liu, H. (2024). Time consistency of dynamic risk measures and dynamic performance measures generated by distortion functions. Stochastic Models. https://doi.org/10.1080/15326349.2024.2353045

Föllmer, H., & Schied, A. (2016). Stochastic Finance: An Introduction in Discrete Time (Fourth.). Berlin, Boston: Walter de Gruyter. https://doi.org/10.1017/CBO9781107415324.004

--- author: Stephen J. Mildenhall title: Spectral Risk Measures are not Weak Acceptance Time Consistent description: '' categories: - notes - mathematics - risk - pmir - insurance - pricing date: '2026-02-17' date-modified: last-modified draft: false image: img/banner.png format: html: toc: false reference-section-title: ' ' # title for references section, passed to pandoc code-tools: true pdf: documentclass: article papersize: a4 fontsize: 12pt keep-tex: true geometry: margin=1in include-in-header: text: | \usepackage{mathtools} \DeclarePairedDelimiter{\set}{\{}{\}} # include-in-header: macros.tex toc: false # pdf-engine: pdflatex # pdf-engine-opt: "-fmt=myquartoformat" pdf-engine: lualatex pdf-engine-opts: - '-interaction=nonstopmode' bibliography: C:/s/TELOS/Biblio/uber-library.bib csl: C:/s/TELOS/Biblio/journal-of-risk-and-uncertainty.csl ---  ![](img/banner.png){width=50%} This note provides a short and motivated proof of Theorem 4.6 from @Bielecki2024, showing that the risk measure associated with a strictly concave distortion on an atomless probability space is not weak acceptance time consistent. Let $(\Omega,\mathcal{F},\mathsf{P})$ be an atomless probability space. Then for every $p\in(0,1)$ there exists $B\in\mathcal{F}$ with $\mathsf{P}(B)=p$, hence $(\Omega,\mathcal{F},\mathsf{P})$ supports Bernoulli$(p)$ variables [@Follmer2016, Appendix A27]. Let $\mathcal{F}_0\subset\mathcal{F}_1\subset\mathcal{F}_2=\mathcal{F}$ be a filtration at times $t\in\set{0,1,2}$. Let $g:[0,1]\to[0,1]$ be a concave distortion with $g(0)=0$ and $g(1)=1$. For a bounded profit and loss variable $X$, the dynamic distortion risk measure of @Bielecki2024 is the conditional Choquet integral $$ \rho_t^g(X) =\int_{0}^{\infty} g(\mathsf{P}(-X>y\mid\mathcal{F}_t))\,dy + \int_{-\infty}^{0}\bigl(g(\mathsf{P}(-X>y\mid\mathcal{F}_t))-1\bigr)\,dy. $$ Weak acceptance time consistency (WATC) for $\rho$ means: for $s>t$, $$ \rho_s^g(X)\le 0 \ \Rightarrow\ \rho_t^g(X)\le 0. $$ For a nonnegative bounded loss $L\ge 0$, define the induced price functional $$ \pi_t^g(L):=\rho_t^g(-L)=\int_{0}^{\infty} g(\mathsf{P}(L>y\mid\mathcal{F}_t))\,dy, $$ since the second integral vanishes for $X=-L$. By cash additivity, $\rho_t^g(m-L)=\pi_t^g(L)-m$. Hence WATC is equivalent to: for all bounded $L\ge0$ and all constants $m$, if $\pi_s^g(L)\le m$ a.s., then $\pi_t^g(L)\le m$ a.s. Intuitively, WATC can fail even in a two-step tree. Consider a mixture of two conditional loss distributions, one low-spread and one high-spread, with the mixture component revealed at $t=1$. Choose them so that the time-1 conditional distortion price is the same on both branches, so $\rho_1(m-L)=0$ statewise. At $t=0$, however, $\rho_0$ applies the distortion to the unconditional mixture distribution, which exposes more extreme tail behavior than either conditional law alone. The theorem below formalizes this mechanism and shows that any non-trivial concave distortion yields such a counterexample, and hence fails WATC. :::{#thm-4-6 .unnumbered} #### Nontrivial concave distortions fail WATC [@Bielecki2024 Theorem 4.6] Let $g$ be a concave distortion with $g(0)=0$, $g(1)=1$. Assume $g$ is neither the identity ($g(s)=s$) not the max distortion ($g_{\max}(s)=\set{{s>0}}$ indicator function). Then there exists a two-step filtered two-branch tree embedded in $(\Omega,\mathcal{F},\mathsf{P})$, and a bounded $X$ such that $$ \rho_1^g(X)\le 0 \ \text{a.s.}\quad\text{but}\quad \rho_0^g(X)>0, $$ so $\rho^g$ is not weakly acceptance time consistent. ::: ::: {.proof} By assumption, there exist $r,s\in(0,1)$ such that $g(sr)>s g(r)$. (Otherwise $g(sr)=s\,g(r)$ for all $r,s$, forcing $g(u)=u$. Here we also rely on $g\ne g_{\max}$ for strict inequality.) Fix such $(r,s)$ and set $w:=g(r)\in(0,1)$. Next, we build a tree and match time-1 prices. Let $S$ be a Bernoulli random variable with $\mathsf{P}\set{S=1}=s$, and let $\mathcal{F}_1=\sigma(S)$. On each state, let the loss $L$ take two values with the same conditional probabilities: $$ \begin{aligned} \begin{cases} \mathsf{P}\set{L=b_1\mid S=0}&=r \\ \mathsf{P}\set{L=a_1\mid S=0}&=1-r \end{cases} \qquad\mathrm{and}\qquad \begin{cases} \mathsf{P}\set{L=b_2\mid S=1}&=r \\ \mathsf{P}\set{L=a_2\mid S=1}&=1-r \end{cases} \end{aligned} $$ with $a_2<a_1<b_1<b_2$. For a two-point loss $Y\in\set{a,b}$ with $\mathsf{P}\set{Y=b}=r$, one has $$ \pi^g(Y)=a+(b-a)w. $$ Hence, for any $m>0$ and any choice of $a\in(0,m)$, setting $$ b=\frac{m-(1-w)a}{w} $$ gives $\pi^g(Y)=m$. Choose $0<a_2<a_1<m$ and define $b_1,b_2$ by this formula; then $a_2<a_1<b_1<b_2$ and $$ \pi_1^g(L)=m $$ in each state $S=0,1$. We now show the time-0 price strictly exceeds $m$. Unconditionally, $L$ has four atoms with probabilities $$ s(1-r), \ (1-s)(1-r),\ (1-s)r,\ \ sr $$ on the ordered values $a_2<a_1<b_1<b_2$. Writing out $\pi_0^g(L)$ as the integral of survival probabilities on the three intervals $(a_2,a_1)$, $(a_1,b_1)$, $(b_1,b_2)$ yields $$ \pi_0^g(L)=m+(a_1-a_2)D, $$ where $$ D:=g(1-s(1-r))-1+\frac{1-w}{w}\,g(sr). $$ Concavity gives the chord bound $g(1-s(1-r))\ge 1-s+s w$ (mixing $1$ and $r$), and the choice of $(r, s)$ gives $g(sr)>s w$. Substituting, $$ D>(1-s+s w)-1+\frac{1-w}{w}\,s w=0, $$ so $\pi_0^g(L)>m$. Finally, let $X:=m-L$. Then $$ \begin{aligned} \rho_1^g(X)&=\pi_1^g(L)-m=0\le 0 \\ \rho_0^g(X)&=\pi_0^g(L)-m>0, \\ \end{aligned} $$ showing that WATC fails. ::: :::{#rem-4-6 .unnumbered} 1. @Bielecki2024 parameterize distortions via a mixing measure $\mu$ (Kusuoka/spectral representation). This is an equivalent encoding of the same spectral functional. 2. The counterexample requires $g(r)\in(0,1)$ and a strict chord inequality at some tail level. It fails only for the two degenerate endpoint cases: the identity and the max distortion $g_{\max}$ (which yields essential supremum for nonnegative losses). 3. The argument uses only concavity, so it covers discontinuous distortions (e.g. with $g(0+)>0$) and does not require absolute continuity. ::: **References.**

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