Dynamic consistency of expected utility under non-classical (quantum) uncertainty

Abstract

Quantum cognition in decision making is a recent and rapidly growing field. In this paper, we develop an expected utility theory in a context of non-classical (quantum) uncertainty. We replace the classical state space with a Hilbert space which allows introducing the concept of quantum lottery. Within that framework, we formulate axioms on preferences over quantum lotteries to establish a representation theorem. We show that demanding the consistency of choice behavior conditional on new information is equivalent to the von Neumann–Lüders postulate applied to beliefs. A dynamically consistent quantum-like agent may violate dynamic recursive consistency, however. This feature suggests interesting applications in behavioral economics as we illustrate in an example of persuasion.

Appendices

Appendix 1: Elementary facts about Hilbert spaces

Hilbert space

Let $\mathbb {R}$ and ${\mathbb {C}}$ denote the fields of real and of complex numbers. For a complex number z, $\bar{z}$ denotes its complex conjugate.

Definition 1

Let H be a vector space over the field $\mathbb {C}$ . An Hermitian form on H is a mapping $(.,.):H\times H\rightarrow \mathbb {C}$ such that: (a) it is linear in the first argument; (b) $(v,w)= \overline{(w,v)}$ for any $v,w\in H\$(in particular, (v, v) is a real number); (c) $(v,v)\ge 0$ for any $v\in H$, and $(v,v)=0$ only for $v=0$ .

Vectors v and w are called orthogonal if $(v,w)=0$; in this case $(w,v)=0$ as well.

A Hilbert space is a vector space H endowed with an Hermitian form, which is complete relatively to the norm $|v|=\sqrt{(v,v)}$. In order to avoid unnecessary difficulties and subtleties we assume further that H has finite dimension; then, H automatically is complete.

When discussing lotteries and measurements we shall not be dealing so much with vectors in H as with special operators (linear mappings from H to H) called Hermitian operators.

Hermitian operators

Definition 2

A linear operator $A:H\rightarrow H$ is called Hermitian, if $(Av,w)=(v,Aw)$ for any $v,w\in H$.

Clearly, $\left( Av,v\right)$ is a real number for any $v\in H$. Hermitian operator A is called nonnegative if $\left( Av,v\right) \ge 0$ for any v. For Hermitian operators A and B, we write $A\ge B$ if $A-B$ is nonnegative. The identity operator E ($Ev=v$ for every $v\in H$) is Hermitian.

A most important, for the purpose of this paper, class of Hermitian operator consists of projectors. A projector is an idempotent Hermitian operator, that is $PP=P$. Since $(Pv,v)=(PPv,v)=(Pv,Pv)\ge 0$, any projector is nonnegative. Each projector P defines a vector subspace $V=\text {Im} P\subset H$, consisting of vectors v such that with $Pv=v$. The kernel of the projector consists of vectors orthogonal to V, $\text {Ker}P=V^{\perp }$ . The set of projectors can be identified with the set of (closed) subspaces of H.

Any linear combination of Hermitian operators with real coefficients is an Hermitian operator. In other words, the set $\mathbf {Herm}(H)$ of Hermitian operators is a real vector space. The crucial importance of projectors is underlined by the following central theorem.

Spectral theorem. Let A be an Hermitian operator. Then, there exists a family of projectors $P_{i}$ and real numbers  $a_{i}$ such that: (1) $P_{i}P_{j}=0$ for $i\ne j$, (2)  $\sum _{i}P_{i}=E$, and (3) $A=\sum _{i}a_{i}P_{i}$.

In other words, in some orthogonal basis the operator A can be represented by a diagonal matrix (with real coefficients). The coefficients $a_i$ are eigenvalues of the operator A. The set of numbers $a_i$ is called the spectrum of the operator A. Clearly A is nonnegative if and only if all coefficients $a_{i}$ are nonnegative. An operator A is a projector if and only if its spectrum SpecA consists of 0 or 1.

Each nonnegative operator A has a (unique) nonnegative square root $\sqrt{A }$ (or $A^{1/2}$, such that $(\sqrt{A})^2=A$). If $A=\sum _i a_i P_{i}$ is a spectral representation of A, then $\sqrt{A}=\sum _i \sqrt{a_i} P_{i}$ .

Trace of operators

For arbitrary (not necessarily Hermitian) linear operator $A:H\rightarrow H$, it is possible to talk about its trace $\mathbf {Tr}\left( A\right)$. More precisely, for any quadratic matrix $A=(a_{ij})$, the trace $\mathbf {Tr}(A)$ is defined as $\sum _{i}a_{ii}$, the sum of its diagonal elements. A remarkable property of the trace is its “commutativity”: $\mathbf {Tr}(AB)=\mathbf {Tr}(BA)$ for any quadratic matrix A and B. This in particular implies that the trace of an operator is independent of the choice of basis, thereby allowing for an unambiguous definition of the trace of a linear operator.

For example, $\mathbf {Tr}(E)=\dim H$. More generally, if P is an (orthogonal) projector (on subspace $V=ImP$) then $\mathbf {Tr}(P)=\dim V$. Due to the spectral theorem, we obtain that the trace of Hermitian operator $A=\sum a_{i}P_{i}$ is equal to $\sum a_{i}\text {rk}(P_{i})$ and, in particular, it is a real number. The trace of nonnegative operator A is nonnegative and is strictly positive if $A\ne 0$.

For two Hermitian operators A and B, define $(A,B)_{Herm}=\mathbf {Tr}(AB)$ . We assert that this “scalar product” gives a structure of Euclidean space on the real vector space $\mathbf {Herm} (H)$. This follows from the Lemma below.

Lemma

• (a) $(A,B)_{Herm}$ is a real number;

• (b) $(A,B)_{Herm}=(B,A)_{Herm}$;

• (c) $(A,A)_{Herm}\ge 0$ and is equal to  0 only if $A=0$.

Proof

Due to the commutativity of the trace, $2\mathbf {Tr}(AB)= \mathbf {Tr}(AB)+\mathbf {Tr}(BA)=\mathbf {Tr}(AB+BA)$. It is easy to understand that the operator $AB+BA$ is Hermitian, hence its trace is real. This proves (a).

• (b) follows from the commutativity of the trace.

• (c) follows from the nonnegativity of the operator $A^{2}$.

$\square$

Appendix 2: Proof of Theorem 1

Let $\preceq$ be a nice preference relation on the set $\mathbf {QL}(H)$. We shall be working with roulette-valued Q-lotteries, that is with expressions of the form $\sum _{i}l_{i}\otimes P_{i}$, where $(P_{i},\ i\in I)$ is a PDU, and $(l_{i},\ i\in I)$ is a collection of roulette lotteries on X. Let $\mathbf {QL}_{\mathcal {P}}(H,\Delta (X))$ denote the set of all roulette valued Q-lotteries with base $\mathcal {P}$. We first provide a result that shows that mixtures of canonical Q-lotteries in $\mathbf {QL}(H)$ and mixtures of Q-lotteries in $\mathbf {QL}_{\mathcal {P}}(H,\Delta (X))$ are compatible in a sense we make clear below.

Consider a lottery in the canonical form $\sigma =\sum _{x}x\otimes P_{x},$ where $x\in X,P_{x}$ are nonnegative Hermitian operators which add up to E,  $\sum _{x}P_{x}=E.$ The mixture $\alpha \sigma +\left( 1-\alpha \right) \tau ,$ where $\tau =\sum _{x}x\otimes Q_{x}$ and $\alpha \in [0,1]$, is given as $\sum _{x}x\otimes \left( \alpha P_{x}+\left( 1-\alpha \right) Q_{x}\right) .$ On the other hand, a roulette valued Q-lottery writes $\sigma =\sum _{i}l_{i}\otimes P_{i}$, where $\mathcal {P}=(P_{i},i\in I)$ is a measurement device (the base of the lottery) and $l_{i}\in \Delta \left( X\right) .$ A mixture of such lotteries is defined by the following formula: $\alpha \sigma +(1-\alpha )\tau =\sum _{i}\left( \alpha l_{i}+\left( 1-\alpha \right) r_{i}\right) P_{i}$, where $\tau =\sum _{i}r_{i}\otimes P_{i}$ is another Q-lottery in $\mathbf {QL}_{\mathcal {P}}(H,\Delta (X))$ and $\alpha \in [0,1]$.

We next define the canonization mapping $can:\mathbf {QL}_{\mathcal {P} }\rightarrow \mathbf {QL}_{c},$ which maps lottery $\sum _{i}l_{i}\otimes P_{i}$ (where the roulette lottery $l_{i}$ has the form $\sum _{x}x\otimes l_{i}\left( x\right) ,$ i.e., $l_{i}$ gives value x with probability $l_{i}\left( x\right) )$ into the canonical lottery $\sum _{x}x\otimes \left( \sum _{i}l_{i}\left( x\right) P_{i}\right) .$

Lemma 2

The mapping “can ” preserves the operation of mixture; that is, for any $\sigma ,\tau \in \mathbf {QL}_{\mathcal {P}}(H,\Delta (X))$ and $\alpha \in [0,1]$, $can(\alpha \sigma +\left( 1-\alpha \right) \tau )=\alpha can\left( \sigma \right) +\left( 1-\alpha \right) can\left( \tau \right)$.

Proof

Assume we have two $\mathcal {P-}$based lotteries $\sigma =\sum _{i}l_{i}\otimes P_{i}$ and $\tau =\sum _{i}r_{i}\otimes P_{i}$ and some $\alpha \in \left[ 0,1\right] .$ We want to show that $can(\alpha \sigma +\left( 1-\alpha \right) \tau ) =\alpha can\left( \sigma \right) +\left( 1-\alpha \right) can\left( \tau \right) .$

The left hand side is equal to $can\left( \sum _{i}\left( \alpha l_{i}+\left( 1-\alpha \right) r_{i}\right) \otimes P_{i}\right) =$

$$\begin{aligned}&\sum _{x}x\otimes \sum _{i}\left( \alpha l_{i}+\left( 1-\alpha \right) r_{i}\right) \left( x\right) P_{i}=\sum _{x}x\otimes \sum _{i}\left( \alpha l_{i}\left( x\right) +\left( 1-\alpha \right) r_{i}\left( x\right) \right) P_{i}\\&\quad =\sum _{x}x\otimes \sum _{i}\left( \alpha \sum _{i}l_{i}\left( x\right) +\left( 1-\alpha \right) \sum _{i}r_{i}\left( x\right) \right) P_{i}. \end{aligned}$$

The right hand side is $\alpha \sum _{x}x\otimes \left( \sum _{i}l_{i}\left( x\right) P_{i}\right) +\left( 1-\alpha \right) \left( \sum _{x}x\otimes \sum _{i}r_{i}\left( x\right) P_{i}\right)$

$=\alpha can\left( \sigma \right) +\left( 1-\alpha \right) can\left( \tau \right)$. $\square$

Returning to the proof of Theorem 1, let $\preceq _{\Delta }$ denote the derived preference relation on the set $\Delta (X)$ of roulette lotteries. The assertion of the theorem is true if the preference $\preceq$ is trivial. Indeed, we can take u to be a constant and take an arbitrary functional $\beta$. So, from now on, we assume that the preference $\preceq$ is nontrivial. That is $\tau \prec \sigma$ for some Q-lotteries $\sigma =\sum _{i}l_{i}\otimes P_{i}$ and $\tau =\sum _{j}m_{j}\otimes Q_{j}$.

Claim 0

Let $l^{*}$ be the best lottery among $(l_{i})$ and $m ^{*}$be the worst among (m$_{j})\$ with respect to the derived weak order $\preceq _{\Delta }$. Then, $\sigma \preceq l^{*}\otimes E$ and $m_{*}\otimes E\preceq \tau _{{}}$.

Proof

Consider the “constant” lottery $\sigma ^{*}=\sum _{i}l^{*}\otimes P_{i}$. Due to A4, $\sigma \preceq \sigma ^{*}$, due to A0, $\sigma ^{*}\approx l^{*}\otimes E$ and due to the transitivity of $\preceq$ (see A1), we conclude that $\sigma \preceq l^{*}\otimes E$. The proof $m_{*}\otimes E\preceq \tau$ is analogous. $\square$

From Claim 0, it follows that the derived preference $\preceq _{\Delta }$ is non-trivial, that is $l_{*}\prec _{\Delta }l^{*}$ for some ordinary lotteries $l_{*}$ and $l^{*}$. We fix such lotteries $l_{*}\prec _{\Delta }l^{*}$; a function $u:\Delta (X)\rightarrow \mathbb {R}$ is said to be normalized if $u(l_{*})=0$ and $u(l^{*})=1$.

Fix now some measurement device $\mathcal {P }=(P_i, \ i\in I)$, and let $\mathbf {QL}_\mathcal {P}(H,\Delta (X))$ denote the set of all roulette valued Q-lotteries with base $\mathcal {P}$. We first extend $\preceq$ into a preference relation $\preceq _\mathcal {P}$ defined on $\mathbf {QL}_\mathcal {P} (H,\Delta (X))$ by setting $\sigma \preceq \tau$ if and only if $can(\sigma )\preceq can(\tau )$ for any $\sigma ,\tau \in \mathbf {QL}_\mathcal {P }(H,\Delta (X))$.

Claim 1

There exists a normalized affine function $u_{ \mathcal {P}}$ on $\Delta (X)$ and a function $\beta _{\mathcal {P}}$ on the set of outcomes I ($\beta _{\mathcal {P}}(i)\ge 0$ and $\sum _{i}\beta _{ \mathcal {P}}(i)=1$) such that the preference $\preceq _{\mathcal {P}}$ is represented by the function $U_{\mathcal {P}}$, $U_{\mathcal {P} }(\sum _{i}l_{i}\otimes P_{i})=\sum _{i}u_{\mathcal {P}}(l_{i})\beta _{\mathcal { P}}(i)$. Moreover, both $u_{\mathcal {P}}$ and $\beta _{\mathcal {P}}$ are unique.

Proof

Each Q-lottery $\sigma =\sum _{i}l_{i}\otimes P_{i}$ can be considered as a “horse” lottery $f:I\rightarrow \Delta (X)$, where $f(i)=l_{i}$. Moreover, due to axioms A1–A4 and Lemma 2, the relation $\preceq _{\mathcal {P}}$ satisfies all the Anscombe–Aumann axioms. Therefore, by theorem 13.2 in Fishburn (1970), we obtain an affine utility function $u_{\mathcal {P}}$ on $\Delta (X)$ and a probability measure $\beta _{\mathcal {P}}\in \Delta (I)$ that achieve the representation stated in Claim 1. The uniqueness of $\beta _{\mathcal {P}}$ is also given by this theorem. The uniqueness of $u_{\mathcal {P}}$ follows from normalization of $u_{\mathcal {P}}$. $\square$

Claim 2

The functions $u_\mathcal {P }$ are independent of $\mathcal {P }$ (and we denote them as u).

Proof

Due to A0, each of the functions $u_{\mathcal {P}}$ represents the derived preference $\preceq _{\Delta }$ on $\Delta (X)$. Therefore, they are positive affine transformations of each other. Normalization gives that they are in fact equal to each other. $\square$

Claim 3

For any Q-lottery $\sigma$, there exists an ordinary lottery $l\in \Delta (X)$ such that $\sigma$ is equivalent to l , that is $\sigma \approx l\otimes E$.

Proof

Let $\sigma =\sum _{i}l_{i}\otimes P_{i}$, and let $l^{b}$ ( $l^{w}$) be a best (a worst) lotteries among $(l_{i},i\in I)$. Due to A4, we have $\sum _{i}l^{w}\otimes P_{i}\preceq \sigma \preceq \sum _{i}l^{b}\otimes P_{i}$, and all these Q-lotteries have the same base $\mathcal {P} =(P_{i},i\in I)$. Therefore, we can apply Claim 1, which gives inequalities

$$\begin{aligned} u(l^{w})\le \sum _{i}u(l_{i})\beta _{\mathcal {P}}(i)\le u(l^{b}). \end{aligned}$$

Hence, $\sum _{i}u(l_{i})\beta _{\mathcal {P}}(i)=\alpha u(l^{w})+(1-\alpha )u(l^{b})=u(l)$ for some $\alpha \in [0,1]$, where $l=\alpha l^{w}+(1-\alpha )l^{b}$. By Claim 1, we have $\sigma \approx \sum _{i}l\otimes P_{i}\approx l\otimes E$. $\square$

Due to Claim 1, the function $U_{\mathcal {P}}$ allows comparing Q-lotteries with base $\mathcal {P}$. But we assert that it allows comparing Q-lotteries with different bases as well.

Claim 4

Let $\sigma =\sum _i l_i\otimes P_i$ be a Q-lottery with a base $\mathcal {P }=(P_i, i\in I)$, and let $\tau =\sum _j m_j\otimes Q_j$ be a Q-lottery with a base $\mathcal {Q }=(Q_j, j\in J)$. Then, $\sigma \preceq \tau$ if and only if $U_\mathcal {P }(\sigma )\le U_\mathcal {Q } (\tau )$.

Proof

Due to Claim 3, lottery $\sigma$ is equivalent to some lottery $l\otimes E$, or to the lottery $\sum _{i}l\otimes P_{i}$. Therefore, $U_{\mathcal {P}}(\sigma )=U_{\mathcal {P}}(\sum _{i}l\otimes P_{i})=u(l)$. Similarly, $\tau$ is equivalent to $m\otimes E$, and $U_{\mathcal {Q}}(\tau )=u(m)$. Now

$$\begin{aligned} \sigma \preceq \tau \Leftrightarrow l\otimes E\preceq m\otimes E\Leftrightarrow U_{\mathcal {P}}(\sigma )=u(l)\le u(m)=U_{\mathcal {Q}}(\tau ). \end{aligned}$$

$\square$

Let us return now to the functions $\beta _\mathcal {P }$. We assert that $\beta _\mathcal {P }(i)$ depends only on the operator $P_i$, not of $\mathcal { P }$ and i.

Claim 5

Let $\mathcal {P }=(P_1,\ldots ,P_n)$ and $Q=(Q_1, \ldots Q_k)$ be two measurement devices (bases), and $P_1=Q_1=R$. Then, $\beta _ \mathcal {P } (1)=\beta _\mathcal {Q }(1)$.

Proof

Consider Q-lottery $\sigma =l^{*}\otimes P_{1}+\sum _{i=2}^{n}l_{*}\otimes P_{i}$ with base $\mathcal {P}$. Its $\mathcal {P}$-utility $U_{\mathcal {P}}(\sigma )$ is equal to $\beta _{ \mathcal {P}}(1)$. Now let us form the auxiliary base $\mathcal {R}=(R,E-R)$ and the following Q-lottery $\rho =l^{*}\otimes R+l_{*}(E-R)$. Since $E-R=P_{2}+\ldots +P_{n}$, the lottery $\rho$ is equivalent to $\sigma$ (see Axiom A0). Therefore, $\mathcal {R}$-utility $U_{\mathcal {R}}(\rho )$ (which equals $\beta _{\mathcal {R}}(1)$) is, by Claim 4, equal to $\beta _{\mathcal { P}}(1)$. The same applies to $\mathcal {Q}$ and gives the equality $\beta _{ \mathcal {R}}(1)=\beta _{\mathcal {Q}}(1)$. Together with the equality $\beta _{\mathcal {R}}(1)=\beta _{\mathcal {P}}(1)$, we obtain the equality $\beta _{ \mathcal {P}}(1)=\beta _{\mathcal {Q}}(1)$. $\square$

As a consequence, we can speak about the number $\beta (P)$ for any “event” P, that is for any Hermitian operator P, $0\le P\le E$. $\beta (P)$ is $\mathcal {R}$-utility $U_{ \mathcal {R}}$ of the following Q-lottery $l^{*}\otimes P+l_{*}\otimes (E-P)$. Correspondingly, the utility of an arbitrary Q-lottery $\sigma =\sum _{i}l_{i}\otimes P_{i}$ can be rewritten as:

$$\begin{aligned} U(\sigma )=\sum _{i}u(l_{i})\beta (P_{i}). \end{aligned}$$

Obviously, $\beta (0)=0$ and $\beta (E)=1$. Moreover, $\beta (P)\ge 0$ for any “event” P, and $\sum _{i}\beta (P_{i})=1$ provided $\sum _{i}P_{i}=E$.

Claim 6

If $0\le P$, $0\le Q$, and $P+Q\le E$, then $\beta (P+Q)=\beta (P)+\beta (Q)$.

Proof

Indeed, consider the Q-lottery $l^{*}\otimes P+l^{*}\otimes Q+l_{*}\otimes (E-P-Q)$. Its utility is $\beta (P)+\beta (Q)$. On the other hand, due to A0, this lottery is equivalent to the lottery $l^{*}\otimes (P+Q)+l_{*}\otimes (E-P-Q)$, whose utility is $\beta (P+Q)$. $\square$

Claim 6 implies that $\beta$ can be extended to a (unique) linear functional $\beta$ on the vector space $\texttt {Herm}(H)$. Obviously, $\beta (A)\ge 0$ for $A\ge 0$, and $\beta (E)=1$. That is $\beta$ is a belief functional. This completes the proof of Theorem 1.

Appendix 3: Proof of Theorem 2

Proof of Assertion (a). Here, we can work with lotteries in the form of Hermitian operators. The utility U(A) of such an operator A is equal to $\mathbf {Tr}(AB)$ and the utility under the condition P is equal to $U(A|P)= \mathbf {Tr}(APBP)$ (up to the factor $\mathbf {Tr}(PBP)$).

Proof of property A5. We assume that $A|W=0$ and we have to show that $U(A|P)=0$. Note that $A|P=0$ is equivalent to $PAP=0$. Now $U(A|P)= \mathbf {Tr}(APBP)=\mathbf {Tr}(PAPB)=\mathbf {Tr}(0B)=0$.

Proof of property A6. Here, we assume that A commute with P and that $(E-P)A=0$ (that is $A=PA=AP$). We have to show that U(A|P) is equal to U(A). But $U(A|P)=\mathbf {Tr}(APBP)=\mathbf {Tr}(PAPB)=\mathbf {Tr} (AB)=U(A)$, because $PAP=A$.

Proof of Assertion (b). We define (for an arbitrary Q-lottery $\sigma =\sum _{i}x_{i}\otimes P_{i}$ and an event W given by a projector P) some special (adapted to P) Q-lottery that we denote $ad_{W}(\sigma )$. Roughly speaking, $ad_{W}(\sigma )=\sum _{i}x_{i}\otimes PP_{i}P$. However, the sum $\sum _{i}PP_{i}P$ is equal to P, not to E. Therefore, we add to this sum a tail-end $x_{*}\otimes (E-P)$. Here, $x_{*}$ is a prize with zero utility. The final formula yields

$$\begin{aligned} ad_{W}(\sigma )=\sum _{i}x_{i}\otimes PP_{i}P+x_{*}\otimes (E-P). \end{aligned}$$

Lemma 3

• (a) $\sigma |W=ad_{W}(\sigma )|W$;

• (b) the adapted lottery $ad_W(\sigma )$ is compatible with P;

• (c) for any Q-lottery $\sigma$ we have $ad_{W}(\sigma )|W^{\perp }=x_{*}\otimes (E-P)$.

Proof

• (a) It is obvious that $P_{i}|W=PP_{i}P|W$. Moreover, $(E-P)|W=P(E-P)=0$.

• (b) For any i, we have $PPP_iP=PP_iPP$, since $PP=P$. Moreover, $P(E-P)=(E-P)P=0$.

• (c) It is clear that $(E-P)PP_iP=0$ for any i.

$\square$

Proposition 4

Suppose that a preference relation $\preceq _{W}$ on the set $\mathbf {QL}(H)$ is a weak order and satisfies the axioms A5 and A6. Then, it is given by the following explicit formula (where $\sigma$ and $\tau$ are Q-lotteries on H):

$$\begin{aligned} \sigma \preceq _{W}\tau \;\text { if and only if }\; ad_{W}(\sigma )\preceq ad_{W}(\tau ). \end{aligned}$$

(1)

Proof

Indeed, due to the axiom A5 and Lemma 3, we have $\sigma \approx _{W}ad_{W}(\sigma )$ and $\tau \approx _{W}ad_{W}(\tau )$. Applying axiom A6 to the lotteries $ad_{W}(\sigma )$ and $ad_{W}(\tau )$ (which is possible due to points 2) and 3) of Lemma 3), we obtain that $ad_{W}(\sigma )\preceq ad_{W}(\tau )$ if and only if $ad_{W}(\sigma )\preceq _{W}ad_{W}(\tau )$. The rest follows from the transitivity of $\preceq _{W}$ . $\square$

We recall that the (unconditional) utility of lottery $ad_{W}(\sigma )$ is equal to $\mathbf {Tr}(Sh(ad_{W}(\sigma ))B)$. If $A=Sh(\sigma )$ then $Sh(ad_{W}(\sigma ))=PAP+0(E-P)=PAP$. Therefore, $U(ad_{W}(\sigma ))=\mathbf { Tr}(PAPB)=\mathbf {Tr}(APBP)=\mathbf {Tr}(PBP)\mathbf {Tr}(AB^{up})$, which is (up to the factor $\mathbf {Tr}(PBP)$) the utility of $\sigma$ with respect to updated belief operator $B^{up}=PBP/\mathbf {Tr}(PBP)$. This completes the proof of Theorem 2.

Appendix 4: Proof of Proposition 1

We here prove an assertion that generalizes Proposition 1 when we substitute projector P with an arbitrary “fuzzy-projector” F, that is an operator $0\le F\le E$ (see Remark 4 of Sect. 4). The posterior $B^{up}$ is given as $\sqrt{F}B \sqrt{F}/\mathbf {Tr}(\sqrt{F}B\sqrt{F})=\sqrt{F}B\sqrt{F}/\mathbf {Tr}(FB)$. We shall denote it as B|F.

Proposition $1^{\prime }$. The distance between a prior B and the posterior B|F is $O((1-\mathbf {Tr}(FB))^{1/4})$.

In other words, if $\varepsilon =1-\mathbf {Tr}(FB)$ then the distance between B and B|F is $O(\varepsilon ^{1/4})$.

Proof

Choose an orthonormal basis of H in which the operator F (as well as $\sqrt{F}$) has diagonal form

$$\begin{aligned} F=\texttt {diag}(f_{1},\ldots ,f_{n}). \end{aligned}$$

In this basis, operator B is represented by Hermitian matrix $(b_{ij})$, where i and j run over 1 to $n=\dim H$. The matrix $\sqrt{F}B\sqrt{F}$ has coefficients $\sqrt{f_{i}}b_{ij}\sqrt{f_{j}}$. The matrix $\sqrt{F}B \sqrt{F}$ differs of $B|F=\sqrt{F}B\sqrt{F}/Tr(FB)$ by less than $O(\varepsilon )$. Thus, we need to compare B and $\sqrt{F}B\sqrt{F}$ and to show that the distance between them is $O(\varepsilon ^{1/4})$. Or equivalently we need to show that (for any i and j) distance between $b_{ij}$ and $\sqrt{f_{i}}b_{ij}\sqrt{f_{j}}$ is $O(\varepsilon ^{1/4})$ .

Let us divide the set of indices i from $\{1,\ldots ,n\}$ into two groups. Say that an index i is non-essential, if $b_{ii}\le \sqrt{\varepsilon }$, and is essential in the opposite case. $\square$

Lemma 4

If i is non-essential, then $|b_{ij}|\le \varepsilon ^{1/4}$.

Proof

A sub-matrix of the matrix B, formed by the rows and columns from the set $\{i,j\}$, is Hermitian and therefore has non-negative determinant $b_{ii}b_{jj}-|b_{ij}|^{2}$. That is $\sqrt{\varepsilon }\ge b_{ii}b_{ij}\ge |b_{ij}|^{2}$. $\square$

Due to Lemma 4, if i or j is non-essential then the coefficient $b_{ij}$ of the matrix B as well as the corresponding coefficient $b_{ij}\sqrt{f_{i} }\sqrt{f_{j}}$ of the matrix $\sqrt{F}B\sqrt{F}$ is $\le \varepsilon ^{1/4}$. Therefore, the distance between them is $\le 2\varepsilon ^{1/4}$.

Hence, we can suppose that the both indexes i and j are essential.

Lemma 5

If an index i is essential then $f_i=1+O(\varepsilon ^{1/2})$.

Proof

We have $\mathbf {Tr}(FB)=1-\varepsilon$, that is $\sum _{i}f_{i}b_{ii}=1-\varepsilon$. Moreover, the trace of B is equal to 1, that is $\sum _{i}b_{ii}=1$. Substracting we obtain that $\sum _{i}b_{ii}(1-f_{i})=\varepsilon$, therefore each term of this sum is less or equal to $\varepsilon$, $b_{ii}(1-f_{i})\le \varepsilon$. In particular, if i is essential then $b_{ii}>\varepsilon ^{1/2}$ and $1-f_{i}<\varepsilon ^{1/2}$, that is $f_{i}=1+O(\varepsilon ^{1/2})$. $\square$

As a corollary we obtain (for an essential i) that $\sqrt{f_i} =1+O(\varepsilon ^{1/2})$.

Let us return to evaluation of the distance between $b_{ij}$ and $b_{ij} \sqrt{f_{i}}\sqrt{f_{j}}$ in the case when i and j are essential indexes. It is clear that $b_{ij}-b_{ij}\sqrt{f_{i}}\sqrt{f_{j}} =b_{ij}(1-(1+O(\varepsilon ^{1/2}))(1+O(\varepsilon ^{1/2})))$. The module of this number is no more than $|b_{ij}|$ (which is $\le 1$) multiplied by $Q(\varepsilon ^{1/2})$. Therefore, the distance is no more than $\le O(\varepsilon ^{1/2})\le O(\varepsilon ^{1/4})$ which proves Proposition $1'$. $\square$