Subjective probability, confidence, and Bayesian updating

Abstract

I define subjective probabilities for an ambiguity averse agent who is given an exogenous information set $\varDelta$ containing the true probability law on the state space S. The agent in my model evaluates every uncertain prospect via a mixture of the least favorable scenario in $\varDelta$ and her unique subjective belief p. This parametric utility structure is characterized by four standard axioms—order, continuity, monotonicity, and Independence—where the last two conditions are modified via the information set $\varDelta$. Any pair of preferences $\succeq _1$ and $\succeq _2$ in my model can be compared in terms of confidence—the weights attached to the subjective beliefs $p_1$ and $p_2$, respectively. The corresponding behavioral condition extends Epstein’s comparative ambiguity aversion that leaves some rankings $\succeq _1$ and $\succeq _2$ unrelated. Moreover, I relax the well-known dynamic consistency principle and characterize the Bayesian updating rule for the belief p conditional on any non-null event in S.

Appendix: Proofs

1.1 Proof of Theorem 1

Fix a non-empty set $\varDelta \subset \mathcal {P}$ and a utility index $u\in \mathcal {U}$. For any $f\in \mathcal {H}$, let

$$\begin{aligned} M(f) = \inf _{q\in \varDelta } u(f(q)) = \min _{q\in \mathrm {cl}(\mathrm {co}\, D)} u(f(q)). \end{aligned}$$

Suppose that $\succeq$ is represented by

$$\begin{aligned} U(f) = (1-\varepsilon )u(f(p)) + \varepsilon M(f) \end{aligned}$$

for some $\varepsilon \in [0,1]$ and $p\in \mathrm {cl}(\mathrm {co}\, \varDelta )$, $u\in \mathcal {U}$. Order and Continuity are obvious. Note that $l\in L(f, \varDelta )$ implies $u(l) \le M(f) \le u(f(p))$, and hence, $u(l) \le U(f)$. Thus $\varDelta$-Monotonicity holds. To verify $\varDelta$-Independence, take $\alpha \in [0,1]$, acts $f,g,h\in \mathcal {H}$, and a lottery $l\in \mathcal {L}$ such that $f \succeq g$, $\alpha {f}+(1-\alpha ) h$ is more secure than $\alpha {f}+(1-\alpha ) l$, but $\alpha {g}+(1-\alpha ) h$ is less secure than $\alpha {g}+(1-\alpha ) l$. Then

$$\begin{aligned}&a_1 = U\left( \alpha {f}+(1-\alpha ) l\right) - U\left( \alpha {g}+(1-\alpha ) l\right) \ge 0 \\&a_2 = M\left( \alpha {f}+(1-\alpha ) h\right) - M\left( \alpha {f}+(1-\alpha ) l\right) \ge 0 \\&a_3 = M\left( \alpha {g}+(1-\alpha ) l\right) - M\left( \alpha {g}+(1-\alpha ) h\right) \ge 0 \\&U\left( \alpha {f}+(1-\alpha ) h\right) - U\left( \alpha {g}+(1-\alpha ) h\right) = a_1 + \varepsilon (a_2+ a_3) \ge 0 \end{aligned}$$

and hence, $\alpha {f}+(1-\alpha ) h\succeq \alpha {g}+(1-\alpha ) h$.

Turn to sufficiency. Suppose that $\succeq$ satisfies Order, Continuity, $\varDelta$-Monotonicity, and $\varDelta$-Independence. By Herstein–Milnor’s Theorem, the ranking of lotteries has an expected utility representation $u\in \mathcal {U}$ that is unique up to a positive linear transformation. If u is constant, then (3.2) is trivial. Assume that u is not constant.

Assume wlog that $\varDelta$ is convex and closed. Otherwise, replace $\varDelta$ by its closed convex hull. The security levels M(f) and all the proofs below are unaffected.

Recall that f is more secure than g if and only if $M(f) \ge M(g)$.

Lemma 1

For all $\alpha \in [0,1]$ and $f, g, h\in \mathcal {H}$, $SP(\alpha , f, h) \ge SP(\alpha , g, h)$ if and only if there is $l\in \mathcal {L}$ such that $\alpha {f}+(1-\alpha ) h$ is more secure than $\alpha {f}+(1-\alpha ) l$ and $\alpha {g}+(1-\alpha ) h$ is less secure than $\alpha {g}+(1-\alpha ) l$.

Proof

For all $\alpha \in [0,1]$, $f, g, h\in \mathcal {H}$, and $l\in \mathcal {L}$,

$$\begin{aligned}&M\left( \alpha {f}+(1-\alpha ) l\right) = \alpha {M(f)}+(1-\alpha ) u(l)\\&M\left( \alpha {g}+(1-\alpha ) l\right) = \alpha {M(g)}+(1-\alpha ) u(l). \end{aligned}$$

By definition of security premia,

$$\begin{aligned} SP(\alpha , f, h) - SP(\alpha , g, h)&= [ M\left( \alpha {f}+(1-\alpha ) h\right) - M\left( \alpha {f}+(1-\alpha ) l\right) ] \\&\quad + [ M\left( \alpha {g}+(1-\alpha ) l\right) - M\left( \alpha {g}+(1-\alpha ) h\right) ]. \end{aligned}$$

Therefore, if $\alpha {f}+(1-\alpha ) h$ is more secure than $\alpha {f}+(1-\alpha ) l$ and $\alpha {g}+(1-\alpha ) h$ is less secure than $\alpha {g}+(1-\alpha ) l$, then $SP(\alpha , f, h) \ge SP(\alpha , g, h)$. Conversely, suppose that $SP(\alpha , f, h) \ge SP(\alpha , g, h)$. Take $l\in \mathcal {L}$ such that $M\left( \alpha {f}+(1-\alpha ) h\right) = M\left( \alpha {f}+(1-\alpha ) l\right)$. Then $\alpha {f}+(1-\alpha ) h$ is more secure than $\alpha {f}+(1-\alpha ) l$ and

$$\begin{aligned} M\left( \alpha {g}+(1-\alpha ) l\right) - M\left( \alpha {g}+(1-\alpha ) h\right) = SP(\alpha , f, h) - SP(\alpha , g, h) \ge 0, \end{aligned}$$

that is, $\alpha {g}+(1-\alpha ) h$ is less secure than $\alpha {g}+(1-\alpha ) l$.

The preference $\succeq$ satisfies all conditions in Theorem 1 in Gilboa and Schmeidler Gilboa and Schmeidler (1989). In particular, for all $\alpha \in [0,1]$ and $f, g\in \mathcal {H}$,

$$\begin{aligned} SP(\alpha , f, g) \ge 0 = SP(\alpha , g, g). \end{aligned}$$

By Lemma 1 and $\varDelta$-Independence, $\succeq$ satisfies Uncertainty Aversion. Thus, there is a unique set convex and closed set $\varPi \subset \mathcal {P}$ such that $\succeq$ is represented by

$$\begin{aligned} U(f) = \min _{q\in \varPi } u(f(q)). \end{aligned}$$

(5.2)

Assume that S and X are finite, and $\varSigma = 2^S$ (the general case is treated separately). For any $a\in \mathbb {R}^S$, let

$$\begin{aligned} V\left( a\right) = \min _{q\in \varDelta } q\cdot a \quad \text {and}\quad W\left( a\right) = \min _{q\in \varPi } q\cdot a. \end{aligned}$$

(5.3)

If $W = V$, then $\succeq$ is extremely cautious, $\varPi = \varDelta$, and the utility representation (3.2) holds for $\varepsilon = 1$ and any $p\in \varDelta$.

Thus, assume that $W \ne V$. Lemmas 2 and 3 in Kopylov (2009) establish that

• there are unique $0\le \varepsilon < 1$ and $p\in \varDelta$ such that

  $$\begin{aligned} W(a) = \varepsilon V(a) + (1-\varepsilon ) p\cdot a \end{aligned}$$

  (5.4)

  for all $a\in \mathbb {R}^S$, and

• p is the only probability measure in $\mathcal {P}$ such that for all $f, g\in \mathcal {H}$, if f is more secure than g and $f(p) \succ g(p)$, then $f \succ g$.

All the notation and the proofs apply as is with the axiom of Cautious Independence renamed to $\varDelta$-Independence.

Representation (5.4) can be rewritten in the required form (3.2). Moreover, if $\succeq$ is not extremely cautious, then this representation is unique up to a positive linear transformation of u, and p is the only probability measure that satisfies the condition (3.3).

Extend the construction of the utility representation (3.2) for an arbitrary state space S with an algebra of events $\varSigma$. For any $A\in \varSigma$, let

$$\begin{aligned} \pi (A) = \min _{q\in \varPi }q(A) \quad \text {and}\quad \pi _*(A) = \min _{q\in \varDelta }q(A). \end{aligned}$$

Consider two cases.

Case 1. $\pi (A) = \pi _* (A)$ for all events $A\in \varSigma$. Let $\varepsilon = 1$ and take any $p\in \varDelta$.

Case 2. $\pi (A) > \pi _* (A)$ for some $A\in \varSigma$. Define $\varepsilon <1$ by

$$\begin{aligned} \varepsilon = \frac{1-\pi (A) -\pi (A^c)}{1- \pi _* (A) - \pi _*(A^c)}. \end{aligned}$$

For all events $B\in \varSigma$, take

$$\begin{aligned} p(B) = \frac{\pi (B) - \varepsilon \pi _*(B)}{1-\varepsilon }. \end{aligned}$$

In both cases, for any finite subalgebra $\varSigma ' \subset \varSigma$ such that $A\in \varSigma '$, the finite case of Theorem 1 implies that p is finitely additive on $\varSigma '$, and the preference $\succeq$ restricted to $\varSigma '$ measurable acts is represented by (3.2) with the triple $(u, \varepsilon , p)$. Thus, p is finitely additive on all of $\varSigma$, and $\succeq$ is represented by (3.2) with the triple $(u, \varepsilon , p)$ on all of $\mathcal {H}$. The inclusion $p\in \mathrm {cl}(\mathrm {co}\, \varDelta )$ follows from the fact that p is the limit of a net $p_{\varSigma '}\in \mathrm {cl}(\mathrm {co}\, \varDelta )$ in the weak* closed set $\mathcal {P}$.

1.2 Proofs of Theorems 2 and 3

Let preferences $\succeq _1$ and $\succeq _2$ have utility representations (3.2) with tuples $(\varepsilon _1, p_1, u_1)$ and $(\varepsilon _2, p_2, u_2)$. Each of the conditions (i), (ii), or (iii) of Theorem 2 implies that $\succeq _1$ equals $\succeq _2$ on the domain $\mathcal {L}$ of all lotteries, and hence, that the risk utility indices $u_1$ and $u_2$ can be taken equal $u_1 = u_2 = u$.

Let $\varPi _1 = \varepsilon _1 \varDelta + (1-\varepsilon _1)\{p_1\}$ and $\varPi _2 = \varepsilon _2 \varDelta + (1-\varepsilon _2)\{p_2\}$. GMM show that for all $f, g\in \mathcal {H}$,

$$\begin{aligned} \begin{aligned} f \succeq _1^* g \quad \Leftrightarrow \quad f(q) \succeq _1 g(q) \quad \text {for all }q\in \varPi _1\\ f \succeq _2^* g \quad \Leftrightarrow \quad f(q) \succeq _2 g(q) \quad \text {for all }q\in \varPi _2. \end{aligned}\end{aligned}$$

(5.5)

Each of the conditions of Theorem 2 is equivalent to the inclusion $\varPi _1 \subset \varPi _2$. For (i), it is established by GMM, for (ii) by Ghirardato and Marinacci (2002), and for (iii), it is shown directly via

$$\begin{aligned}&p_2 \in \tfrac{\varepsilon _1- \varepsilon _2}{1-\varepsilon _2}\mathrm {cl}(\mathrm {co}\, \varDelta ) + \tfrac{1 - \varepsilon _1}{1-\varepsilon _2} \{p_1\} \quad \Leftrightarrow \quad \\&\quad (1-\varepsilon _2)\{p_2\} + \varepsilon _2 \mathrm {cl}(\mathrm {co}\, \varDelta ) \subset (1-\varepsilon _1)\{p_1\} + \varepsilon _1 \mathrm {cl}(\mathrm {co}\, \varDelta ). \end{aligned}$$

Turn to Theorem 3. Suppose that $\varDelta$ is convex and closed and is not an interval (if the convex closed hull of $\varDelta$ is an interval, then $\varDelta$ is also an interval).

Let the preferences $\succeq _1$ and $\succeq _2$ have representations (3.2) with tuples $(u_1, p_1, \varepsilon _1)$ and $(u_2, p_3, \varepsilon _2)$, respectively. Take x, y such that $x \succ _1 y$ and $x \succ _2 y$. Let $u = u_1 = u_2$ on $\mathcal {L}(x, y)$.

Suppose that $\varepsilon _2 \le \varepsilon _1$. Take any $f, g\in \mathcal {H}$ such that $\tfrac{f + g}{2} \in \mathcal {L}(x, y)$. Then

$$\begin{aligned} \tfrac{f(p_1) + g(p_1)}{2} = \tfrac{f + g}{2}(p_1) = \tfrac{f + g}{2}(p_2) = \tfrac{f(p_2) + g(p_2)}{2} . \end{aligned}$$

Thus, either $u(f(p_2))) \ge u(f(p_1))$ or $u(g(p_2)) \ge u(g(p_1))$. Wlog, assume that $u(f(p_2)) \ge u(f(p_1))$. Let

$$\begin{aligned} M(f) = \min _{q\in \varDelta } u(f(q)) \le u(f(p_1)) \le u(f(p_2)). \end{aligned}$$

Then,

$$\begin{aligned} U_2(f) = \varepsilon _2 M(f) + (1-\varepsilon _2) u(f(p_2))\ge & {} \varepsilon _1 M(f) + (1-\varepsilon _1) u(f(p_2))\\\ge & {} \varepsilon _1 M(f) + (1-\varepsilon _1) u(f(p_1)) = U_1(f). \end{aligned}$$

Thus, $U_1(f) \ge u(l)$ implies that $U_2(f) \ge u(l)$ as well. It follows that $\succeq _2$ is more confident than $\succeq _1$.

Turn to sufficiency. Wlog assume that S is finite and $\varSigma = 2^S$. (Otherwise, take a finite algebra $\varSigma ' \subset \varSigma$ such that $\varDelta$ restricted to $\varSigma '$ is not an interval. Identify $S'$ with the elements of the finest partition generated by $\varSigma '$. Replace $(S, \varSigma )$ with $(S', 2^{S'})$.)

Suppose that $\succeq _2$ is more confident than $\succeq _1$, but $\varepsilon _2 > \varepsilon _1$. Consider two cases.

Case 1. $p_1 = p_2 = p$. Take $q, r\in \varDelta$ such that $p, q, r\in \mathbb {R}^S$ are linearly independent. Then, there exist $a, b \in \mathbb {R}^S$ and $\alpha , \beta \in \mathbb {R}$ such that

$$\begin{aligned} r - q = \alpha (p - r) + a \quad \text {and}\quad (p-r) \cdot a = 0\\ r- q = \beta (p- q) + b \quad \text {and}\quad (p-q) \cdot b = 0. \end{aligned}$$

It follows that $(r- q) \cdot a > 0$ and $(r-q ) \cdot b >0$. Thus $r \cdot a = p\cdot a > q\cdot a$ and $r \cdot b > p \cdot b = q\cdot b$, and

$$\begin{aligned} r \cdot c > p \cdot c > q\cdot c \end{aligned}$$

where $c = a+b$. Take $r', q' \in \varDelta$ such that

$$\begin{aligned} r' \cdot c \ge p' \cdot c \ge q' \cdot c \end{aligned}$$

for all $p' \in \varDelta$. Take a constant vector $\gamma \in \mathbb {R}^S$ such that

$$\begin{aligned} \varepsilon _1 (q'\cdot c) + (1-\varepsilon _1) (p\cdot c) = \varepsilon _1 (r'\cdot (\gamma -c )) + (1-\varepsilon _1) (p \cdot (\gamma -c )). \end{aligned}$$

Wlog let u be such that the vectors c and $\gamma -c$ belong to $[u(y), u(x)]^S$. Then, there exist acts $f, g: S \rightarrow \mathcal {L}(x,y)$ such that $u\circ f = c$ and $u\circ g=\gamma -c$. By construction, $\tfrac{f + g}{2}\in \mathcal {L}(x,y)$ and $U_1(f) = U_1 (g) = u(l)$ for some $l \in \mathcal {L}(x,y)$. Note that $p \cdot c > q'\cdot c$ and $p \cdot (\gamma -c) > r' \cdot (\gamma -c)$. As $\varepsilon _1 < \varepsilon _2$, then

$$\begin{aligned} U_2(f)&= \varepsilon _2 (q'\cdot c) + (1-\varepsilon _2) (p\cdot c) < \varepsilon _1 (q'\cdot c) + (1-\varepsilon _1) (p\cdot c) = U_1(f)\\ U_2(g)&= \varepsilon _2 (r'\cdot (\gamma -c)) + (1-\varepsilon _2) (p \cdot (\gamma -c)) < \varepsilon _1 (r'\cdot (\gamma -c))\\&\quad + (1-\varepsilon _1) (p \cdot (\gamma -c))= U_1(g). \end{aligned}$$

Hence, $f \succeq _1 l \succ _2 f$ and $g\succeq _1 l\succ _2 g$. Thus $\succeq _2$ is not more confident than $\succeq _1$.

Case 2. $p_1 \ne p_2$. Take $r\in \varDelta$ such that $p_1, p_2, r\in \mathbb {R}^S$ are linearly independent. Then there is $c\in \mathbb {R}^S$ and $\alpha \in \mathbb {R}$ such that

$$\begin{aligned} r - p_1 = \alpha (p_1 - p_2) + c \quad \text {and}\quad c \cdot (p_1 - p_2) = 0. \end{aligned}$$

Thus, $r \cdot c > p_1 \cdot c = p_2 \cdot c$. If there is $q\in \varDelta$ such that $p_1 \cdot c > q \cdot c$, then one can construct f and g analogously to Case 1 and show that $\succeq _2$ is not more confident than $\succeq _1$.

Assume that $p' \cdot c \ge p_1\cdot c$ for all $p'\in \varDelta$. Take $r' \in \varDelta$ such that $r' \cdot c \ge p'\cdot c$ for all $p'\in \varDelta$ as well. Let $d = c + \delta (p_1 - p_2)$ for arbitrary small $\delta >0$. Then, $(p_1 - p_2) \cdot d > 0$, and hence, $p_1 \cdot d > p_2 \cdot d$. Analogously to Step 1, construct acts $f, g\in \mathcal {H}$ such that $u \circ f = d$, $u \circ g = \gamma - d$, $\tfrac{f + g}{2} \in \mathcal {L}(x, y)$, and $U_1(f) = U_1(g) = u(l)$ for some $l \in \mathcal {L}(x, y)$. As $\varepsilon _2 > \varepsilon _1$, and $p_1 \cdot d > p_2 \cdot d$, then $U_2(f) < U_1(f) = u(l)$. As $\varepsilon _2 > \varepsilon _1$ and $r' \cdot (-c) < p_1 \cdot (-c) = p_2 \cdot (-c)$, then $\delta$ can be taken sufficiently small so that $U_2(g) < U_2(g) = u(l)$ as well. Thus, $\succeq _2$ is not more confident than $\succeq _1$.

1.3 Proofs of Theorem 4, Theorem 5, and Theorem 6

Suppose that E is a non-null event, preferences $\succeq$ and $\succeq _E$ satisfy Axioms 1–4 with information sets $\varDelta$ and $\varDelta |E$, respectively, and $\varDelta$-Dynamic Consistency holds.

Turn to Theorem 4. By Theorem 1, $\succeq$ and $\succeq _E$ have representations (3.2) with components $(\varepsilon , p, u)$ and $(\lambda , p_E, u)$ where $p\in \mathrm {cl}(\mathrm {co}\, \varDelta )$ and $p_E\in \mathrm {cl}(\mathrm {co}\, \varDelta |E)$, respectively. If $\succeq _E$ is extremely cautious, then $p_E$ is arbitrary and can be taken equal to p|E. Suppose that $\lambda <1$.

Assume that $p_E \ne p|E$. Let

$$\begin{aligned} V_E(a) = \min _{q\in \varDelta |E} q \cdot a. \end{aligned}$$

I claim that there are acts $f, g\in \mathcal {H}$ such that

$$\begin{aligned} p_E \cdot u(f) < p_E\cdot u(g), \quad (p|E)\cdot u(f) > (p|E)\cdot u(g), \text { and }V_E(u(f)) = V_E(u(g)). \end{aligned}$$

(5.6)

To construct such f and g, take an event $A \subset S$ such that $p_E(A) > (p|E)(A)$. Let $\pi _*(A) = \min _{q\in \varDelta |E}q(A)$ and $\pi _*(A^c) = \min _{q\in \varDelta |E}q(A^c)$. Take vectors $a, b\in \mathbb {R}^S$ such that

$$\begin{aligned} a_s = {\left\{ \begin{array}{ll} 1 - \pi _*(A) &{} \quad \text {if }s\in A \\ - \pi _*(A) &{} \quad \text {if }s\not \in A \end{array}\right. } \quad \text {and}\quad b_s = {\left\{ \begin{array}{ll} -\pi _*(A^c) &{} \quad \text {if }s\in A \\ 1-\pi _*(A^c) &{} \quad \text {if }s\not \in A. \end{array}\right. } \end{aligned}$$

By construction, $p_E\cdot a > (p|E)\cdot a$, $(p|E)\cdot b > p_E \cdot b$, $p_E\cdot a \ge V_E(a) = 0$, and $p_E\cdot b \ge V_E(b) = 0$, If $p_E\cdot a = p_E \cdot b$, then take $f, g\in \mathcal {H}$ such that $u(g) = a$ and $u(f) = b$ (the range of u is assumed to contain $[-1, 1]$.) If $p_E\cdot a \ne p_E \cdot b$, then take $f, g\in \mathcal {H}$ such that $u(g) = (p_E\cdot b) a$ and $u(f) = (p_E\cdot a) b$.

Take $l\in \mathcal {L}$ such that $u(l) = V_E(u(f)) = V_E(u(g))$. Then, for all $q\in \mathrm {cl}(\mathrm {co}\, \varDelta )$,

$$\begin{aligned} u(fEl(q))= & {} q(E) u(f(q|E)) + (1-q(E))u(l)\\\ge & {} q(E) V_E(f) + (1-q(E))u(l) = u(l). \end{aligned}$$

As u(f(q|E)) can be arbitrarily close to $V_E(f)$,

$$\begin{aligned} M(fEl) = \inf _{q\in \varDelta } u(fEl(q)) = u(l). \end{aligned}$$

Similarly, $M(gEl) = u(l)$.

It follows from (5.6) and (3.2) that fEl is less secure than gEl, f is more secure than g on E, $fEl \succeq gEl$ because $(p|E)\cdot u(f) > (p|E)\cdot u(g)$, but $g \succ _E f$ because $p_E \cdot u(f) < p_E\cdot u(g)$ and $\lambda <1$.

This is a contradiction with $\varDelta$-Dynamic Consistency. Thus, $p_E = p|E$.

To prove Theorem 6, proceed analogously, but take l such that $u(l) \le \min _{s\in S} u(f(s))$ and $u(l) \le \min _{s\in S} u(g(s))$. As E is surprising, then $\inf _{q\in \varDelta } q(E) = 0$, and hence, $M(fEl) = M(gEl) = u(l)$. Therefore, $p_E \ne p|E$ contradicts $\varDelta$-Dynamic Consistency as well.

To prove Theorem 5, observe first that $f E l \succeq l$ holds in my model if and only if $\min _{q\in \varPi |E} u(f(q)) \ge u(l)$. Let $\succeq '_E$ be the preference represented by $\min _{q\in \varPi |E} u(f(q))$. Then, property (4.2) is identical to the comparative ambiguity aversion (3.7) applied to rankings $\succeq _1 = \succeq '_E$ and $\succeq _2 = \succeq _E$. It follows that (4.2) holds if and only if $\varPi _E \subset \varPi |E$, that is,

$$\begin{aligned} \lambda \mathcal {P}(E) + (1-\lambda ) \{p|E\} \subset (\varepsilon \mathcal {P}+ (1-\varepsilon )\{ p\}) | E. \end{aligned}$$

Direct computation shows that

$$\begin{aligned} (\varepsilon \mathcal {P}+ (1-\varepsilon ) \{p\}) | E = \frac{\varepsilon }{(1-\varepsilon )p(E) + \varepsilon } \mathcal {P}(E) + (1-\varepsilon ) \frac{p(E)}{(1-\varepsilon ) p(E) + \varepsilon } \{p|E\}. \end{aligned}$$

The inequality

$$\begin{aligned} 1- \lambda \ge (1-\varepsilon ) \frac{p(E)}{(1-\varepsilon ) p(E) + \varepsilon } \end{aligned}$$

follows.