Aggregating Information

For every $p,q$, ${\mathbb{E}}_q[(p-X{)}^2]=q(p-1{)}^2+(1-q)p^2=(p-q{)}^2+q(1-q)$: the Brier penalty is strictly proper, the identity is proved by expanding the squares and checked in exact arithmetic on $1681$ rational pairs. The absolute-error penalty satisfies ${\mathbb{E}}_q|p-X|=q+p(1-2q)$, affine in $p$: it is minimised at $p=0$ for $q<\frac{1}{2}$, at $p=1$ for $q>\frac{1}{2}$, and at every $p\in [0,1]$ when $q=\frac{1}{2}$ — never uniquely at an interior belief. At $q=\frac{2}{5}$ the honest report expects $\frac{12}{25}$ and the report $0$ expects $\frac{2}{5}$: a forecaster evaluated by absolute error is paid to round every uneven belief to the nearer certainty. The logarithmic penalty is also strictly proper (Good 1952); it is cited, not computed, since its values are not rational.

The forecast path is $\frac{1}{2}\to \frac{3}{4}\to \frac{9}{10}\to \frac{3}{4}$ and the realised payments telescope: the sponsor's total cost is $(p_0-X{)}^2-(p_K-X{)}^2$ — here $\frac{3}{16}$ if the event occurs and $-\frac{5}{16}$ (a gain) if not — and never exceeds $(p_0-X{)}^2$, which is at most $\frac{1}{4}$ for $p_0=\frac{1}{2}$ (and $\max \{p_0^2,(1-p_0{)}^2\}$ in general). By the identity of Proposition 10.2 (p. 414), a trader's expected payment under belief $q$ is $(p_{i-1}-q{)}^2-(p_i-q{)}^2$, maximised exactly at the honest $p_i=q$: a one-shot trader has nothing to gain from distortion, whatever the others do.

On agreeing good signals the Bayes pool reports $\frac{9}{10}$ where the average reports $\frac{3}{4}$; on disagreement both report $\frac{1}{2}$. The exact expected Brier scores are $\frac{3}{16}$ for one expert, $\frac{21}{128}$ for the average, and $\frac{3}{20}$ for the Bayes pool: averaging two experts with conditionally independent signals recovers five eighths of the improvement that their combined evidence allows and forgoes the other three eighths, because two agreeing moderate probabilities are stronger evidence than either alone, and an average cannot say so. The comparison assumes the stated independence; perfectly correlated experts would make the Bayes pool overconfident, and the rule, not the slogan, carries the assumption.

The identity holds exactly on the table: forecaster A scores $\frac{4}{25}=\frac{1}{600}-\frac{49}{600}+\frac{6}{25}$; the climatology forecaster scores $\frac{6}{25}=0-0+\frac{6}{25}$. Climatology is perfectly calibrated — in every group the issued probability equals the observed frequency — and has zero resolution: it never distinguishes one event from another. Forecaster A is slightly miscalibrated and much more resolving, and wins. Calibration is a virtue and an empty one on its own; skill lives in resolution, net of miscalibration. For the Brier score the three parts add with these signs; the statement is about this decomposition and this table, not about every proper score.

(a) For any finite joint law of the closing price $p_K$ and $X$, the sponsor's expected payment is

(times $b$ in general): the subsidy pays for resolution and is refunded by miscalibration.

(b) In the insider variant every report is honest, the insider moves the price to $X$, and the analyst's signal is worth $\mathbb{E}[(Q-\pi {)}^2]=\frac{1}{16}$ when the price is unmoved, where $Q$ is its posterior. The analyst buys the signal if and only if $(1-\rho )\frac{1}{16}\ge \frac{1}{32}$, that is $\rho \le {\rho }^{\ast }=\frac{1}{2}$. At every $\rho$ the closing price is calibrated, $\mathrm{Rel}(p_K)=0$; its expected Brier score is

the sponsor's expected cost equals the resolution, which is $\frac{1}{4}$ minus that score; the insider's expected profit is $\rho \cdot \frac{1}{4}$; the analyst's expected net profit is $(1-\rho )\frac{1}{16}-\frac{1}{32}\ge 0$ while it researches and $0$ after.

At ${\rho }^{\ast }$ the Brier score jumps from $\frac{3}{32}$ to the limit $\frac{1}{8}$ and returns to $\frac{3}{32}$ only at ${\rho }^{\circ }=\frac{5}{8}$: on $(\frac{1}{2},\frac{5}{8})$ the price is less informative than at ${\rho }^{\ast }$, although its Brier score $(1-\rho )\frac{1}{4}$ still falls as $\rho$ rises.

(c) In the stake variant, a price-setter with $q<\tau$ reports $\tau$ if $K>b(\tau -q{)}^2$ and $q$ otherwise. For $0<g\le \tau$, honesty at every belief at least $g$ below $\tau$ therefore holds if and only if $b\ge b^{\ast }(g)=K/g^2$, and at scale $b$ every belief in $[0,1]\cap (\tau -\sqrt{K/b},\tau )$ is reported as $\tau$. In the base values the band is $(\frac{3}{8},\frac{1}{2})$ (a belief of $\frac{2}{5}$ gains $\frac{9}{1600}$ by reporting $\frac{1}{2}$), and $b^{\ast }=\frac{1}{4},1,\frac{25}{16},4,\frac{25}{4}$ for $g=\frac{1}{4},\frac{1}{8},\frac{1}{10},\frac{1}{16},\frac{1}{20}$; since the sponsor's payment never exceeds $b\max \{p_0^2,(1-p_0{)}^2\}$ (Proposition 10.4 (p. 415)), honesty at distance $g$ is bought with a worst-case liability of $K/(4g^2)$ when $p_0=\frac{1}{2}$.

(a) At $\pi =\frac{1}{2}$ the first quotes are $\frac{1}{2}\pm \mu (2a-1)/2$, so the spread is $\mu (2a-1)$; an informed order gains $(1-\mu )(2a-1)/2$ and an uninformed order loses $\mu (2a-1)/2$ in expectation; the dealer's expected profit is zero, so the informed traders' total gain equals the uninformed traders' total loss. At base values the quotes are $\frac{5}{8}$ and $\frac{3}{8}$, the spread $\frac{1}{4}$, the gain $\frac{3}{8}$ and the loss $\frac{1}{8}$; at $a=\frac{3}{4}$ the spread is $\frac{1}{8}$, the gain $\frac{3}{16}$ and the loss $\frac{1}{16}$.

(b) At $a=1$ and any prior $\pi$, an informed order gains $2\pi (1-\pi )(1-\mu )/(1-{\mu }^2(2\pi -1{)}^2)$ in expectation. At $\pi =\frac{1}{10}$ and $\mu =\frac{1}{4}$ the quotes are $\frac{5}{32}$ and $\frac{1}{16}$ and the gain is $\frac{9}{64}$.

For $a=1$, every $N$ and every $\mu$, the pre-statement price is calibrated, so that $\mathbb{E}[p_{N}X]=\mathbb{E}[p_N^2]$, and

At $\pi =\frac{1}{2}$ and $N=1$ the run-up share is ${\mu }^2$; at base values it is $\frac{1}{16}$ for $N=1$ and $\frac{2}{17}$ for $N=2$. At $\pi =\frac{1}{2}$, for every $a>\frac{1}{2}$, the statement-conditioned run-up $R_S$ does not depend on $a$ and equals the share at $a=1$, while the run-up share itself is $R_X=(2a-1{)}^2{R}_S$ and, for $\mu >0$, falls as the statement is believed less: at $a=\frac{3}{4}$, $\mu =\frac{1}{4}$ and $N=1$, $\mathbb{E}[p_1\mid X=1]=\frac{65}{128}$ and $\mathbb{E}[p_1\mid S=h]=\frac{33}{64}$, so $R_X=\frac{1}{64}$ while $R_S=\frac{1}{16}$; at $a=1$ the two coincide, and at $a=\frac{1}{2}$, $R_X=0$ and $R_S$ is undefined. The informed traders' expected total gain over a window is $\frac{3}{32}$ for $N=1$ and $\frac{99}{544}$ for $N=2$; an informed second order gains $\frac{6}{17}$, less than the first order's $\frac{3}{8}$, because the price has already learned.

(a) The probability that an audit identifies at least one informed order among $n$ orders is $1-(1-\mu s{)}^n$. At base values it is $\frac{1}{40}$ per order, $\frac{79}{1600}$ for $n=2$ and $0.3973$ for $n=20$ (truncated); on the grid of step $\frac{1}{100}$ the smallest sampling rate giving at least $\frac{1}{2}$ over twenty orders is $\frac{7}{50}$, which gives $0.5096$.

(b) The probability that a flagged order is informed is $\mu t/(\mu t+(1-\mu )f)$: $\frac{16}{19}$ at base values, and $\frac{16}{115}$ at $\mu =\frac{1}{100}$, where $\frac{23}{400}$ of all orders are flagged.

(a) In the proportional variant the informed trader's expected net gain is $G(1-s(1+k))$: for every positive gain $G>0$ it trades if and only if $s<1/(1+k)$, a threshold that no other parameter of the one-slot contract moves, while at $G=0$ the tie rule selects no trade at every sampling rate; at $k=4$ the threshold is $\frac{1}{5}$.

(b) In the fixed variant the equilibrium informed share is ${\mu }^{\ast }=\lambda$ if $(1-\lambda )/2\ge sF$, ${\mu }^{\ast }=1-2sF$ if $(1-\lambda )/2<sF<\frac{1}{2}$, and ${\mu }^{\ast }=0$ if $sF\ge \frac{1}{2}$; the spread equals ${\mu }^{\ast }$ and an uninformed order loses ${\mu }^{\ast }/2$. For $F>0$ and $\lambda >0$, expected identifications per order, ${\mu }^{\ast }s$, are largest at $s=\min \{1,\max \{(1-\lambda )/(2F),1/(4F)\}\}$: when the inner expression exceeds $1$ the identifications still increase at $s=1$ and the maximum sits at the boundary; at $\lambda =0$ no informed order exists and identifications vanish at every sampling rate. At base values the kink is at $s=\frac{1}{15}$, where identifications peak at $\frac{1}{30}$; trading is fully deterred from $s=\frac{2}{15}$; at $s=\frac{1}{20}$ and $s=\frac{1}{10}$ identifications are both $\frac{1}{40}$ while ${\mu }^{\ast }$ is $\frac{1}{2}$ and $\frac{1}{4}$ (Table 10.3 (p. 409)).