Conservativeness in jury decision-making

Abstract

This paper studies the three kinds of conservativeness in a jury decision-making structure: the voting rule, the threshold of reasonable doubt, and the legal information system. In a model of simultaneous voting, Feddersen and Pesendorfer (The American Political Science Review, 92(1), 23-35, 1998) argue that the unanimity rule is the worst-performing voting rule because voters with strategic behaviour mitigate the bias brought about by the voting rule. If this bias can be offset by an opposing (less conservative) bias in the legal information system, we would be able to restore the rationality of informative voting. The new informative equilibrium is strictly better than the original strategic equilibrium in terms of ex post error probability. When a strategic equilibrium survives, we can lower the probability of convicted defendant being innocent by making the information system more informative or by forming a jury with a higher (more conservative) threshold of reasonable doubt.

Appendix: Proofs

Proof of Lemma 1

The ‘more informative’ part

It is well known that Blackwell’s notion of being more informative is equivalent to the existence of a stochastic matrix that transforms one matrix to another. An information structure expressed by a matrix P is more informative than Q if and only if there exists a Markov matrix M such that

$$\begin{aligned} P\times M=Q \end{aligned}$$

Let

$$\begin{aligned}P=\left( \begin{array}{cc} \gamma _2 &{} 1-\delta _2 \\ 1-\gamma _2 &{} \delta _2 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned}Q=\left( \begin{array}{cc} \gamma _1 &{} 1-\delta _1 \\ 1-\gamma _1 &{} \delta _1 \\ \end{array} \right) \end{aligned}$$

Since $\gamma _2+\delta _2>1$, P is invertible. Then,

$$\begin{aligned}M=\frac{1}{\gamma _2+\delta _2-1}\left( \begin{array}{cc} \gamma _1\delta _2-(1-\gamma _2)(1-\delta _1) &{} \gamma _2(1-\delta _1)-\gamma _1(1-\delta _2) \\ (1-\gamma _1)\delta _2-(1-\gamma _2)\delta _1 &{} \gamma _2\delta _1-(1-\gamma _1)(1-\delta _2) \\ \end{array} \right) \end{aligned}$$

We need to evaluate the nonnegativity of each element in the matrix.

$M_{11}\ge 0$ iff $\gamma _1\delta _2-(1-\gamma _2)(1-\delta _1)\ge 0$, which is equivalent to $\frac{\delta _2}{1-\gamma _2}\ge \frac{1-\delta _1}{\gamma _1}$. From $\gamma _2+\delta _2>1$ we derive $\delta _2>1-\gamma _2$ and hence,

$$\begin{aligned} \frac{\delta _2}{1-\gamma _2}>1 \end{aligned}$$

(25)

From $\gamma _1+\delta _1>1$, it is derived that $\gamma _1>1-\delta _1$ and hence,

$$\begin{aligned} \frac{1-\delta _1}{\gamma _1}<1 \end{aligned}$$

(26)

Based on (25) and (26), the condition $M_{11}\ge 0$ always holds.

$M_{22}\ge 0$ iff $\gamma _2\delta _1-(1-\gamma _1)(1-\delta _2)\ge 0$, which is equivalent to $\frac{\gamma _2}{1-\delta _2}\ge \frac{1-\gamma _1}{\delta _1}$. From $\gamma _2+\delta _2>1$, it is derived that $\gamma _2>1-\delta _2$; hence,

$$\begin{aligned} \frac{\gamma _2}{1-\delta _2}>1 \end{aligned}$$

(27)

From $\gamma _1+\delta _1>1$, $\delta _1>1-\gamma _1$ is derived; hence,

$$\begin{aligned} \frac{1-\gamma _1}{\delta _1}<1 \end{aligned}$$

(28)

Based on (27) and (28), the condition $M_{22}\ge 0$ always holds.

$M_{21}\ge 0$ iff $(1-\gamma _1)\delta _2-(1-\gamma _2)\delta _1\ge 0$, which is equivalent to the following:

$$\begin{aligned} \frac{\delta _2}{1-\gamma _2}\ge \frac{\delta _1}{1-\gamma _1} \end{aligned}$$

(29)

$M_{12}\ge 0$ iff $\gamma _2(1-\delta _1)-\gamma _1(1-\delta _2)\ge 0$, which is equivalent to the following:

$$\begin{aligned} \frac{\gamma _2}{1-\delta _2}\ge \frac{\gamma _1}{1-\delta _1} \end{aligned}$$

(30)

Therefore, information system $(\gamma _2,\delta _2)$ is more informative than information system $(\gamma _1,\delta _1)$ if and only if (29) and (30) hold.

The ‘more conservative’ part

By construction,

$$\begin{aligned} \frac{\gamma (\kappa )}{1-\delta (\kappa )}=\frac{F(\kappa |I)}{F(\kappa |G)} \end{aligned}$$

The derivative of this with respect to $\kappa$ yields the following:

$$\begin{aligned} \frac{d}{d\kappa }\left( \frac{\gamma (\kappa )}{1-\delta (\kappa )}\right) =\frac{f(\kappa |I)F(\kappa |G)-F(\kappa |I)f(\kappa |G)}{F(\kappa |G)^2} \end{aligned}$$

Therefore, $\frac{\gamma (\kappa )}{1-\delta (\kappa )}$ is strictly decreasing iff

$$\begin{aligned} \frac{f(\kappa |I)}{f(\kappa |G)}<\frac{F(\kappa |I)}{F(\kappa |G)} \end{aligned}$$

(31)

The model assumes that $f(\eta )$ is a normal density, $G=1$ and $I=0$. By the property of normal distribution, the MLRP holds and for $\theta <\kappa$ and $I=0<G=1$ we have the following:

$$\begin{aligned} \frac{f(\kappa |G)}{f(\kappa |I)}>\frac{f(\theta |G)}{f(\theta |I)} \end{aligned}$$

Put differently,

$$\begin{aligned} \frac{f(\theta |I)}{f(\kappa |I)}>\frac{f(\theta |G)}{f(\kappa |G)} \end{aligned}$$

Integrating both sides of this with respect to $\theta$ from $-\infty$ to $\kappa$ yields the following:

$$\begin{aligned} \frac{F(\kappa |I)}{f(\kappa |I)}>\frac{F(\kappa |G)}{f(\kappa |G)}, \end{aligned}$$

which tells us that the condition (31) holds.

On the other hand,

$$\begin{aligned} \frac{\delta (\kappa )}{1-\gamma (\kappa )}=\frac{1-F(\kappa |G)}{1-F(\kappa |I)} \end{aligned}$$

The derivative of this with respect to $\kappa$ yields the following:

$$\begin{aligned} \frac{d}{d\kappa }\left( \frac{\delta (\kappa )}{1-\gamma (\kappa )}\right) =\frac{-f(\kappa |G)(1-F(\kappa |I)+f(\kappa |I)(1-F(\kappa |G)}{(1-F(\kappa |I))^2} \end{aligned}$$

Therefore,$\frac{\delta (\kappa )}{1-\gamma (\kappa )}$ is strictly increasing iff

$$\begin{aligned} \frac{f(\kappa |I)}{f(\kappa |G)}<\frac{1-F(\kappa |I)}{1-F(\kappa |G)} \end{aligned}$$

(32)

Provided by the property of MLRP, for $\theta '>\kappa$ and $G=1>I=0$,

$$\begin{aligned} \frac{f(\theta '|G)}{f(\theta '|I)}>\frac{f(\kappa |G)}{f(\kappa |I)} \end{aligned}$$

Put differently,

$$\begin{aligned} \frac{f(\theta '|G)}{f(\kappa |G)}>\frac{f(\theta '|I)}{f(\kappa |I)} \end{aligned}$$

and integrating both sides of this inequality with respect to $\theta '$ from $\kappa$ to $\infty$ yields the following:

$$\begin{aligned} \frac{1-F(\kappa |G)}{f(\kappa |G)}>\frac{1-F(\kappa |I)}{f(\kappa |I)} \end{aligned}$$

which is the same condition as (32). $\square$

Proof of Proposition 3

Note that $\left\{ piv,i\right\}$ is a joint event in the following inequality. Signals are conditionally independent, and pivotality is generated by the responsive strategies (a strategy that is a function of a signal). Hence, $\left\{ piv\right\}$ and $\left\{ i\right\}$ are independent events conditional on a true state.

Then, using Bayes’ rule, $\Pr (G|piv,i)$ is given by the following:

$$\begin{aligned} \Pr (G|piv,i)&=\frac{\Pr (piv,i|G)\Pr (G)}{\Pr (piv,i|G)\Pr (G)+\Pr (piv,i|I)\Pr (I)} \end{aligned}$$

(33)

$$\begin{aligned}&=\frac{\Pr (piv|G)\Pr (i|G)\Pr (G)}{\Pr (piv|G)\Pr (i|G)\Pr (G)+\Pr (piv|I)\Pr (i|I)\Pr (I)}\end{aligned}$$

(34)

$$\begin{aligned}&=\frac{\Pr (G|piv)\Pr (piv)(1-\delta )}{\Pr (G|piv)\Pr (piv)(1-\delta )+\Pr (I|piv)\Pr (piv)\gamma }\end{aligned}$$

(35)

$$\begin{aligned}&=\frac{\Pr (G|piv)(1-\delta )}{\Pr (G|piv)(1-\delta )+(1-\Pr (G|piv))\gamma } \end{aligned}$$

(36)

The first equality is a direct application of Bayes’ rule. The second equality comes from conditional independence. The third one uses the equalities of Bayes’ rule, $\Pr (piv|G)\Pr (G)=\Pr (G|piv)\Pr (piv)$ and $\Pr (piv|I)\Pr (I)=\Pr (I|piv)\Pr (piv)$.

Hence,

$$\begin{aligned} q\ge \frac{\Pr (G|piv)(1-\delta )}{\Pr (G|piv)(1-\delta )+(1-\Pr (G|piv))\gamma }, \end{aligned}$$

which gives the following:

$$\begin{aligned} \Pr (G|piv)(1-\delta )-q\Pr (G|piv)(1-\delta )\le q(1-\Pr (G|piv))\gamma \end{aligned}$$

Hence

$$\begin{aligned} \Pr (G|piv)\le \frac{q\gamma }{(1-\delta )(1-q)+q\gamma } \end{aligned}$$

(37)

Define $W_G$ as the probability that a voter votes for conviction, that is, uses strategy c when the defendant is guilty and $W_I$ as the probability that a voter votes for conviction when the defendant is innocent. Let $\nu (s)$ be the probability of voting for conviction in the presence of signal s. Then,

$$\begin{aligned} \Pr (c|G)=W_G=&\Pr (g|G)\nu (g)+\Pr (i|G)\nu (i)=\delta \nu (g)+(1-\delta )\nu (i) \end{aligned}$$

(38)

$$\begin{aligned} \Pr (c|I)=W_I=&\Pr (i|I)\nu (i)+\Pr (g|I)\nu (g)=\gamma \nu (i)+(1-\gamma )\nu (g) \end{aligned}$$

(39)

Based on this definition and the fact that conviction, C, is the event in which the pivotal voter votes for conviction, C is seen as the joint event of pivotality and the use of strategy; c. piv and c are conditionally independent for the same reason that piv and i are conditionally independent.

$$\begin{aligned} \Pr (G|piv,c)&=\frac{\Pr (piv,c|G)\Pr (G)}{\Pr (piv,c|G)\Pr (G)+\Pr (piv,c|I)\Pr (I)} \end{aligned}$$

(40)

$$\begin{aligned}&=\frac{\Pr (piv|G)\Pr (c|G)\Pr (G)}{\Pr (piv|G)\Pr (c|G)\Pr (G)+\Pr (piv|I)\Pr (c|I)\Pr (I)}\end{aligned}$$

(41)

$$\begin{aligned}&=\frac{\Pr (G|piv)\Pr (piv)W_G}{\Pr (G|piv)\Pr (piv)W_G+\Pr (I|piv)\Pr (piv)W_I}\end{aligned}$$

(42)

$$\begin{aligned}&=\frac{\Pr (G|piv)W_G}{\Pr (G|piv)W_G+(1-\Pr (G|piv))W_I} \end{aligned}$$

(43)

That is,

$$\begin{aligned} \Pr (G|C)=\frac{1}{1+\left( \frac{1}{\Pr (G|piv)}-1\right) \frac{W_I}{W_G}} \end{aligned}$$

(44)

Again, the first equality is a direct application of Bayes’ rule. The second equality comes from conditional independence. The third one uses the equalities of Bayes’ rule, $\Pr (piv|G)\Pr (G)=\Pr (G|piv)\Pr (piv)$ and $\Pr (piv|I)\Pr (I)=\Pr (I|piv)\Pr (piv)$.

On the other hand,

$$\begin{aligned}&\delta W_I-(1-\gamma )W_G \end{aligned}$$

(45)

$$\begin{aligned}&=\delta \gamma \nu (i)+\delta (1-\gamma )\nu (g)-(1-\gamma )\delta \nu (g)-(1-\gamma )(1-\delta )\nu (i)\end{aligned}$$

(46)

$$\begin{aligned}&=\left[ \delta \gamma -(1-\gamma )(1-\delta )\right] \nu (i) \end{aligned}$$

(47)

$$\begin{aligned}&=\left[ \gamma +\delta -1\right] \nu (i)\end{aligned}$$

(48)

$$\begin{aligned}&\ge 0 \end{aligned}$$

(49)

The last inequality comes from $2>\gamma +\delta >1$, which indicates that the information structure is informative but not perfectly informative and that $\nu (i)\ge 0$. From this, we have the following:

$$\begin{aligned} \frac{W_I}{W_G}\ge \frac{1-\gamma }{\delta } \end{aligned}$$

(50)

From (37)

$$\begin{aligned} \left( \frac{1}{\Pr (G|piv)}-1\right) \ge \frac{(1-\delta )(1-q)}{q\gamma } \end{aligned}$$

Combining this with (50) yields the following:

$$\begin{aligned} \left( \frac{1}{\Pr (G|piv)}-1\right) \frac{W_I}{W_G}\ge \frac{(1-\delta )(1-q)(1-\gamma )}{q\gamma \delta } \end{aligned}$$

(51)

Combining (51) with (44) yields the following:

$$\begin{aligned} \Pr (G|C)&\le \frac{1}{1+\frac{(1-\delta )(1-q)(1-\gamma )}{q\gamma \delta }} \end{aligned}$$

(52)

$$\begin{aligned}&=\frac{q\delta \gamma }{q\delta \gamma +(1-q)(1-\delta )(1-\gamma )} \end{aligned}$$

(53)

Therefore,

$$\begin{aligned} \Pr (I|C)&\ge \frac{(1-q)(1-\delta )(1-\gamma )}{q\delta \gamma +(1-q)(1-\delta )(1-\gamma )}\end{aligned}$$

(54)

$$\begin{aligned}&=\frac{1}{\left( \frac{q}{1-q}\right) \left( \frac{\delta }{1-\gamma }\right) \left( \frac{\gamma }{1-\delta }\right) +1} \end{aligned}$$

(55)

which completes the proof. $\square$