A note on burdens of proof in civil litigation

Abstract

It has been widely believed that who bears the burden of proof significantly affects the incentives of the legal parties. In particular, Hay and Spier (J Legal Stud 26:413–431, 1997) argues that if legal parties have a commonly accessible body of evidence (perfectly correlated pieces of evidence), the party who bears the burden of proof will present the evidence if and only if the evidence supports his position, while the other party (without the burden) will refrain from presenting it regardless of whether the evidence supports his position. In this paper, I claim that the result will be dramatically changed if the pieces of evidence that each party possesses are not perfectly correlated. I show that each party will present the evidence that supports his position whenever available, regardless of the burden of proof assignment. This implies that allocating the burden of proof does not matter in terms of information elicitation.

Appendix

1.1 Proof of Proposition 1

Claim 1

It is the dominant strategy for the party who bears the burden of proof to present the available evidence.

Proof

Without loss of generality, assume that the burden of proof is placed on the plaintiff. $\square$

First, I will show that the best response of P to presenting evidence is to present $Y_P$. If he does not present $Y_P$ so that D is the only one who presents evidence, the conditional probability that D was negligent can be similarly calculated as

$$\begin{aligned} P\left( x<\overline{x}\mid Y_{P}^{C}, Y_{D}\right) &= \frac{P\left( x< \overline{x}, Y_{P}^{C}, Y_{D}\right) }{P\left( Y_{P}^{C}, Y_{D} \right) } \nonumber \\ &= \frac{P\left( Y_{P}^{C}, Y_{D}\mid x<\overline{x}\right) P\left( x<\overline{x}\right) }{P\left( Y_{P}^{C}, Y_{D}\mid x<\overline{x}\right) P\left( x<\overline{x}\right) + P\left( Y_{P}^{C}, Y_{D}\mid x\ge \overline{x}\right) P\left( x\ge \overline{x}\right) }, \end{aligned}$$

(1)

where $Y_{i}$ is the event that the evidence is available to i and $Y_{i}^{C}$ is the complement of the event.¹⁹

If P presents evidence as well as D, the conditional probability will be

$$\begin{aligned} P\left( x< \overline{x}\mid Y_{P}, Y_D \right) &= \frac{P\left( x< \overline{x}, Y_{P}, Y_{D} \right) }{P\left( Y_{P}, Y_{D}\right) } \nonumber \\ &= \frac{P\left( Y_{P}, Y_{D} \mid x<\overline{x}\right) P\left( x< \overline{x}\right) }{P\left( Y_{P}, Y_{D} \mid x< \overline{x}\right) P\left( x< \overline{x}\right) +P\left( Y_{P}, Y_{D}\mid x\ge \overline{x}\right) P\left( x\ge \overline{x}\right) }. \end{aligned}$$

(2)

Taking inverses yield

$$\left[ P\left( x<\overline{x}\mid Y_{P}^{C}, Y_{D}\right) \right] ^{-1} = 1+ \frac{ P\left( Y_{P}^{C}, Y_{D}\mid x\ge \overline{x}\right) P\left( x\ge \overline{x}\right) }{P\left( Y_{P}^{C}, Y_{D}\mid x<\overline{x}\right) P\left( x<\overline{x}\right) },$$

(3)

$$\left[ P\left( x< \overline{x}\mid Y_{P}, Y_D \right) \right] ^{-1}= 1+ \frac{P\left( Y_{P}, Y_{D}\mid x\ge \overline{x}\right) P\left( x\ge \overline{x}\right) }{P\left( Y_{P}, Y_{D} \mid x<\overline{x}\right) P\left( x< \overline{x}\right) }.$$

(4)

Then, it follows from Condition [CI] that $P(x< \overline{x}\mid Y_{P}, Y_D )> P(x< \overline{x}\mid P(Y_{P}^{C}, Y_{D} )$, since

$$\frac{P\left( Y_P, Y_D \mid x\ge \overline{x}\right) }{P\left( Y_P, Y_D \mid x< \overline{x}\right) }=\frac{ \left( 1-\lambda \right) P\left( Y_D \mid x \ge \overline{x}\right) }{\lambda P\left( Y_D \mid x< \overline{x}\right) }<\frac{\lambda P\left( Y_D \mid x \ge \overline{x}\right) }{\left( 1-\lambda \right) P\left( Y_D \mid x< \overline{x}\right) }=\frac{P\left( Y_{P}^{C}, Y_D \mid x\ge \overline{x}\right) }{P\left( Y_{P}^{C}, Y_D \mid x< \overline{x}\right) }.$$

Second, I will show that the best response of P to not presenting evidence is to present $Y_P$ whenever it is available. If P presents the evidence the judge’s inference will proceed based on the evidence as follows.

$$\begin{aligned} P\left( x<\overline{x}\mid Y_{P}, Y_{D}^{C}\right) &= \frac{P\left( x< \overline{x}, Y_{P}, Y_{D}^{C} \right) }{P\left( Y_{P}, Y_{D}^{C}\right) } \nonumber \\ &= \frac{P\left( Y_{P}, Y_{D}^{C}\mid x< \overline{x}\right) P\left( x< \overline{x}\right) }{P\left( Y_{P}, Y_{D}^{C}\mid x< \overline{x}\right) P\left( x< \overline{x}\right) +P\left( Y_{P}, Y_{D}^{C}\mid x\ge \overline{x}\right) P\left( x\ge \overline{x}\right) }, \end{aligned}$$

(5)

and if he does not, the posterior belief will be

$$\begin{aligned} P\left( x< \overline{x}\mid Y_{P}^{C}, Y_{D}^{C}\right) &= \frac{P\left( x< \overline{x}, Y_{P}^{C}, Y_{D}^{C} \right) }{ P\left( Y_{P}^{C}, Y_{D}^{C}\right) } \nonumber \\ & = \frac{P\left( Y_{P}^{C}, Y_{D}^{C}\mid x< \overline{x}\right) P\left( x< \overline{x}\right) }{P\left( Y_{P}^{C}, Y_{D}^{C}\mid x<\overline{x}\right) P\left( x< \overline{x}\right) +P\left( Y_{P}^{C}, Y_{D}^{C}\mid x\ge \overline{x}\right) P\left( x\ge \overline{x}\right) }. \end{aligned}$$

(6)

Similarly, it can be shown that $P(x< \overline{x}\mid Y_{P}, Y_{D}^{C} )> P(x< \overline{x}\mid P(Y_{P}^{C}, Y_{D}^{C} )$. This means that presenting evidence is the dominant strategy of the plaintiff (with the burden) once the evidence is available.

Claim 2

It is the dominant strategy for the party who does not bear the burden of proof to present the available evidence.

Proof

Even if D does not bear the burden of proof, he can present the evidence voluntarily. Whether he presents $Y_D$ voluntarily or not depends on the relative size of $P(x\ge \overline{x} \mid Y_{D}^{C}, Y_P )$ and $P(x\ge \overline{x}\mid Y_D, Y_P )$ if P presents $Y_P$, and the relative size of $P(x\ge \overline{x} \mid Y_{D}^{C}, Y_{P}^{C})$ and $P(x\ge \overline{x}\mid Y_D, Y_{P}^{C})$ otherwise. Then, due to the symmetry between $Y_P$ and $Y_D$, the proof is immediate. $\square$

1.2 Proof of Proposition 2

I will use a fixed point argument to prove the existence of $\bar{c}$. Suppose there is some $\tilde{c}>0$ such that a legal party who has evidence does not present it for any $c\ge \tilde{c}$. Since $P( c\le \tilde{c})=F(\tilde{c})$ and F(c) is atomless, let $\alpha = P( c\ge \tilde{c})=1-P(c\le \tilde{c})=1-F(\tilde{c})$. That is, $\alpha$ is the probability that a legal party who has evidence does not present it. If P does not present evidence, the judge believes that P either has no evidence or has the evidence but the disclosure cost is high ($c\ge \tilde{c}$). Under [XI] condition, Bayesian updating implies that

$$\begin{aligned} P\left( x<\overline{x}\mid \emptyset , Y_D \right) &= \frac{P\left( x<\overline{x}, \, Y_D \right) }{P\left( {\emptyset}, Y_D \right) } \nonumber \\ &= \frac{\left( 1-\lambda \right) q+\alpha \lambda q}{\left( 1-\lambda \right) q+\lambda (1-q)+\alpha \left\{ \lambda q+\left( 1-\lambda \right) (1-q)\right\} }, \end{aligned}$$

(7)

where $\emptyset$ is the event that no evidence is presented. If P presents evidence, the conditional probability is simply as

$$\begin{aligned} P\left( x<\overline{x}\mid Y_{P}, Y_{D}\right) &= \frac{P\left( x< \overline{x}, Y_{P}, Y_{D} \right) }{P\left( Y_{P}, Y_{D}\right) } \nonumber \\ &= \frac{P\left( Y_{P}, Y_{D}\mid x< \overline{x}\right) P\left( x< \overline{x}\right) }{P\left( Y_{P}, Y_{D}\mid x< \overline{x}\right) P\left( x< \overline{x}\right) +P\left( Y_{P}, Y_{D}\mid x\ge \overline{x}\right) P\left( x\ge \overline{x}\right) } \nonumber \\ &= \frac{\lambda q}{\lambda q+\left( 1-\lambda \right) (1-q)}. \end{aligned}$$

(8)

To compare (7) and (8), I need the following lemma.

Lemma 1

$\frac{Q}{R}>\frac{S}{T}$ implies that $\frac{Q}{R}>\frac{S+\alpha Q}{T+\alpha R}$.

Proof

$\psi \equiv Q(T+\alpha R)-R(S+\alpha Q)=QT-RS$ since $\frac{Q}{R}>\frac{S}{T}$. $\square$

It directly follows from Lemma 1 that $P(x<\overline{x}\mid Y_P, Y_{D})>P(x<\overline{x}\mid {\emptyset}, Y_{D})$ for any $\alpha \in [0,1]$. Let $\Delta (\alpha )=P(x<\overline{x}\mid Y_{P}, Y_{D})-P(x<\overline{x}\mid {\emptyset}, Y_{D})>0$. Note that $\Delta (\alpha (\tilde{c}))$ is increasing in $\alpha$ so decreasing in $\tilde{c}$, since $\alpha$ is decreasing in $\tilde{c}$. Therefore, there exists some value for $\tilde{c}$ such that

$$w\Delta \left( \alpha \left( \tilde{c}\right) \right) =\tilde{c}.$$

(9)

We will denote the value for $\tilde{c}$ (i.e., satisfying Eq. 9) by $\bar{c}$. That is, $\bar{c}$ is a fixed point of the function of $c, w\Delta (\alpha (c))$. Clearly, P strictly prefers presenting evidence if $c<\bar{c}$, and strictly prefers not presenting evidence if $c>\bar{c}$. This implies that if $c<\bar{c}$, presenting evidence is the dominant strategy of P regardless of whether or not D presents $Y_D$. By symmetry, presenting evidence is also the dominant strategy of D if $c<\bar{c}$. Since this is the case regardless of the burden of proof assignment, the proof is completed.