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Correction to Feddersen and Pesendorfer (1998)

Published online by Cambridge University Press:  15 June 2026

SAMI PETERSEN Open the ORCID record for SAMI PETERSEN [Opens in a new window]
SAMI PETERSEN*
Affiliation:
University of California, Berkeley , United States
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Abstract

The proof of Proposition 2 in Feddersen and Pesendorfer (1998, American Political Science Review 92(1): 23–35) is flawed. This note presents an elementary direct proof of the result.

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Type
Replication/Reappraisal
Information
American Political Science Review , First View , pp. 1
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Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (http://creativecommons.org/licenses/by/4.0), which permits unrestricted re-use, distribution and reproduction, provided the original article is properly cited.
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© The Author(s), 2026. Published by Cambridge University Press on behalf of American Political Science Association

In their seminal paper, Feddersen and Pesendorfer (Reference Feddersen and Pesendorfer1998, 32) make a squeeze argument to establish a limit (Proposition 2) using the inequalities

(6) $$ \begin{array}{rll}1+\frac{-1}{n}\ln f\ge {f}^{\frac{-1}{n-1}}\ge 1+\frac{-1}{n-1}\ln f,& & \end{array} $$ StartLayout 1st Row 1st Column 1 plus StartFraction negative 1 Over n EndFraction ln f greater than or equals f Superscript StartFraction negative 1 Over n minus 1 EndFraction Baseline greater than or equals 1 plus StartFraction negative 1 Over n minus 1 EndFraction ln f comma 2nd Column Blank 3rd Column Blank EndLayout

where $ f\in (0,1) $ f element of left parenthesis 0 comma 1 right parenthesis and $ n\ge 2 $ n greater than or equals 2 .Footnote 1 However, only the second inequality holds.Footnote 2 The argument, therefore, establishes just one bound on the quantity of interest.

This bound only serves to demonstrate that, as the jury size grows, the equilibrium probability of convicting an innocent defendant stays bounded away from one; not zero. Still, Proposition 2 is true. I provide a short proof below.

Claim. Let $ p\in (1/2,1) $ p element of left parenthesis 1 divided by 2 comma 1 right parenthesis , $ q\in (1-p,1) $ q element of left parenthesis 1 minus p comma 1 right parenthesis , and $ f:=\frac{(1-q)(1-p)}{qp} $ f colon equals StartFraction left parenthesis 1 minus q right parenthesis left parenthesis 1 minus p right parenthesis Over q p EndFraction . Then

$$ \begin{array}{rll}\underset{n\uparrow \infty }{\lim }{\left(\frac{p}{2p-1}{f}^{-\frac{1}{n-1}}-\frac{1-p}{2p-1}\right)}^{-n}={f}^{\frac{p}{2p-1}}.& & \end{array} $$ StartLayout 1st Row 1st Column limit Underscript n up arrow infinity Endscripts left parenthesis StartFraction p Over 2 p minus 1 EndFraction f Superscript minus StartFraction 1 Over n minus 1 EndFraction Baseline minus StartFraction 1 minus p Over 2 p minus 1 EndFraction right parenthesis Superscript negative n Baseline equals f Superscript StartFraction p Over 2 p minus 1 EndFraction Baseline period 2nd Column Blank 3rd Column Blank EndLayout

Proof. Let $ a:=\frac{p}{2p-1} $ a colon equals StartFraction p Over 2 p minus 1 EndFraction . For $ n\ge 2 $ n greater than or equals 2 , write $ {x}_n:=a\left(\right.{f}^{-\frac{1}{n-1}}-1\left)\right. $ x Subscript n Baseline colon equals a left parenthesis f Superscript minus StartFraction 1 Over n minus 1 EndFraction Baseline minus 1 right parenthesis and

$$ {\displaystyle \begin{array}{rcl}{h}_n:& =& \frac{p}{2p-1}{f}^{-\frac{1}{n-1}}-\frac{1-p}{2p-1}\\ {}& =& 1+a\left({f}^{-\frac{1}{n-1}}-1\right)=1+{x}_n.\end{array}} $$ StartLayout 1st Row 1st Column h Subscript n Baseline colon 2nd Column equals 3rd Column StartFraction p Over 2 p minus 1 EndFraction f Superscript minus StartFraction 1 Over n minus 1 EndFraction minus StartFraction 1 minus p Over 2 p minus 1 EndFraction 2nd Row 1st Column Blank 2nd Column equals 3rd Column 1 plus a left parenthesis f Superscript minus StartFraction 1 Over n minus 1 EndFraction Baseline minus 1 right parenthesis equals 1 plus x Subscript n Baseline period EndLayout

Note that $ {x}_n>0 $ x Subscript n Baseline greater than 0 (when $ n\ge 2 $ n greater than or equals 2 ) since $ a>0 $ a greater than 0 and $ f\in (0,1) $ f element of left parenthesis 0 comma 1 right parenthesis . So,

$$ {\displaystyle \begin{array}{rcl}\ln \left({\left({h}_n\right)}^{-n}\right)& =& -n\ln \left(1+{x}_n\right)\\ {}& =& -n\left[\frac{-\ln f}{n-1}\right]\frac{x_n}{\left[\frac{-\ln f}{n-1}\right]}\frac{\ln \left(1+{x}_n\right)}{x_n}\end{array}} $$ StartLayout 1st Row 1st Column ln left parenthesis left parenthesis h Subscript n Baseline right parenthesis Superscript negative n Baseline right parenthesis 2nd Column equals 3rd Column minus n ln left parenthesis 1 plus x Subscript n Baseline right parenthesis 2nd Row 1st Column Blank 2nd Column equals 3rd Column minus n left bracket StartFraction minus ln f Over n minus 1 EndFraction right bracket StartStartFraction x Subscript n Baseline OverOver left bracket StartFraction minus ln f Over n minus 1 EndFraction right bracket EndEndFraction StartFraction ln left parenthesis 1 plus x Subscript n Baseline right parenthesis Over x Subscript n Baseline EndFraction EndLayout

is well-defined. Using $ {f}^{-\frac{1}{n-1}}={e}^{\frac{-\ln f}{n-1}} $ f Superscript minus StartFraction 1 Over n minus 1 EndFraction Baseline equals e Superscript StartFraction minus ln f Over n minus 1 EndFraction ,

$$ \ln \left({\left({h}_n\right)}^{-n}\right)={\displaystyle \begin{array}{l}\underset{\to \ln f}{\underbrace{-n\left[\frac{-\ln f}{n-1}\right]}}\times \underset{\to \hskip0.3em a}{\underbrace{\frac{a({e}^{\left[\frac{-\ln f}{n-1}\right]}-1)}{\left[\frac{-\ln f}{n-1}\right]}}}\\ {}\times \underset{\to \hskip0.3em 1}{\underbrace{\frac{\ln \left(1+{x}_n\right)}{x_n}}}\hskip0.60em \underset{}{\overset{n\uparrow \infty }{\to }}\hskip0.60em a\ln f\end{array}} $$ ln left parenthesis left parenthesis h Subscript n Baseline right parenthesis Superscript negative n Baseline right parenthesis equals StartLayout 1st Row ModifyingBelow minus n left bracket StartFraction minus ln f Over n minus 1 EndFraction right bracket With bottom brace Underscript right arrow ln f Endscripts times ModifyingBelow StartStartFraction a left parenthesis e Superscript left bracket StartFraction minus ln f Over n minus 1 EndFraction right bracket Baseline minus 1 right parenthesis OverOver left bracket StartFraction minus ln f Over n minus 1 EndFraction right bracket EndEndFraction With bottom brace Underscript right arrow a Endscripts 2nd Row times ModifyingBelow StartFraction ln left parenthesis 1 plus x Subscript n Baseline right parenthesis Over x Subscript n Baseline EndFraction With bottom brace Underscript right arrow 1 Endscripts right arrow Overscript n up arrow infinity Endscripts a ln f EndLayout

since $ {\lim}_{n\uparrow \infty }{x}_n=0 $ limit Underscript n up arrow infinity Endscripts x Subscript n Baseline equals 0 while $ {\lim}_{y\to 0}\frac{\ln (1+y)}{y}=1 $ limit Underscript y right arrow 0 Endscripts StartFraction ln left parenthesis 1 plus y right parenthesis Over y EndFraction equals 1 and $ {\lim}_{y\to 0}\frac{e^y-1}{y}=1 $ limit Underscript y right arrow 0 Endscripts StartFraction e Superscript y Baseline minus 1 Over y EndFraction equals 1 . Therefore,

$$ \begin{array}{rll}\underset{n\uparrow \infty }{\lim }{({h}_n)}^{-n}={f}^a\equiv {f}^{\frac{p}{2p-1}}.& & \end{array} $$ StartLayout 1st Row 1st Column limit Underscript n up arrow infinity Endscripts left parenthesis h Subscript n Baseline right parenthesis Superscript negative n Baseline equals f Superscript a Baseline identical to f Superscript StartFraction p Over 2 p minus 1 EndFraction Baseline period 2nd Column Blank 3rd Column Blank EndLayout

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CONFLICT OF INTEREST

The author declares no ethical issues or conflicts of interest in this research.

ETHICAL STANDARDS

The author affirms this research did not involve human participants.

Footnotes

Sami Petersen, PhD Student, Department of Economics, University of California, Berkeley, United States, sami.petersen@gmail.com

Handling editor: Macartan Humphreys.

1 The middle term in $ (6) $ left parenthesis 6 right parenthesis is printed slightly differently here to avoid ambiguity.

2 In fact, $ {f}^{\frac{-1}{n-1}}={e}^{\frac{-\ln f}{n-1}}>1+\frac{-\ln f}{n-1}>1+\frac{-\ln f}{n} $ f Superscript StartFraction negative 1 Over n minus 1 EndFraction Baseline equals e Superscript StartFraction minus ln f Over n minus 1 EndFraction Baseline greater than 1 plus StartFraction minus ln f Over n minus 1 EndFraction greater than 1 plus StartFraction minus ln f Over n EndFraction .

References

REFERENCE

Feddersen, Timothy, and Pesendorfer, Wolfgang. 1998. “Convicting the Innocent: The Inferiority of Unanimous Jury Verdicts under Strategic Voting.” American Political Science Review 92 (1): 23–35.10.2307/2585926CrossRefGoogle Scholar
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