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                     \begin{center}
            \begin{Large}
    {\bf    Notes on Prosecutor Incentives: The Investment Problem
 } \\
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           \vspace*{12pt}  
March 31, 1998, title change June 9, 1999\\
                    \bigskip
                 J. Mark Ramseyer and    Eric Rasmusen \\
                        \vspace*{.3 in} 
                      
 \end{center}

   \noindent \hspace*{20pt}Ramseyer:   Harvard Law School, Cambridge, Mass. 02138.   \\
        \noindent \hspace*{20pt}Rasmusen:  Indiana University, Kelley School of
Business, BU 456, 1309 E. 10th Street, Bloomington, Indiana,
47405-1701.
  Office: (812) 855-9219.  Fax: 812-855-3354. Email:
Erasmuse@indiana.edu. Web:  Php.indiana.edu/$\sim$erasmuse.
  


		    The notes that follow set up model of a  prosecutor, but an identical model can be used for a firm with a fixed capital budget that is choosing investment projects and how much to invest in each. Then, P(e) is the  project's return for investment e with fixed cost F.  

 I think that if the conjecture could be proved, we'd have a publishable finance paper--and a very worthwhile one.  If anyone can do the proof for us, I'd be happy to sign him on as a third author and send it off to J. of Finance.   



\newpage
  

A prosecutor has a budget of B.   Each case that he   prosecutes incurs a fixed cost of F and a variable cost of e. The probability of conviction is  P(e), where $P(0) =0$,  $P'>0$, $P \in [0,1)$, and $P(\infty) <1$.    Let us also assume that $P''<0$, so  there are decreasing returns to effort in a particular case.  Figure 1 shows how $P$ might vary with $e$.   


 
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\epsffile{pros1.eps} 

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     {\bf  Figure 1:  The Conviction Function} 
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\vspace*{36pt}


     We will use the modelling contrivance of a continuum of cases. This means we will model the situation as each case being trivial compared to the whole.  Potential cases will be indexed by $\theta$, where $\theta$ varies from 0 to 1.   The number of cases of type $\theta$ is represented by a density function $g(\theta)$, where $\int_0^\infty g(\theta) d\theta  =1$, which is to say that  we will normalize so the universe of possible cases has size 1.   Figure 2 shows  one way that  the cases might be distributed.    If the prosecutor decides to prosecute the easiest  1/3 of cases, he takes all the cases  in the interval $[\overline{\theta}, 1]$.  

  Let us now make a substantive assumption.  Each type of case differs in how hard it is to prosecute.  Suppose the prosecutor spends amount $e(\theta)$ on each  case of type $\theta$.  Let  a case of type $\theta$ have a prosecution success function of 
 \begin{equation} \label{e1} 
  P(\theta, e(\theta ) = \theta P(e(\theta)). 
\end{equation}  
    What this assumption says is that  all the P functions have roughly the same shape, just scaled up or down by $\theta$.  This means,  first, that if effort e* is the same for two types of case, the one with the bigger $\theta$ has the bigger probability of success--- and that is true for any e* one might pick.   Second,  if effort e* is the same for types of cases, the  marginal product of effort  is bigger for the case with the bigger $\theta$.   If the prosecutor  chooses to equalize the marginal product of effort in two cases with $\theta=.8$ and $\theta =1$, as in Figure 1, the he would choose $e'$ and $e''$ to equalize the slopes of the $P(e; \theta )$ functions, which means that  $e''>e'$. 

   

 \epsfysize=3in 
   
\epsffile{pros2.eps} 

\begin{center}
     {\bf  Figure 2:  The  Distribution  of Cases} 
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\vspace*{36pt}




  We now come to the prosecutor's problem. He has two kinds of choices. First, he must decide which cases to prosecute.  He will want to prosecute the easier cases first,  which  amounts to choosing a lower cutoff $\overline{\theta}$ for the interval of types  [$\overline{\theta}, 1]$ that he prosecutes.  Second,  he needs to pick the $e(\theta)$ function  which shows how much he spends on each type $\theta$ of case .  

  The  payoff function for the prosecutor is 
 \begin{equation} \label{e2} 
  \int_{\overline{\theta}}^1  g(\theta )    \theta P(e(\theta)) d\theta.
\end{equation}   

 The budget constraint is 
 \begin{equation} \label{e3} 
  \int_{\overline{\theta}}^1  g(\theta )   [e(\theta)   + F]  d\theta  \leq B. 
\end{equation}  

 The Lagrangian for this is 
 \begin{equation} \label{e4} 
      Maximize\;\;  L = \int_{\overline{\theta}}^1  g(\theta )    \theta P(e(\theta)) d\theta  +  \lambda \{ B -    \int_{\overline{\theta}}^1  g(\theta )   [e(\theta)   + F]  d\theta  \} . 
\end{equation}  

Using loose calculus of variations/Maximum principle methods that I will have to refine later,  there are two kinds of first-order conditions.  First, there is choice of the $e(\theta)$ function, which yields
    \begin{equation} \label{e4} 
   g(\theta )  \theta   \frac{d P(e(\theta)) }{de}    - \lambda      g(\theta )   (1)   = 0.    
\end{equation}  

 Rearranging, we get that the marginal product of effort has to be the same for each case prosecuted. For any $\theta$,  
 \begin{equation} \label{e5} 
 \frac{d L}{d e(\theta)} =  \theta   \frac{d P(e(\theta)) }{d e}   =  \lambda    .    
\end{equation}  


 The second condition has to do with choice of  $\overline{\theta}$ It says that 
  \begin{equation} \label{e6} 
 \frac{d L}{d  \overline{\theta}} =    -  g(\overline{\theta})   \overline{\theta} P(e(\overline{\theta}))  - - \lambda      g(\overline{\theta} )   [e(\overline{\theta})   + F]    = 0.  
\end{equation}  

 This implies that 
  \begin{equation} \label{e7} 
   \frac{     \overline{\theta} P(e(\overline{\theta}))    } { e(\overline{\theta})    + F   }  =  \lambda .
\end{equation}  

 In words: the marginal product of labor has to equal the average product of labor for the  marginal case! (Because the $\lambda$ is the same number as in the first of our two first-order conditions.)

Note also that  
   \begin{equation} \label{e7.5} 
 \frac{d L}{d B} =      \lambda ,
\end{equation}  
 which is to say that $\lambda$ equals the marginal value of relaxing the budget constraint. 

\bigskip

 CONJECTURE:  As the budget B increases, the average probability of conviction falls. 

   The average probability  of conviction is  
    \begin{equation} \label{e8} 
   \overline{P}   =  \frac{   \int_{ \theta }^1     P(e(\overline{\theta})) 
 d\theta }   { 1-      \overline{\theta}   }.
   \end{equation}  
  
  Proving the conjecture is difficult for a number of reasons. One big problem is that  corner solutions have to be considered.  Very possibly $P=1$ for   some of the $\theta$.  

  Intuition says that there two things will matter a lot as B increases. First,  how big is F? If it is big, then extra budget would best go to existing cases, and the average will rise, not fall, so maybe the conjecture is wrong.  But maybe if F is big, then  most existing cases are at P=1 anyway, so they won't rise at all, and since  some new projects will be adopted, the average will fall, so the conjecture is correct. 

 Moreover, if F is very small, then the prosecutor will prosecute ALL cases--another corner solution---   and an increase in the budget will INCREASE the average probabilityo f success for sure

  The same goes with $P''$.  If $P''$ is very big and negative, then there are sharply diminishing returns to spending on existing cases.  Then,  the average  probability of success would fall with new spending, because it goes mostly to new cases.  

 If $P''$ is near zero, though, so there are almost constant returns to scale, then  there will be a lot of $P=1$ cases.  Then, the probability of success also falls with new cases.  In the extreme, with  $P''=0$, every prosecuted case will have $P=1$, and the average probability of success will certainly fall with increased spending.  
 
 It may be helpful to note that 
as $B$ gets bigger, $\lambda$ gets smaller, because the marginal payoff value of extra budget gets smaller.   Any proof, though, runs into the difficulty that calculus methods alone are unreliable because of the corner solutions. 



   

Obscure notes for future work:  In the discrete case, almost any small change will just increase the amount spent on old cases and not cause any new cases to be brought.  That is a weird technical effect, though. 
 
 Actually,the corporate investment angle is simpler than that, and doesn't rely on strategic considerations at all. It is the smae intuition that you sent me in notes originally-- that  since when you start with the best projects and then go to the worse ones, if you do more projects, your average return will be lower. Your average return can  still, however, be greater than your cost of capital.  Thus,a firm with a lower ``return on capital'' as measured by accountants might be more profitable. 

  An  agency problem  does start to appear at that point, though.  What if the shareholders don't realize that they make more moeny if the firm borrows more money and invests in worse projects,  the total rturn (and thereturn on equity) rising at the same time as the average return on capital falls?   Then they'll fire the manager. So if his shareholders are dumb, he'll act dumb too. 

      This very much applies to  our prosecutors.  American prosecutors like to boast about the percentage of their cases that get convictions.   This give sthem a big incentive to  dismiss weak cases.  

 In fact, a neato implication of that idea is that local prosecutors, who are elected, should have  higher conviction rates than federal prosecutors, who are appointed.   Another impilcation is that  criminal conviction rates should be much higher than civil plaintiff victory rates--- something very true, and otherwise mysterious.  (It isn't that  criminal juries have to be unanimous--- that should be neutral in determining which percntage of trials end in convictions, since the plaintiff takes the toughness of getting unanimity into account). 


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