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 \textbf{When Does Extra Risk Strictly Increase the Value of Options? }

  

 



 



 

\noindent

  \hspace*{20pt}November 5, 2004.   Eric Rasmusen, Indiana University Foundation Professor,

Department  of Business Economics and Public Policy,

  Kelley School of

Business,Indiana University,  BU 456, 1309 E. 10th Street,

Bloomington, Indiana, 47405-1701.

Office: (812) 855-9219. Fax: 812-855-3354.

Erasmuse@indiana.edu.\newline

Http://www.rasmusen.org. Erasmuse@Indiana.edu.

Http://www.rasmusen.org/papers/options\_rasmusen.pdf.   



 

 



 



Let there be an asset  which has terminal value $x_i$ with probability

$f(x_i)$,  where the  values of $x_i$ with  positive probability are

$x_1< x_2<, ...<x_m$.   Denote  by $V_{call}(f, p)$ the current value of

a  call option  on that asset with strike price $ p$ such that $x_1 < p<

x_m. $   The call option entitles its owner to

buy the asset at price $p$ at the terminal time if he wishes. 



  



 



Consider  a different distribution $g(x)

$ which has the same mean as $f(x)$,  so

\begin{equation} \label{e1}

E x = \sum_{i=1}^m f(x_i) x_i =   \sum_{i=1}^m g(x_i) x_i + \sum_{i=m+1

}^{n} g(x_i) x_i,

\end{equation}

 where $x_{m+1}<x_{m+2}<  ...< x_n$ are  points in the support of $g$

but not $f$.



 



\noindent

The value of a  call option with strike price $p$ is 

\begin{equation} \label{e2}

V_{call}(f, p)  =   {\displaystyle \sum_{i=1}^m   Max \{0, f(x_i) (x_i-

p)\}}   

  =   {\displaystyle  \sum_{i=j}^m     f(x_i) (x_i-p) \; where \; j:

x_{j-1} <p< x_{j }}\\

\end{equation}



 

 



 

\noindent

{\bf  Proposition 1 (Merton [1970] Theorem 8, p. 149): }{\it   If

$g$ is riskier than $f$, then $V_{call}(f, p)  \leq  V_{call}(g,p)$.}



 

   





\noindent

    {\bf  Definition 2 (new): }{\it  Distribution $g(x)$ is   {\bf

pointwise riskier} than $f(x)$  iff $f$ and $g$  have the same mean and

there

exist points $\underline{x} $

and $\overline{x} $ in $(x_1, x_m)$ such that \\ 

(a)  if $x < \underline{x}$ then $g(x)  \geq f(x)$, and if $f(x)>0$ then

$g(x)  > f(x)$;\\

(b) if  $x   \in [ \underline{x}, \overline{x}] $ then   $g(x)  \leq

f(x)$,  and if $f(x)>0$ then $g(x)  < f(x)$;\\

  (c) if $x  >  \overline{x}$ then $g(x) \geq f(x)$, and if $f(x)>0$

then

$g(x)  > f(x)$.}





 





 \noindent

{\bf  Proposition 2: }{   If $g$ is pointwise riskier than $f$, then 

 for any $p$,    $V_{call}(f, p) < V_{call}(g,p)$.} 



 

  



\noindent 

  {\bf  Definition 3 (new): }{\it  Distribution $g(x)$ is   {\bf extreme

riskier} than $f(x)$  iff \\

  (a)  either   $f(x_1) < g(x_1)$,  or     $g(x)

>0$ for some $x<x_1$;   and \\

  (b)either $f(x_m) < g(x_m)$   or    $g(x) >0$ for some $x>x_m$.  

 }



  



 {\bf Alternative  Definition 3': }{\it  

  $g$   is extremum-riskier than  $f$  if  it has 

  more   probability mass  at an atom or over an  interval  at  or beyond both extrema. If the initial lower and upper points with positive probability or density are $x_1$ and $x_m$, then it must be that $G(x_1+\epsilon)>   F(x_1+\epsilon)$   

 and 

$1- G(x_m-\epsilon)<   1- F(x_m-\epsilon)$   for arbitrarily small $\epsilon$.}







\noindent

    {\bf  Proposition 3:  }{\it  Consider two distributions $f$ and

$g$.  A necessary condition for it to be true that for  any $p$,

$V_{call}(g, p) > V_{call}(f,p)$   is that  $g$   be extreme-riskier

than  $f$. }



 



 



 

 

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