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\begin{LARGE}



  





\begin{center}

\begin{LARGE}

 

  \textbf{When Does Extra Risk Strictly Increase  an Option's Value? }

\\[0pt]





\end{LARGE}



\bigskip



November 7, 2004 \\[0pt]



\bigskip



Eric Rasmusen \\[0pt]



\bigskip



\textit{Abstract}

\end{center}



It is well known that  risk increases the value of options. This paper makes that precise in a new way. The conventional theorem  says that the value of an option does not fall if the   underlying option becomes riskier in the conventional sense of the mean-preserving spread. This paper uses two new definitions of ``riskier'' to show      that the value of an option strictly increases  (a) if the underlying asset becomes     ``pointwise riskier,''  and  (b)  only if the underlying asset becomes  ``extremum riskier.''



\noindent

 {\small \hspace*{20pt} 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,

http://www.rasmusen.org/papers/options\_rasmusen.pdf.  }



\noindent {\small I thank Maria Arbatskaya,  Peter Rangazas,  Steve Russell,  Yacheng Sun, and

participants

in talks at the 2003 BLISS Conference and DePaul University for their

comments.

  }





 







\newpage







   A call option is the right to buy

the asset at a strike price, $P$, and when  extra risk  increases the

probability that the  market price exceeds $P$, the value of the option

increases.  This has been well-known at least since Merton (1973).

Copeland and Weston (3rd edition, p. 243) say 







\begin{quotation}

 

 

``The holder of a call

option will prefer more variance in the price of the stock to less. The

greater the variance, the greater the probability that the stock price

will exceed the exercise price, and this is of value to the call

holder.''



 

\end{quotation}





  Suppose the strike price is \$50 and the

current price of the asset  is \$40.



 If the probability of the price

being between \$10 and \$15 or   \$45 and \$49 increases, while the probability it is

between \$38 and \$42 falls, the asset has become riskier, but there is

no effect on the option value, because the probabilities of asset values

above the strike price of \$50 have not changed.





\newpage

   

 \noindent

  {\bf Applications of Option Value: }



1. Real options. An investment project with greater uncertainty is more worth starting, because you can always abandon it if its expected value does not increase. 



2. Search models. A more scattered distribution is worth searching more, because you don't have to accept and pay for the results of a given search. 



3. Auction models. If there is a cost to entering an auction or to acquiring information about an auction, that cost is more worth incurring if the value of the item being auction is more uncertain, since you can always decide not to bid. 



4. What else? Come back at the end. 











\newpage



\noindent 

\textbf{II. The Model}



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. $ 



 This rules out strike prices of $x_1$ or below and $x_m$ and

above, because they would lead to riskless options which would    be

exercised  always or never. It does allow a strike price that does not

happen to equal any of the $x_i$.



 The call option entitles its owner to

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



We will

assume the discount rate is zero.



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}

\begin{array}{lll}

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{array}

\end{equation}



 

\newpage



 



\noindent

{\bf  Definition 1a: }{\it A {\bf mean-preserving spread}  consists of

three   numbers   $s(y_1)$, $s(y_2)$, and $s(y_3)$   for  $y_1<y_2<y_3$

such that }

\begin{equation} \label{e2a}

 s(y_1)y_1 +  s(y_2) y_2 + s(y_3)y_3 =0, \; {\rm (the \; mean \; is \;

preserved)}

\end{equation}

\begin{equation} \label{e2c}

 s(y_1)  +  s(y_2) + s(y_3)y  =0, \; {\rm (the \;  new \;probabilities

\;  sum \; to  \;

zero)}

\end{equation}

 { and }

\begin{equation} \label{e2b}

 s(y_1)>0,\;   s(y_2)<0,\;    s(y_3)>0 \;   \;

{\rm (the \;  probability \; is \;spread)}

\end{equation}







 

    



Thus  we arrive at  Definition 1b, the definition of risk originated  in

Rothschild \& Stiglitz (1970).



 

\noindent

{\bf  Definition 1b: }{\it Distribution $g(x)$ is {\bf riskier}  than

$f(x)$  iff $g(x)$ can be reached from $f(x)$ 

by a  sequence of mean-preserving spreads.    }



\newpage



   An example to show that increased variance can

increase utility for a risk-averse person is the following.



 Let the

utility be $ U = x$ for $x \leq 10$ $U = 10 + x/2$ for $x  \geq 10$,

which is weakly concave. 



 Suppose wealth is initially distributed as

(.8-7, .2-12), which has mean 8, variance $   4 (= .8*1^2+.2*4^2)$, and

utility $ 7.8 (=.8*7 + .2* 11)$.



 If the distribution is changed to (.2-

0, .8-10),  the  mean  is still 8, the  variance increases to

$16 (=.2*8^2+.8*2^2) $, and utility rises to  $8 (=.2 *0 + .8* 10)$.





Kurtosis, which increases when moving weight to the tails of the

distribution,  is equally unreliable for ranking  the riskiness of

distributions; it  starts as   $    52 (= .8*1^4+.2*4^4)$   in this

example and rises  to $832 (=.2*8^4+.8*2^4).$   



\newpage



 





\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  Proof: }  

I need   to show  that adding a mean-preserving spread to $f(x)$ does

not reduce the value of the put or call option. Having done that, we

know by induction that adding more mean-preserving spreads would not

reduce the value, so if $g$ is riskier its options are not worth less.

 From  the definition of  the mean-preserving spread and the formula for

the value of a call from equation (\ref{e2}),

\begin{equation} \label{e2d}

\begin{array}{ll}

V_{call}(f, p) & \leq  V_{call}(g,p)\\

 \\

  {\displaystyle  \sum_{i=j}^m     f(x_i) (x_i-p)    \leq } &

  {\displaystyle   \sum_{i=j}^m

f(x_i) (x_i-p)  +  s(y_1)Max (y_1-p,0) } \\

&  +  s(y_2) Max (y_2-p,0) + s(y_3) Max (y_3-p,0) \\ 

  where \; j: x_{j-1} <p< x_{j }  &   \\

\end{array}

\end{equation}

 which implies that 

\begin{equation} \label{e2d1}

\begin{array}{l}

0  \leq    

  s(y_1)Max (y_1-p,0) +  s(y_2) Max (y_2-p,0) + s(y_3)

Max (y_3-p,0).   \\

\end{array}

\end{equation}



 



 Note first that  by equation (\ref{e2a})   the  spread is mean-

preserving, so  $s(y_1)y_1 +  s(y_2) y_2 +

s(y_3)y_3 =0$, and  by equation (\ref{e2c})     the spread's

probabilities add to zero,  so  $[s(y_1)  +  s(y_2)   + s(y_3)]=0$.

Together, these imply that

\begin{equation} \label{e101}

\begin{array}{l}

    s(y_1) (y_1-p) +  s(y_2)  (y_2-p ) + s(y_3)(y_3-p ) \\

  \\

=   s(y_1)

(y_1) +  s(y_2)  (y_2) + s(y_3)(y_3 )\\

    -  [s(y_1)  +  s(y_2)   + s(y_3)]p =    0,    \\

\end{array} 

 \end{equation}

a result that will be used below. 



\newpage



\noindent

   (i) Suppose  $p \leq  y_1$, so  inequality (\ref{e2d1}) becomes 

\begin{equation} \label{e100}

\begin{array}{l}

0  \leq      s(y_1) (y_1-p) +  s(y_2)  (y_2-p ) + s(y_3)(y_3-p ).   \\

\end{array} 

\end{equation}

      Equation (\ref{e101}) tells us that this is true as  an equality. 

 



\noindent

  (ii) Suppose  $p \geq  y_3$, so  inequality (\ref{e2d1}) becomes 

\begin{equation} \label{e104}

\begin{array}{l}

0  \leq      s(y_1) (0) +  s(y_2)  (0 ) + s(y_3)(0 ).   \\

\end{array} 

\end{equation}

 This is obviously true  as an equality. 



    



\noindent

(iii)  Suppose  that $p \in (y_1, y_3)$.  Then, since $Max (y_1-p,0)=

0$ and $Max (y_3-p,0) =  y_3-p$, we can rewrite expression (\ref{e2d})

as

\begin{equation} \label{e2f}

0 \leq  0  +  s(y_2) Max (y_2-p,0) + s(y_3) (y_3-p )   

\end{equation}



\noindent

  (a)  If $Max (y_2-p,0) =0$, then inequality (\ref{e2f}) is true as a

strict inequality, since

$s(y_3) >0$ and $y_3>p$.



\noindent

  (b) If  $Max (y_2-p,0) = y_2-p$, then

inequality (\ref{e2f}) is true if 

\begin{equation} \label{e105}

 s(y_2)  (y_2-p) + s(y_3) (y_3-p ) \geq 0   

\end{equation}

   Equation  (\ref{e101}) tells   us that $s(y_1)  (y_1-p) +s(y_2)

(y_2-p) + s(y_3) (y_3-p )

= 0$, so since  $s(y_1)>0$ and, in case (iii),  $  (y_1-p) <0$, it

follows that (\ref{e2f}) is true as a strict inequality. QED.



  

 

 

\newpage





\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  Proposition 1a (false): }{\it   If $g$ is riskier than $f$, then 

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



\noindent

{\bf Disproof. }     Consider a call option with an exercise price of

4.5  and the asset price distribution shown in  Figure 1.   $V_{call}(f,

4.5)  =   V_{call}(g,4.5)$, even though $g$ is

riskier than $f$.  The increase in risk has no effect   because only

changes in the probabilities of terminal values greater than 4.5 would

matter to the value of the call, and there are no such changes in the

example.  



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\begin{center}

{\bf Figure 1: A  Counterexample: Risk Does Not Increase Option Value

}

\end{center}



 \newpage



\begin{center}

 {\bf  WEAK INEQUALITY PROPOSITIONS ARE DISSATISFYING}

\end{center}



\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  Proposition 1$'$ (Merton [1970] Theorem 8, p. 149): }{\it  Adding risk to an asset does not reduce the value of a call option on it. }



 \noindent

{\bf  Proposition 1$''$: }{\it   If there is a  supernova in the Andromeda Galaxy, that does not reduce the   value of a call option on an asset. }





Merton's proposition is more meaningful than Proposition 1$''$ for a couple of reasons. 



A.  Risk is something  we might think was relevant to asset value. It is actually surprising that we get even the weak result, because we are used to thinking that risk reduces the value of assets. Even leaving the asset value unchanged is mildly interesting-- though not too surprising if the  investor is risk neutral, as here.



 B. Behind Merton's proposition is reasoning which carries us a lot further than the proposition itself. The proposition doesn't say much, but if you understand the intuition or logic behind the proposition, you have other, more interesting propositions floating cloudlike in your mind  around the Merton proposition. 



   









 \newpage





\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  Proposition 1b: }{    If $g$ is riskier than $f$, then there

there exists some exercise price such that the associated call  option

is more valuable under $g$ than under $f$   but no exercise price such

that a call option is more valuable under $f$:\\

    $ \exists p': V_{call}(f, p')  <   V_{call} (g,p')$;     but     there does not exist  

$p'': V_{call}(f, p'' )  >   V_{call}(g,p'' )$.   



\noindent

{\bf  Proof: }  \\

 The proof of Proposition 1 showed that if  $p \in (y_1, y_3)$, then the

value of the call or put strictly increases. Thus, simply pick $p'$   in

$(y_1, y_3)$ for one of the spreads that makes $g$ riskier than $f$.



 That there exists no value $p''$  for which option value declines is a

direct corollary of Proposition 1. QED.





 \newpage

 

 Another approach is to find a definition of risk under which something

like Proposition 1b is true, and the value of the option  does increase

with ``risk'' regardless of the strike price. See Figure 2 below. 





\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

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





 





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\newpage



   Since  pointwise

riskiness and second-order stochastic dominance both    can    be

defined in terms of  functions that cross a limited number of times, the reader

may wonder if pointwise riskiness is the same as  the strict second-order

stochastic dominance.





 Distribution $G$  strictly second-order stochastically dominates

$F$ if $  \int_0^t G(x)dx \leq  \int_0^t  F(x)dx      $ for all

$t$ and the inequality is strict for all   values of $t$ such that $G(t)>0$ and

$G(t)<1$. 



 Even if  $G$ dominates $F$ in this way, however, it  might not be pointwise riskier.  It could be that $     \int_0^t G(x)dx < \int_0^t  F(x)dx   $ at some $t$  even though  $g(t) = f(t)$ (which pointwise riskiness would not allow except at the two crossing point values of $x$), because the   values of $F(x)$  for lower values of $x$ are enough greater than the values of $G(x)$ that their integral remains greater even once $t$ is included.



 Similarly, strict second-order stochastic dominance, unlike pointwise riskiness, does not imply extremum riskiness.  It could be that $     \int_0^T F(x)dx < \int_0^T G(x)dx   $ at  $T$, the maximimal value of $x$, so $F$ does  dominate $G$,  even though  $g(T) < f(T)$, so $G$ is not extremum riskier than $F$.  In that case, not every call on $G$ would be more valuable than a call with the same strike price on $F$.   



 

 \newpage



\noindent

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

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 $x>x_m$.

 }







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\begin{center}

{\bf Figure 2:    Pointwise and Extremum Riskiness   }

\end{center}











 The distribution  in part (c) of Figure 2 is extremum-riskier than the distributions in parts (a) and (b). 



\newpage







\noindent

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

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 $x>x_m$.

 }



 The distribution in part (b) of Figure 3 is extremum-riskier than the

distribution

in part (a).



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\begin{center}

{\bf Figure 3:  Extremum Risk   versus Risk

}

\end{center}





\newpage



\noindent

  {\bf  Definition $3'$ (ALTERNATIVE): }{\it  Cumulative distribution $G(x)$ is   {\bf extremum

riskier } than $F(x)$  iff, for arbitrarily small $\epsilon$,  \\

  (a)  $G(x_1+\epsilon)>   F(x_1+\epsilon)$ \\

 and \\

  (b)$G(x_m-\epsilon)< F(x_m-\epsilon)$.

 }



This will cover atoms, discrete distributions, and continuous distributions. 





 From Definition $3'$ we can derive Lemma X for a discrete distribution. 



\noindent

  {\bf  Lemma X: }{\it If distribution $G(x)$ is   {\bf extremum

riskier } than $F(x)$ then  \\

  (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 $x>x_m$.

 }



\noindent

  {\it  Proof: } Proof of Part (a).  Let $\epsilon =0$. Since $G(x_1+\epsilon)>   F(x_1+\epsilon)$, it follows that $G(x_1 )>   F(x_1)$. Since $x_1$ is the bottom of $F$'s support, $F(x_1) = f(x_1)$. (i)  If $x_1$ is the bottom of $G$'s support too, then it must be that $g(x_1) > f(x_1)$. (ii) If $x_1$ is above  the  bottom of $G$'s support,  $g(x) >0$ for some $x<x_1$. (iii)  If $x_1$ is below the bottom of $G$'s support, then there it is impossible that $G(x_1 )>   F(x_1)$, so we can rule out that case. 





\newpage





\noindent

  {\bf  Definition $3'$ (ALTERNATIVE): }{\it  Cumulative distribution $G(x)$ is   {\bf extremum

riskier } than $F(x)$  iff, for arbitrarily small $\epsilon$,  \\

  (a)  $G(x_1+\epsilon)>   F(x_1+\epsilon)$ \\

 and \\

  (b)$G(x_m-\epsilon)< F(x_m-\epsilon)$.

 }



 



\noindent

  {\bf  Lemma X: }{\it If distribution $G(x)$ is   {\bf extremum

riskier } than $F(x)$ then  \\

  (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 $x>x_m$.

 }







Proof of Part (b). Since $x_m$ is the top of $F$'s support, $F(x_m) = 1$, and for small $\epsilon$, $F(x_m-\epsilon)  = 1-f (x_m)$. 

 (i)  If $x_m$ is the top  of $G$'s support too, then $G(x_m-\epsilon)  = 1-g (x_m)$ and it must be that $g(x_m) > f(x_m)$  since $G$ is extremum riskier and  $G(x_m-\epsilon)< F(x_m-\epsilon)$.   (ii) If $x_m$ is below  the  top of $G$'s support,  $g(x) >0$ for some $x>x_m$.  (iii)  If $x_m$ is above the top of $G$'s support, then $G(x_m-\epsilon)=1$, which is greater than   it is $F(x_m-\epsilon) = 1- f(x_m)$, so we can rule out that case. 



 Note that Lemma X is a lemma, not an equivalent definition of extremum riskiness. Lemma X (Definition 3) in the November 1, 2004 draft of the paper) would be a weaker definition of extremum riskiness-- a weaker necessary condition for options to increase in value. 







\newpage







\noindent

    {\bf  Proposition 2:  }{\it  Consider two distributions $f$ and $g$.  A

necessary condition for it to be true that $V_{call}(g, p) > V_{call}(f,p)$ for

any strike price $p$    is that  $g$ be extremum-riskier than  $f$. }



\noindent

{\bf  Proof: }  \\

 (i)    If $f(x)$ and $g(x)$ are identical for all $x>p$ or for all $x<p$,

then $V_{call}(g, p) =  V_{call}(f, p)$.



 If  $f$ and $g$ are identical for

all $x>p$, then clearly the call value of equation (\ref{e2}) must be equal

for $f$ and $g$, since then $f$ and $g$ are identical for $x \geq x_j$, and

only  such values of $x$ enter into equation (\ref{e2}), reproduced below.

\begin{equation} \label{e200}

\begin{array}{lll}

V_{call}(f, p) = & {\displaystyle  \sum_{i=j}^m     f(x_i) (x_i-p) \; where \;

j:

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

\end{array}

\end{equation}



  If, on the other hand,   $f$ and $g$ are identical for all  $x<p$, then

since their unconditional means, the expected values taken over all possible

values of $x$,  are equal,

and their means conditional  on $x<p$ are equal,   their means conditional

 on $x>p$ must also  be equal. This, too  implies that

the values of   calls on $f$ and $g$ are equal.



Thus, for  $V_{call}(g, p) > V_{call}(f,p)$  to be  true, it is  necessary

that $g(x) > f(x)$ for some  but not all $x>p$.



Since this is true for every $p$, it must be true for $p= x_m-\epsilon$, for

any small value $\epsilon$. Thus, for some $x  \geq  x_m$, $g(x) > f(x)$. But

this can be true only if either $g(x_m) > f(x_m)$,  or if $g(x)>0$ for some

$x>x_m$. That is condition (b) in Definition 3.



\newpage



\noindent

    {\bf  Proposition 2:  }{\it  Consider two distributions $f$ and $g$.  A

necessary condition for it to be true that $V_{call}(g, p) > V_{call}(f,p)$ for

any strike price $p$    is that  $g$ be extremum-riskier than  $f$. }







Similarly, if $f(x)$ and $g(x)$ are identical for all $x<p$, then $V_{call}(f,

p) = V_{call}(f, p)$. Thus, if $V_{call}(g, p) > V_{call}(f,p)$  is  to be

true, it must also be true that  for any $p$, for some $x<p$ it must be true

that  $g(x) > f(x)$. Since this is true for every $p$, it must be true for $p=

x_1+\epsilon$, for any small value $\epsilon$. Thus,  we need that for  some

$x   \leq   x_1$ it is true that $g(x) > f(x) $. But this can be true only  if

either $g(x_1) > f(x_1)$ or if $g(x)>0$ for some $x<x_1$. That is condition

(a) in Definition 3.



 Thus, for the value of the call to be greater under $g(x)$ for all $p$, it is necessary that $g$   satisfy the conditions for being extremum riskier than $f$. QED.



\newpage



{\bf Why does extreme riskiness require that $g$ add probability at  {\it both}  extremes, not just the maximum?}



 



 The example, shown in in Figure 4, in which 

   $g$ is made  riskier than $f$ by shifting probability away from $x=2$, the mean, to $x = 1.33$ and $x = 4.67$. So doing    makes $g$  

  put more

probability  than $f$ on the upper extremum of $x=4.67$, but no more

probability on the  lower extremum of $x=0.$  



 A call with a strike price above 1.33 will increase in value as a result of changing the distribution from $f$ to $g$. But think about a 

   call with a strike price of 1.  It will have equal value under $f$

and $g$, because the mean of the distribution conditional on $x$ being greater than 1 has not changed. Or,  looked at a bit  differently, the probability of the  state  of the world ($x=0$) in which the call is {\it not} exercised has not changed at all. 



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\begin{center}

{\bf Figure 4:   Why Extremum Risk Needs Spread at Both Extrema

}

\end{center}



\newpage

   



\bigskip



  {\bf   Why is extreme riskiness  just a necessary condition, and not sufficient?  }



 In Figure 3,  $g(x)$ has more probability at the extremes than $f(x)$ does-- the probability of each extreme is .25 instead of .20--- but it is not riskier.



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\begin{center}

{\bf Figure 3:  Extremum Risk   versus Risk

}

\end{center}









 If the strike price is 4.5, then the call's value is higher under distribution $g(x)$, because  the outcome $x= 5$ occurs with probability .25 instead of  $f(x)$'s .20. $V_{call}(f, 4.5) = .20(5-4.5)= .10 < V_{call}(g, 4.5) =  .25 (5-4.5) = .125$.    If the strike price is 3.5, however, the call's value is higher under distribution $f(x)$, because under $g(x)$   the outcomes  $x=4$ and $x= 5$ together occur  with probability .25 instead of .40 and  $V_{call}(f, 3.5)  =  .20(4-3.5) + .20 (5- 3.5) = .40 > V_{call}(g, 3.5) =   .00(4- 3.5) + .25 (5-3.5) =  .375$.



\newpage





   





\noindent

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

sufficient condition for it to be true that $V_{call}(g,p)> V_{call}(f, p)   $

for  any strike price $p$  is that  \\

 (a) $g$   is  extremum-riskier than  $f$;

and \\

 (b) $g$ is riskier than $f$. }



\noindent

{\bf  Proof: }  \\

 From Proposition  1  we know that if condition (b) is true, then $V_{call}(g,

p) \geq V_{call}(f, p)$, that is,    Proposition 3's  inequality is true at

least weakly.

Thus, all that we need    show is that

condition (a)  makes   the inequality   strict.



Suppose Proposition 3 is false, and there does exist some value $p$ such that

$V_{call}(g,

p)= V_{call}(f, p)$.  Recall that equation (\ref{e1}) says that $g(x)

$ has the same unconditional mean as $f(x)$:

  $$

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,

  $$

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

but not $f$. If $V_{call}(g,

p)= V_{call}(f, p)$, then it is also true

  that

$$

\begin{array}{l}

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

(x_i-p) + \sum_{i=m+k}^n   g(x_i)

(x_i-p),}\\

  \\

 \; where \; j:

x_{j-1} <p< x_{j } and  k:

x_{m+k-1} <p< x_{m+k },    \\

\end{array}

$$

so  $f$ and $g$ have the same means conditional on $x$ being

greater than $p$.





\newpage







  $$

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,

  $$

  

$$

\begin{array}{l}

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

(x_i-p) + \sum_{i=m+k}^n   g(x_i)

(x_i-p),}\\

  \\

 \; where \; j:

x_{j-1} <p< x_{j } and  k:

x_{m+k-1} <p< x_{m+k },    \\

\end{array}

$$



Putting these two equations together implies that

\begin{equation} \label{e231}

 \begin{array}{l}

  {\displaystyle  \sum_{i=1}^{j-1}  f(x_i) (x_i-p)=  \sum_{i=1}^{j-1}     g(x_i)

(x_i-p) + \sum_{i=m+1}^{m+k-1}   g(x_i)

(x_i-p),}\\

 \\

 \; where \; j:

x_{j-1} <p< x_{j } and  k:

x_{m+k-1} <p< x_{m+k },

\end{array}

\end{equation}

  so  $f$ and $g$ have the same means conditional on $x$ being

less than $p$.



 Since $g$ is extremum-riskier, though, we know from Definition 3 that  either

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

>0$ for some $x<x_1$-- that $g$ puts more weight on the far left tail of the

distribution.  Thus, equation (\ref{e231}) is false-- $g$'s mean conditional on

$x<p$ is lower than $f$'s. In turn, this implies  that  equation (\ref{e231}) ia

false, and no $p$ exists for which $V_{call}(g,

p) = V_{call}(f, p)$. Every call on $g$ is strictly more valuable than it would

be on $f$.    Q.E.D.



\newpage





 

 \noindent

    {\bf  Proposition 3a (false):  }{\it  Consider two distributions $f$ and

$g$.  A necessary and sufficient condition for it to be true that $V_{call}(f,

p) < V_{call}(g,p)$    for  any strike price $p$  is that \\

 (a) $g$   is extremum-riskier than  $f$; and \\

 (b) $g$ is riskier than $f$.  }



\noindent

 {\it Disproof:}  Conditions (a) and (b) are jointly   sufficient, as

Proposition 3 says. Condition (a) by itself is necessary, as Proposition 2 says.

Thus, what we need to show to disprove Proposition 3a  is that there exist

distributions $f$ and $g$  such that Condition (b) is violated but nonetheless

$V_{call} (f, p) < V_{call}(g,p)$ for any $p$. That is, we must show that $g$'s

options are always more valuable, but $g$ is not riskier than $f$.



Consider the example in Figure 5. Distribution $g$ is extremum-riskier than

distribution $f$, but it is not riskier, because it has more probability at the

mean, $x=5$ (in fact, $g(5) =0$).  The  distributions $f$ and $g$  cannot be

ordered by risk.  









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\newpage





 

 \noindent

    {\bf  Proposition 3a (false):  }{\it  Consider two distributions $f$ and

$g$.  A necessary and sufficient condition for it to be true that $V_{call}(f,

p) < V_{call}(g,p)$    for  any strike price $p$  is that \\

 (a) $g$   is extremum-riskier than  $f$; and \\

 (b) $g$ is riskier than $f$.  }





 The value of a call option on an asset

with density $f$ and   strike price $p \in (2,8)$ is, from equation

(\ref{e2}),

\begin{equation} \label{e20}

\begin{array}{ll }

V_{call}(f, p) & =   Max \{0, .25 (2- p)   \}  +   Max \{0, .25 (4- p)

\}   \\

 & + Max \{0, .25 (6- p)   \} + Max \{0, .25 (8- p)   \}    \\

\end{array}

\end{equation}

and the  value of a call option on an asset with density $g$ and      strike

price $p$   is

\begin{equation} \label{e21}

\begin{array}{lll}

V_{call}(g, p)& = & Max \{0, .25 (1- p)   \}  

\} \\

 & &  +   Max \{0, .40 (5- p)  + Max \{0, .30 (9- p)   \}    \\

\end{array}

\end{equation}



  The possible values of $p$ go from $p=2$ to $p=8$, where the endpoints are

not possible (as the option would then be always or never exercised). We will

split this up into four intervals and examine each in turn.







\newpage





  The possible values of $p$ go from $p=2$ to $p=8$, where the endpoints are

not possible (as the option would then be always or never exercised). We will

split this up into four intervals and examine each in turn.



\noindent

{\it Lemma 1:}  $V_{call}

(f, p) < V_{call}(g,p)$ for  $p \in (2,4]. $



\noindent

 {\it Proof:}  Then $V_{call}(f, p)  =         .25 (4- p) + .25 (6- p)     +

.25 (8- p) = .25 (18-p)=    4.5 - .75 p.  $ On the other hand, $V_{call}(g, p)

=    .40 (5- p) +   .30 (9- p)    =  4.7 - .70p,  $  which is greater, so $g$

has the more valuable options.





 \noindent

{\it Lemma 2:} $V_{call}

(f, p) < V_{call}(g,p)$ for  $p \in (4,5]. $



\noindent

 {\it Proof:}  Then $V_{call}(f, p)  =    .25 (6- p)    .25 (8- p)  = 3.5 -

.50p.    $  On the other hand, $V_{call}(g, p)  =    .40 (5- p) + .30 (9- p)

=  4.7 - .70 p $.  It is true that $ 3.5 - .50p <4.7 - .70 p$ if $.20p <  1.2$,

which is true if $p < 6$, and in particular if   $ p \in [4, 5]$, so $g$ has the

more valuable options.



\newpage





 \noindent

{\it Lemma 3:}$V_{call}

(f, p) < V_{call}(g,p)$ for  $p \in [5,6]. $



\noindent

 {\it Proof:}

  Then $V_{call}(f, p)  =       .25 (8- p) = 2 - .25p.    $  On the other hand,

$V_{call}(g, p)  =     .30 (9- p)    = 2.7 - .30p.   $  It is true that $ 2 -

.25p < 2.7 - .30p$ if $.05p <  .7$, which is true if $p < 14$, and in particular

if   $ p \in [5, 6]$, so $g$ has the more valuable options.





 \noindent

{\it Lemma 4:} $V_{call}

(f, p) < V_{call}(g,p)$ for  $p \in [6,8). $



\noindent

 {\it Proof:}

  Then

$V_{call}(f, p)  =  0.   $  On the other hand,

$V_{call}(g, p)  =   .30 (9- p)   = 2.7 - .30 p. $ It is true that $0  <

2.7 - .30p$ if   if $p < 9$, and in particular  is true  if   $ p \in

[6, 8]$, so $g$ has the more valuable options.





 Thus, for any $p \in (2,8)$, the set for which an option on $f$ is

risky,  $g$ has more valuable options, which was to be proved.



\newpage





\noindent

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

sufficient condition for it to be true that $V_{call}(g,p)> V_{call}(f, p)   $

for  any strike price $p$  is that  \\

 (a) $g$   is  extremum-riskier than  $f$;

and \\

 (b) $g$ is riskier than $f$. }



  

  {\it Conjecture 1:} If $g$ is extremum riskier than $f$ but not riskier, then there exists a strike price $p$ such that $V_{call}(g, p) < V_{call}(f,p)$.



If true, Conjecture 1 would tell us that Proposition 3 is a  narrowest possible sufficient condition. 





\newpage





 For many  applications, it is convenient  to specify a simple sufficient

condition for one option

to be riskier than another.  



\noindent

{\bf  Proposition 4: }{\it  If $g$ is pointwise riskier than $f$, then

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



\noindent

{\bf  Proof: } If $g$ is pointwise riskier than $f$, then it is also  riskier

and extremum riskier. It is riskier because we can move from $f$ to $g$ by a

series of mean- preserving spreads that take probability away from the middle

interval $[ \underline{x}, \overline{x}]$   and move it to the extremes.  It

is extremum riskier because $x_1< \underline{x}$ and $ x_m>\overline{x}$, so

$g$   puts more probability on $x_1$ and $x_m$ than $f$ does.  It follows from

Proposition 3, proved above,  that calls on $g$ will be more valuable than

calls on $f$. Q.E.D.



 









 \end{LARGE}

\end{document}

