0$, when $p-y_3<0$, as in case (iii), it must be that $ s(y_2) (p-y_2 ) + s(y_3) (p-y_3 ) > 0$, and (\ref{new3}) is true even if $y_2

V_{call}(f, p) $ and $ V_{put}(g,p) < V_{put}(f, p) $ for any strike price $p$.} \noindent {\bf Disproof. } Consider a call option with an exercise price of 4.5 and the asset price distributions 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 matter to the value of the call, and there are no such changes in the example. (Similarly, $V_{put} (f, 4.5) = V_{put}(g,4.5)$.) \begin{figure} \centering %\includegraphics[width=80mm]{options1.jpg} \caption{Figure 1: Risk Does Not Increase Option Value} \end{figure} Propositions 1 and 1a differ only in the weakness of the inequality. That is enough, however, for ``Proposition 1a: Option value increases with risk'' to be false. Instead, we are left with ``Proposition 1: Option value does not fall with risk,'' which although true, is very weak. That kind of statement can be made of any variable outside the model: ``Option value does not fall with wealth,'' or ``Option value does not fall with unemployment,'' or ``Option value does not fall with the temperature in Bloomington.'' The statement ``Option value does not fall with risk,'' however, though it does translate the mathematical notation of Proposition 1, is unnecessarily weak. We can instead say that ``Option value does not fall with risk, and for at least one value of the strike price it increases.'' Propostion 1b expresses this in mathematical notation. \noindent {\bf Proposition 1b: }{\it If $g$ is riskier than $f$, then there exists some exercise price $p'$ such that the associated call and put options are more valuable under $g$ than under $f$, but no exercise price $p''$ such that they are more valuable under $f$:\\ $\;\;\;\;\; \exists\; p': V_{call}(f, p') < V_{call}(g, p') \; and \; V_{put}(f, p') < V_{put}(g, p') $; \\ but \\ $ \;\;\;\;\;\not\exists \;p'': V_{call}(f, p'' ) > V_{call}(g,p'' ) \; or \; V_{put}(f, p') > V_{put}(g, p') $. } \noindent {\bf Proof: } Proposition 1's proof showed that if $p \in (y_1, y_3)$, where $y_1$ and $y_3$ are from one of the spreads that makes $g$ riskier than $f$, then the values of the call strictly increases. Thus, simply pick $p'$ from inside $(y_1, y_3)$. That there exists no value $p''$ for which option value declines is a direct corollary of Proposition 1. QED. \bigskip \noindent \textbf{III. New Definitions of Risk} Another approach is to find a definition of risk under which something like the false Proposition 1b becomes true, and the value of the option does strictly increase with ``risk'' regardless of the strike price. \noindent {\bf Definition 2: }{\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 $g(x) > f(x)$.} Definition 2 says that $g(x)$ is pointwise riskier than $f(x)$ if it takes probability away from {\it each point} in the middle of the distribution and adds probability to {\it each point} at the two ends, while preserving the mean. Distribution $g(x)$ in Figure 2 is an example. Definition 2 also allows $g(x)$ to add probability to points outside the interval $[x_1, x_m]$--- that is, beyond the two extremes of the support of $f(x)$. Pointwise riskiness captures something of the same intuition as the idea of the mean-preserving spread-- that probability is to be moved from the middle to the ends of the distribution. If $g$ is pointwise riskier than $f$, it is also riskier in the conventional sense. Note that the change from $f$ to $g$ need not be symmetric around the mean of the distribution. Every point on the ``sides'' must gain probability, but not necessarily the same amount of probability, nor must more extreme points gain more probability than less extreme ones. Note, too, that the definition applies directly to continuous distributions, where it has the additional implication that the densities of $f$ and $g$ cross twice, at $\underline{x}$ and $\overline{x}$, as density $f$ does with $g$ and $h$ in Figure 4 below. \begin{figure} \centering %\includegraphics[width=80mm]{options2.jpg} \caption{Figure 2: Pointwise and Extremum Riskiness} \end{figure} Pointwise riskiness will be sufficient but not necessary for option value to increase with risk for all strike prices, as we will see later once we have derived other results useful in proving sufficiency.\footnote{\label{strict} Since pointwise riskiness and strict second order stochastic dominance both can be defined in terms of functions that cross a limited number of times, the reader may wonder if they are the same. Distribution $F$ strictly second-order stochastically dominates $G$ if $ \int_0^t G(x)dx > \int_0^t F(x)dx $ for all values of $t$ such that $G(t)>0$ and $G(t)<1$. It could be, however, that $G$ is pointwise riskier than $F$ without $F$ strictly second-order dominating $G$. Suppose, for example, that $F$ is uniform, with $F(1) = 0.25, F(2) = 0.5, F(3)=0.75, F(4) = 1$ and $G$ moves weight from the middle to the tails and is pointwise riskier so $G(1) = 0.30, G(2) = 0.5, G(3)=.7, G(4) = 1$. If we define $D_F(t) \equiv \int_0^t F(x)dx$ (and similarly for $G)$ then $D_F(1) = 0.25, D_F(2) = 0.75, D_F(3)=1.5$ and $D_G(1) = .30, D_G(2) = .8, D_G(3)=1.5$. Since $D_F(3)= D_G(3)$, $F$ does not strictly dominate $G$. $F$ does weakly dominate $G$, as we would expect since $G$ is riskier in the conventional sense. } Distribution $h(x)$ in Figure 2 is an example in which $h$ is not pointwise riskier than $f$, but $ V_{call}(h,p) >V_{call}(f, p) $ for all $p$ nonetheless. Pointwise riskiness is stronger than the standard risk of Definition 1b; if $g$ is pointwise riskier than $f$ it is riskier too, but $g$ could be riskier without being pointwise riskier.\footnote{ Other strengthened definitions of riskiness have been proposed in the context of comparative statics--- for example, how an agent's investment portfolio responds to an increase in risk. Often the response to an increase in standard risk is ambiguous,and clear predictions can only be made for increases in stronger, nonstandard, forms of riskiness. Gollier (1995) has shown what definition is both necessary and sufficient for making predictions about the behavior of any risk-averse agent when riskiness increases. As one might expect from such a strong result, the definition of riskiness he uses is complicated enough to be difficult to use in applications. Another approach, transforming the random variable, was pioneered in Sandmo (1971). In this approach, the distribution function $F(x)$ is transformed to a new distribution by replacing $x$ with $t(x)$ for a suitable function $t$ that keeps the mean constant but increases risk. Sandmo (1971) uses a transformation in which $t(x)-x$ is linear. Meyer \& Ormiston (1989) generalize to the ``simple transformation,'' in which $t(x)-x$ can be any monotonic function. This approach provides a way to think about ``stretching out'' distributions, but it will increase the support of the distribution, something not necessarily true of pointwise riskiness. } \bigskip Our other new definition of risk is one which will be {\it necessary} for extra risk to increase option value: extremum risk. \noindent {\bf Definition 3: }{\it Distribution $g(x)$ is {\bf extremum riskier } than $f(x)$ iff $G(x_1+\epsilon)> F(x_1+\epsilon)$ and $ G(x_m-\epsilon)< F(x_m-\epsilon)$ for arbitrarily small $\epsilon>0$. } This is stated in terms of the cumulative distributions of $f$ and $g$, but in discrete distributions it has easy-to-understand implication that (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 $x > x_m$. Definition 3 is slightly stronger than this, however, because it says that if $g$ extends to more extreme values of $x$ it must also increase the total probability of values of $x$ beyond $x_1$ and $x_m$. In Figure 2 (above), distribution $h(x)$ is extremum riskier than $f(x)$ or $g(x)$. In Figure 3 (below), $g(x)$ is both riskier and extremum riskier than $f(x)$, but not riskier, since the probability of both the mean and the extreme values have increased. \begin{figure} \centering %\includegraphics[width=80mm]{options3.jpg} \label{options1.jpg } \caption{Figure 3: Extremum Riskiness versus Conventional Risk} \end{figure} The definition of extremum riskiness applies to continuous as well as discrete distributions. Cumulative distributions must be used because if $f$ is a continuous density then each of the extrema has zero probability, even if positive density, and to change the value of an option it is necessary to change probabilities over an interval of $f$'s support, not just over one point. The density $g$ must put more probability on the intervals $[-\infty, x_1+\epsilon] $ and $[x_m- \epsilon, \infty]$. If the density $f$'s support is unbounded (which cannot happen with a discrete distribution) then the numbers $x_1$ and $x_m$ are no longer the bounds of $f$'s support. We can continue to define them, however, as the bounds of the interval in which the strike price lies, so $x_1

V_{call} (f,p)$
for any strike price $p$ is that $g$ be extremum-riskier than $f$. That $g$ be extremum-riskier than $f$ is also a necessary condition for $V_{put}(g, p) > V_{put}(f,p)$
for any strike price $p$. }
\noindent
{\bf Proof: }
Let us begin with calls. Since we must show that for any call, $V_{call}(g, p) > V_{call} (f,p)$, that must be true for $p= x_1+\epsilon$ and $p=x_m -\epsilon$, for arbitrarily small $\epsilon>0$. We need to show that $V_{call}(g, p)- V_{call}(f, p)>0$, so from equation (\ref{e1})'s notation for $g$'s support and equation (\ref{e2}) for call value, we must show that
\begin{equation} \label{e200}
\begin{array}{l}
{\displaystyle
\beta \left( \sum_{i=1}^m g(x_i) Max \{0, (x_i- p)\} + \sum_{i= m+1}^{n} g(x_i) Max \{0, (x_i- p)\} \right) - \beta \sum_{i=1}^m f(x_i) Max \{0, (x_i- p)\} >0 } \\
\end{array}
\end{equation}
where $x_{m+1} p$ but not on every
$x_i$ individually. Thus, it is possible that $g(x_k) < f(x_k)$ for some
particular value of $x_k >p$ in a way that makes it impossible to rank $f$ and
$g$ by risk, but for that to be outweighed by $g$'s greater weight on most
high values of $x_i$. We can generalize this to the idea that an
option can be more valuable for {\it any} price $p$ even though risk does not
rise. We can find a $g$ that puts so much weight on its
extrema compared to $f$ that $g$'s expected values over $x_i >p$ will be greater
even if it puts more weight on the mean of $x$ too.
\begin{figure}
\centering
%\includegraphics[width=80mm]{options6.jpg}
\caption{Figure 6: Conventional and Extremum Riskiness Are Not Jointly Necessary
to Increase Option Value }
\end{figure}
Now let us leave extremum riskiness and look back to the second new
definition of ``riskier'': ``pointwise riskiness''. In
applications, it is convenient to specify a simple sufficient condition for
options on one distribution to have higher value than those on another. Proposition 4 says that pointwise
riskiness is such a condition.
\noindent
{\bf Proposition 4: }{\it If $g$ is pointwise riskier than $f$, then for any
$p$, $V_{call}(g,p) >V_{call}(f, p) $ and $ V_{put}(g,p)> V_{put}(f, 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 that calls and puts on $g$ will be more valuable than
calls and puts on $f$. Q.E.D.
\bigskip
We have already found one sufficient condition for options on $g$
to be more valuable than options on $f$, the combination of riskiness
and extremum riskiness in Proposition 3. Proposition 3, in
fact, provides a tighter sufficient condition than Proposition 4. If $g$ is pointwise riskier than $f$
it is always both riskier and extremum riskier--- but $g$ can be riskier
and extremum riskier without being pointwise riskier.
Pointwise riskiness is nonetheless a useful concept, because it is simpler and more
intuitive than the combined conditions.
\bigskip
\noindent
\textbf{V. Concluding Remarks}
If distribution $g$ is riskier than distribution $f$, then any call option on
an asset whose value has distribution $g$ will be at least as valuable as the
equivalent option on an asset with distribution $f$. But the option on $g$ might
not be {\it more} valuable, because the values might be equal. This paper has
developed a necessary condition for all call options on an asset whose value
has distribution $g$ to be strictly more valuable than the equivalent option on
an asset with distribution $f$, and two sufficient conditions for it, differing
in strength and convenience. The necessary condition is that $g$ be
``extremum riskier'': it must put more probability on the extreme values of the
asset. One sufficient condition is that $g$ be not only extremum riskier, but
also riskier under the conventional definition of risk-- that $g$ can be reached
from $f$ by a series of mean-preserving spreads. A second sufficient condition,
more restrictive but simpler, is that $g$ be ``pointwise riskier'': asset
values in the middle of $g$ have higher probability than under $f$, and asset
values outside the middle have lower probability.
\newpage
\noindent
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\end{raggedright}
\end{document}