Asset Pricing：Preference over Risk and Uncertainty

Asset Pricing：Preference over Risk and Uncertainty

Suppose there are $X^j,X^k,X^j=[x_j^1,x_j^2,?,x_j^S]$ , how to rank?

1. State-by-State Dominance：given two random variables $X,Y$ defined over the state space $(\Omega ,F,P)$, we say that $Y$ state-by-state dominates $X$ if $?\omega \in \Omega ,X(\omega )\le Y(\omega )$.

   $X^j?X^k?x_s^j\ge x_s^k,?s$?

   for example：$?\begin{array}{c}3\\ 1\\ 2\end{array}?$ state-by-state dominates $?\begin{array}{c}2\\ 1\\ 0\end{array}?$

   Problem : 不符合完备性，如：$?\begin{array}{c}3\\ 1\\ 2\end{array}?、?\begin{array}{c}4\\ 0\\ 1\end{array}?$??无法比较。

2. Mean-Variance preference：assume that we like more expected return and less volatility.

   suppose we rank investments according to which Sharpe ratio is highest:
   $$\begin{gathered} S_x=\frac{E[R_x]?R^f}{\sigma (R_x)}\in R \\ >?? \end{gathered}$$

3. Stochastic Dominance : Ranking over Probability Distribution

   Definition（First Order Stochastic Dominance，FOSD）:

   Let $F_A,F_B$? denotes two CDFs of two random investments, defined on [a,b], we say that $F_A$? FOSD $F_B$? if $?x\in [a,b],F_A(x)\le F_B(x)$?.

   State-by-State Dominance $\to$? FOSD

   Definition（Second Order Stochastic Dominance，SOSD）:

   Let $F_A,F_B$ denotes two CDFs of two random investments, defined on [a,b], we say that $F_A$ SOSD $F_B$ if $?x\in [a,b],{\int }_a^x[F_B(t)?F_A(t)dt]\ge 0$.

   FOSD $\to$ SOSD

4. Mean-Preserving Spread :

   $X_A$? is MPS of $X_B$ if $X_A=X_B+\epsilon ,\epsilon$ is independent of A and B, $E(\epsilon )=0,\sigma (\epsilon )>0$?

   $X_A$ is risker than $X_B$ : $Var(X_A)=Var(X_B)+Var(\epsilon ),E[\epsilon |X_A]=E[\epsilon ]=0$

   Proposition : $X_A?F_A,X_B?F_B$? , defined on [a,b], $E[X_A]=E[X_B]$?, then $F_A$? SOSD $F_B$? iff $X_B$?? is MPS of $X_A$??

   proof（长篇警告！）：

   1. SOSD $\to$? MPS：

      首先证明Lemma：$X_B$ MPS $X_A??$ V(·) concave , $E[V(X_A)]\ge E[V(X_B)]$?

      我们知道 Jessan’s inequality : for any concave function V(x) , $E[V(x)]\le V[E(x)]$?

      同时有：Law of iterated Expectation : $E[x]=E[E[x|y]]$?

      $X_B$? MPS $X_A$:
      $$\begin{gathered} E[V(X_B)]=E[V(X_A+\epsilon )]=E[E[V(X_A+\epsilon )|X_A]] \\ \le E[V(E(X_A+\epsilon |X_A))]=E[V(X_A)] \end{gathered}$$
      Next 证明：$?$ V(·) concave , $E[V(X_A)]\ge E[V(X_B)]\to X_B$ MPS $X_A$?

      First we proof：if $?V(?)$??? concave, $E[V(X_A)]\ge E[V(X_B)]\to T(t)\equiv {\int }_a^t[F_B(s)?F_A(s)]ds\ge 0,?t,T(b)=0$?

      Consider the subset of increasing, concave functions $V_z(x)\equiv \min (x,z)$. The derivative of $V_z(x)$ is 1 for $x<z$ and 0 for $x>z$. Integrating by parts:
      $$\begin{gathered} 0\le E[V_z(X_A)]?E[V_z(X_B)]={\int }_a^b{V}_z(t)[d{F}_A(t)?d{F}_B(t)] \\ =V_z(t)[F_A(t)?F_B(t)]{|}_a^b?{\int }_a^b{V}_z^{\prime }(t)[F_A(t)?F_B(t)]dt \\ =?{\int }_a^z[F_A(t)?F_B(t)]dt\equiv T(z) \end{gathered}$$
      As this holds for all utility functions indexed by z, $T(z)$ must be nonnegative for all z.

      Next, we proof : $T(t)\equiv {\int }_a^t[F_B(s)?F_A(s)]ds\ge 0,?t,T(b)=0\to$? $X_B$? MPS $X_A$?

      Assume $E_A(x)=E_B(x)=0$?. Define :
      $$\begin{gathered} F_{AI}(z)\equiv {\int }_a^z{F}_A(x)dx,F_{BI}(z)\equiv {\int }_a^z{F}_B(x)dx \\ {F}_{AI}(a)=F_{BI}(a)=0,F_{AI}(b)=F_{BI}(b)=b \\ {F}_{BI}(z)\ge F_{AI}(z) \end{gathered}$$
      Assume that the latter is strictly true for $z\in (a,b)$????. ? If not the following construction can be applied separately to each interval in which the strict inequality does holds. Because $F_{AI}、F_{BI}$?? are integrals of increasing, nonnegative functions, they are both increasing, convex functions as shown in the figure.
      

      Now pick any point $z_0$??? on $F_{BI}$??? and consider a tangent line, $L$???, extended until it intersects $F_{AI}$??? at points $z^{\prime },z^{\prime \prime }$???. These intersections must occur in the interval [a,b] because $F_{AI}=F_{BI}$??? at those two endpoints. Define $H_1(z)=F_{AI}(z)$??? for $z\le z^{\prime },z\ge z^{\prime \prime }$??? and $H_1(z)=L(z)$??? for $z^{\prime }<z<z^{\prime \prime }$???.

      $H_1$ is the integral of some cumulative distribution function because it is increasing and convex. Furthermore, as its derivative is constant between $z^{\prime },z^{\prime \prime }$?, the cumulative distribution to which is belongs is also constant over that interval. This means that there is no probability mass within the interval $(z^{\prime },z^{\prime \prime })$ for the distribution $H_1$. In other words, the probability mass has been spread to the points $z^{\prime },z^{\prime \prime }$. The kinks in $H_1$ at those points mark the atoms of added probability. Furthermofe:
      $$E[X_A?z]={\int }_a^{b}t[d{F}_A(t)?d{H}_1(t)]=t[F_A(t)?H_1(t)]{|}_a^b?{\int }_a^b[F_A(t)?H_1(t)]dt=0$$
      So the underlying $H_1$??? distribution has the same mean as the $F_A$?? distribution. Thus, $H_1$?? is a simple mean-preserving spread of $F_A$??.

      $H_2$???? can be constructed by picking a point between B and $z_0$???? and a point between $z_0$???? and A, then running their tangents lines until they intersect with $H_1$????. $H_2$???? then represents a probability distribution that differs from $H_1$???? by two simple mean preserving spreads. Continuing in this fashion $H_n\to F_{BI}$???? pointwise. So in the limit, the distribution $F_A$???? is the same as the distribution $F_B$???? plus a number of simple mean-preserving spreads. If $F_A,F_B$????? are discrete distributions, then the match can be achieved with a finite number of simple mean-preserving spreads.

      Lemma得以证明。

      然后我们来证明：$?x\in [a,b],{\int }_a^x{F}_B(t)dt\ge {\int }_a^x{F}_A(t)dt\to ?V(?)$? concave on [a,b], $E[V(X_A)]\ge E[V(X_B)]$?，即 ${\int }_a^{b}V(t)d{F}_A(t)\ge {\int }_a^{b}V(t)d{F}_B(t)(??)$??

      Define $S_A(x)={\int }_a^x{F}_A(t)dt,S_B(x)={\int }_a^x{F}_B(t)dt,F_A(b)=F_B(b)=1,F_A(a)=F_B(a)=0$???

      $$\begin{gathered} {\int }_a^{b}V(t)d{F}_B(t)=[V(t)F_B(t)]{|}_a^b?{\int }_a^b{F}_B(t)dV(t)=V(b)?{\int }_a^b{F}_B(t)V^{\prime }(t)dt=V(b)?{\int }_a^b{V}^{\prime }(t)d{S}_B(t) \\ =V(b)?[V^{\prime }(t){\int }_a^t{F}_B(w)dw]{|}_a^b+{\int }_a^b({\int }_a^t{F}_B(w)dw)d{V}^{\prime }(t) \\ =V(b)?V^{\prime }(b){\int }_a^b{F}_B(w)dw+{\int }_a^b({\int }_a^t{F}_B(w)dw)V^{\prime \prime }(t)dt \end{gathered}$$

      ${\int }_a^b{F}_B(w)dw=[F_B(w)w]{|}_a^b?{\int }_a^{b}w{f}_B(w)dw=b?E(X_B),E(X_B)=E(X_A)$?

      ${\int }_a^{b}V(t)d{F}_B(t)=V(b)?V^{\prime }(b)(b?E(X_B))+{\int }_a^b({\int }_a^t{F}_B(w)dw)V^{\prime \prime }(t)dt$

      ${\int }_a^{b}V(t)d{F}_A(t)=V(b)?V^{\prime }(b)(b?E(X_A))+{\int }_a^b({\int }_a^t{F}_A(w)dw)V^{\prime \prime }(t)dt$?

      $(??)$? implies ${\int }_a^b({\int }_a^t{F}_A(w)dw)V^{\prime \prime }(t)dt\ge {\int }_a^b({\int }_a^t{F}_B(w)dw)V^{\prime \prime }(t)dt$

      SOSD $\to {\int }_a^t{F}_A(w)dw\le {\int }_a^t{F}_B(w)dw,V^{\prime \prime }(t)<0\to (??)$?? 成立

      所以SOSD $\to$$?V(?)$ concave , $E[V(X_A)]\ge E[V(X_B)]$ $?$ $X_B$ MPS $X_A$

   2. MPS $\to$? SOSD：

      conduct : $V_w(t)=\{\begin{array}{l}t,t\le w\\ w,t\ge w\end{array},t\in [a,b]$?? （一族符合条件的函数）?

      $$\begin{gathered} {\int }_a^b{V}_w(t)d{F}_A(t)={\int }_a^{w}td{F}_A(t)+{\int }_w^{b}wd{F}_A(t)=[t{F}_A(t)]{|}_a^w?{\int }_a^w{F}_A(t)dt+w[F_A(t)]{|}_w^b \\ =w{F}_A(w)?{\int }_a^w{F}_A(t)dt+w[1?F_A(w)] \\ =w?{\int }_a^w{F}_A(t)dt \end{gathered}$$

      $X_A$ MPS $X_B$ $?$ $?V(?)$ concave on [a,b], $E[V(X_A)]\ge E[V(X_B)]$，即 ${\int }_a^{b}V(t)d{F}_A(t)\ge {\int }_a^{b}V(t)d{F}_B(t)$?

      ${\int }_a^{b}V(t)d{F}_A(t)\ge {\int }_a^{b}V(t)d{F}_B(t)\to w?{\int }_a^w{F}_A(t)dt\ge w?{\int }_a^w{F}_B(t)dt\to {\int }_a^w{F}_B(t)dt\ge {\int }_a^w{F}_A(t)dt$

      证毕。