Asset Pricing:Valuation
Asset Pricing:Valuation
complete market: M = R S → q ( z ) M=R^S\to q(z) M=RS→q(z)
incomplete market: M ⫅ R S → Q ( z ) M\subseteqq R^S\to Q(z) M⫅RS→Q(z) (Valuation functional)
估值泛函(Valuation functional) 是对收益定价泛函的推广:定义如下:
Q : R S → R ∀ z ∈ M , Q ( z ) = q ( z ) Q:R^S\to R\\\forall z\in M,Q(z)=q(z) Q:RS→R∀z∈M,Q(z)=q(z)
Fundamental Theorem of Finance:
- securities price exclude Arbitrage iff ∃ Q ( z ) \exist Q(z) ∃Q(z) is strictly positive ⇔ \Leftrightarrow ⇔ if z>0, Q(z)>0
- weaker form : securities price exclude Strong Arbitrage iff ∃ Q ( z ) \exist Q(z) ∃Q(z) is positive.
充分性:由下列定理可得:
- Theorem : q ( z ) q(z) q(z) is linear and strictly positive iff NA.
- Theorem : q ( z ) q(z) q(z) is linear and positive iff NSA.
因为存在严格为正(正)的估值泛函可以得出存在严格为正(正)的收益定价泛函,收益定价泛函是对估值定价的限制。余下证明必要性。
将收益定价泛函从asset span推广到整个商品空间,可以通过将 q q q 每次增加一维得到。
- First step:选择不属于asset span M M M 的未定权益 z ^ \hat z z^,将 q q q 推广到由 M M M 和 z ^ \hat z z^ 张成的子空间上,其维数等于 M M M 的dimension+1。通过对未定权益 z ^ \hat z z^ 设定一个值 π \pi π 就可以将收益定价泛函推广。为保留严格正(正),所选的 π \pi π 须满足所有大于 z ^ \hat z z^ 的收益的价格严格大于(大于) π \pi π,所有小于 z ^ \hat z z^ 的收益的价格严格小于(小于) π \pi π,这些约束给出了 π \pi π 所在的区间。The extension is the payoff pricing functional for security markets consisting of J J J securities with payoffs { x 1 , ⋯ , x J } \{x_1,\cdots,x_J\} {x1,⋯,xJ} and prices { p 1 , ⋯ , p J } \{p_1,\cdots,p_J\} {p1,⋯,pJ} and a security with payoff z ^ \hat z z^ and price π \pi π.
- Second step:选择不属于第一步中 J + 1 J+1 J+1 种证券的张成空间的未定权益,重复第一步,经过 S − J S-J S−J 步后,成功推广到整个商品空间。
How to do the First step?
未定权益 z ∈ R S z\in R^S z∈RS 的upper bound:
q u ( z ) ≡ min h { p h : h X ≥ z } q_u(z)\equiv\min_h\{ph:hX\geq z\} qu(z)≡hmin{ph:hX≥z}
if h h h does not exist, then q u ( z ) = + ∞ q_u(z)=+\infin qu(z)=+∞
lower bound:
q l ( z ) ≡ max h { p h : h X ≤ z } q_l(z)\equiv\max_h\{ph:hX\leq z\} ql(z)≡hmax{ph:hX≤z}
if h h h does not exist, then q l ( z ) = − ∞ q_l(z)=-\infin ql(z)=−∞
Example : x 1 = ( 1 , 2 ) , z = ( 1 , 1 ) , s = 2 , P 1 = 1 x_1=(1,2),z=(1,1),s=2,P_1=1 x1=(1,2),z=(1,1),s=2,P1=1, M = { ( h , 2 h ) : h ∈ R } M=\{(h,2h):h\in R\} M={(h,2h):h∈R}
q l ( z ) = max h { p h : h X ≤ z } = max h { h : ( h , 2 h ) ≤ ( 1 , 1 ) } , q l ( z ) = 1 2 q_l(z)=\max_h\{ph:hX\leq z\}=\max_h\{h:(h,2h)\leq(1,1)\},q_l(z)=\dfrac{1}{2} ql(z)=maxh{ph:hX≤z}=maxh{h:(h,2h)≤(1,1)},ql(z)=21
q u ( z ) = min h { p h : h X ≥ z } = min h { h : ( h , 2 h ) ≥ ( 1 , 1 ) } , q u ( z ) = 1 q_u(z)=\min_h\{ph:hX\geq z\}=\min_h\{h:(h,2h)\geq(1,1)\},q_u(z)=1 qu(z)=minh{ph:hX≥z}=minh{h:(h,2h)≥(1,1)},qu(z)=1
Proposition 1 : if NSA, then ∀ z ∈ M , q u ( z ) = q l ( z ) = q ( z ) \forall z\in M,q_u(z)=q_l(z)=q(z) ∀z∈M,qu(z)=ql(z)=q(z)
proof:for z ∈ M z\in M z∈M, q u ( z ) ≤ q ( z ) , q l ( z ) ≥ q ( z ) q_u(z)\leq q(z),q_l(z)\geq q(z) qu(z)≤q(z),ql(z)≥q(z)
suppose ∃ z ∈ M , q u ( z ) < q ( z ) \exist z\in M,q_u(z)<q(z) ∃z∈M,qu(z)<q(z), then q u ( z ) < + ∞ q_u(z)<+\infty qu(z)<+∞, ∃ h ′ \exist h' ∃h′ satisfy:
h ′ X ≥ z p h ′ < q ( z ) h'X\geq z\\ph'<q(z) h′X≥zph′<q(z)
令 h h h 满足 h X = z , p h = q ( z ) hX=z,ph=q(z) hX=z,ph=q(z),则 ( h ′ − h ) (h'-h) (h′−h) 为SA,与假设矛盾,所以 q u ( z ) = q ( z ) q_u(z)=q(z) qu(z)=q(z)。类似的可以证明 q l ( z ) = q ( z ) q_l(z)=q(z) ql(z)=q(z)。
Proposition 2 : if NSA, then ∀ z ∈ R S , z ∉ M , q u ( z ) ≥ q l ( z ) \forall z\in R^S,z\notin M,q_u(z)\geq q_l(z) ∀z∈RS,z∈/M,qu(z)≥ql(z), ∀ z ∈ R S , q u ( z ) > − ∞ , q l ( z ) < + ∞ \forall z\in R^S,q_u(z)>-\infty,q_l(z)<+\infty ∀z∈RS,qu(z)>−∞,ql(z)<+∞
proof:
by controdition, suppose ∃ z ∈ R S , q u ( z ) < q l ( z ) \exist z\in R^S,q_u(z)<q_l(z) ∃z∈RS,qu(z)<ql(z), then q u ( z ) < + ∞ , q l ( z ) > − ∞ q_u(z)<+\infty,q_l(z)>-\infty qu(z)<+∞,ql(z)>−∞,and ∃ h ′ , h ′ ′ , s . t . \exist h',h'',s.t. ∃h′,h′′,s.t.
h ′ X ≤ z ≤ h ′ ′ X , p h ′ > p h ′ ′ h'X\leq z\leq h''X,ph'>ph'' h′X≤z≤h′′X,ph′>ph′′
but 投资组合 ( h ′ ′ − h ′ ) (h''-h') (h′′−h′) 满足:
p ( h ′ ′ − h ′ ) < 0 , ( h ′ ′ − h ′ ) X ≥ 0 p(h''-h')<0,(h''-h')X\geq0 p(h′′−h′)<0,(h′′−h′)X≥0
因此为SA,矛盾。Assume no redundant security without loss of generality。
by controdiction, Assume ∃ z ∈ R S , q u ( z ) = − ∞ \exist z\in R^S,q_u(z)=-\infty ∃z∈RS,qu(z)=−∞, then ∃ { h n } , s . t . \exist \{h^n\},s.t. ∃{hn},s.t.
h n X ≥ z , lim n → ∞ p h n = − ∞ → lim ∣ ∣ h n ∣ ∣ = + ∞ h^nX\geq z,\lim_{n\to\infty}ph^n=-\infty\\\to\lim||h^n||=+\infty hnX≥z,n→∞limphn=−∞→lim∣∣hn∣∣=+∞
consider 有界序列 { h n / ∣ ∣ h n ∣ ∣ } → h ^ ≠ 0 \{h^n/||h^n||\}\to\hat h\neq0 {hn/∣∣hn∣∣}→h^=0:
h n X ≥ z , lim n → ∞ p h n = − ∞ h n ∣ ∣ h n ∣ ∣ X ≥ z ∣ ∣ h n ∣ ∣ n → ∞ : h ^ X ≥ 0 , p h ^ ≤ 0 h^nX\geq z,\lim_{n\to\infty}ph^n=-\infty\\\frac{h^n}{||h^n||}X\geq\frac{z}{||h^n||}\\n\to\infty :\hat hX\geq0,p\hat h\leq0 hnX≥z,n→∞limphn=−∞∣∣hn∣∣hnX≥∣∣hn∣∣zn→∞:h^X≥0,ph^≤0
h ^ ≠ 0 \hat h\neq0 h^=0 , there is no redundant security, so h ^ X ≠ 0 → h ^ X > 0 → h ^ \hat hX\neq0\to\hat hX>0\to \hat h h^X=0→h^X>0→h^ is arbitrage 矛盾类似的可以证明 q l ( z ) < + ∞ q_l(z)<+\infty ql(z)<+∞
Proposition 3 : if NA, ∀ z ∈ R S , z ∉ M , q u ( z ) > q l ( z ) \forall z\in R^S,z\notin M,q_u(z)>q_l(z) ∀z∈RS,z∈/M,qu(z)>ql(z)
proof:根据Propositon 2 :NA → \to → NSA → \to → q u ( z ) ≥ q l ( z ) q_u(z)\geq q_l(z) qu(z)≥ql(z),只需证明: ∀ z ∉ M , q u ( z ) ≠ q l ( z ) \forall z\notin M,q_u(z)\neq q_l(z) ∀z∈/M,qu(z)=ql(z)
by controdiction, Assume ∃ z ∉ M , q u ( z ) = q l ( z ) \exist z\notin M,q_u(z)=q_l(z) ∃z∈/M,qu(z)=ql(z), then 根据Propositon 2 的第二部分的证明得: q u ( z ) < + ∞ , q l ( z ) > − ∞ q_u(z)<+\infty,q_l(z)>-\infty qu(z)<+∞,ql(z)>−∞. So ∃ h ′ , h ′ ′ , s . t . \exist h',h'',s.t. ∃h′,h′′,s.t.
h ′ X ≤ z ≤ h ′ ′ X p h ′ = p h ′ ′ = q u ( z ) h'X\leq z\leq h''X\\ph'=ph''=q_u(z) h′X≤z≤h′′Xph′=ph′′=qu(z)
前面的不等号不能取等,因为 z z z 不在资产张成空间中→ ( h ′ ′ − h ′ ) X > 0 , p ( h ′ ′ − h ′ ) = 0 \to(h''-h')X>0,p(h''-h')=0 →(h′′−h′)X>0,p(h′′−h′)=0, h ′ ′ − h ′ h''-h' h′′−h′ is Arbitrage, 矛盾
(推广)Extension:
给定未定权益 z ^ ∉ M \hat z\notin M z^∈/M,定义 N N N:
N = { z + λ z ^ : z ∈ M , λ ∈ R } N=\{z+\lambda\hat z:z\in M,\lambda\in R\} N={z+λz^:z∈M,λ∈R}
N ⫅ R S , d i m ( N ) = d i m ( M ) + 1 N\subseteqq R^S,dim(N)=dim(M)+1 N⫅RS,dim(N)=dim(M)+1, N N N 包含 M , z ^ M,\hat z M,z^。
if NSA(等价地,if payoff functional q ( z ) q(z) q(z) is positive),then by Proposition 2 and 3, q u ( z ) ≥ q l ( z ) q_u(z)\geq q_l(z) qu(z)≥ql(z) . if z ^ ∉ M , q u ( z ^ ) > q l ( z ^ ) \hat z\notin M,q_u(\hat z)>q_l(\hat z) z^∈/M,qu(z^)>ql(z^) .
选择有限值 π \pi π 满足:
q l ( z ^ ) ≤ π ≤ q u ( z ^ ) q_l(\hat z)\leq\pi\leq q_u(\hat z) ql(z^)≤π≤qu(z^)
将 q q q 推广为 N N N 上的线性泛函,定义 Q : N → R Q:N\to R Q:N→R:
Q ( z + λ z ^ ) ≡ q ( z ) + λ π Q(z+\lambda\hat z)\equiv q(z)+\lambda\pi Q(z+λz^)≡q(z)+λπ
下面证明所定义的 Q Q Q 是所期望的 q q q 的正推广。
Proposition 4 : Q : N → R Q:N\to R Q:N→R is positive if q : M → R q:M\to R q:M→R is positive
proof:let y ∈ N → ∃ z ∈ M , λ ∈ R , s . t . y\in N\to \exist z\in M,\lambda\in R,s.t. y∈N→∃z∈M,λ∈R,s.t.
y = z + λ z ^ y=z+\lambda\hat z y=z+λz^
λ > 0 \lambda>0 λ>0.
y ≥ 0 → z ^ ≥ − z λ → q l ( z ^ ) ≥ q l ( − z λ ) y\geq0\to\hat z\geq-\dfrac{z}{\lambda}\to q_l(\hat z)\geq q_l(-\dfrac{z}{\lambda}) y≥0→z^≥−λz→ql(z^)≥ql(−λz) ( q l ( ⋅ ) q_l(·) ql(⋅) is increasing. q u ( ⋅ ) q_u(·) qu(⋅) is increasing)
− z λ ∈ M -\dfrac{z}{\lambda}\in M −λz∈M, by Propositon 1 → q l ( − z λ ) = q ( − z λ ) → q l ( z ^ ) ≥ q ( − z λ ) \to q_l(-\dfrac{z}{\lambda})=q(-\dfrac{z}{\lambda})\to q_l(\hat z)\geq q(-\dfrac{z}{\lambda}) →ql(−λz)=q(−λz)→ql(z^)≥q(−λz)
因为 π ≥ q l ( z ^ ) → π ≥ q ( − z λ ) → λ π + q ( z ) ≥ 0 \pi\geq q_l(\hat z)\to\pi\geq q(-\dfrac{z}{\lambda})\to\lambda\pi+q(z)\geq0 π≥ql(z^)→π≥q(−λz)→λπ+q(z)≥0
由于 Q ( y ) = Q ( z + λ z ^ ) = q ( z ) + λ π ≥ 0 Q(y)=Q(z+\lambda\hat z)=q(z)+\lambda\pi\geq0 Q(y)=Q(z+λz^)=q(z)+λπ≥0,所以得到 Q ( y ) ≥ 0 Q(y)\geq0 Q(y)≥0。
λ < 0 \lambda<0 λ<0
y ≥ 0 → z ^ ≤ − z λ → q u ( z ^ ) ≤ q u ( − z λ ) y\geq0\to\hat z\leq-\dfrac{z}{\lambda}\to q_u(\hat z)\leq q_u(-\dfrac{z}{\lambda}) y≥0→z^≤−λz→qu(z^)≤qu(−λz)
by Propositon 1 → q u ( − z λ ) = q ( − z λ ) → q u ( z ^ ) ≤ q ( − z λ ) \to q_u(-\dfrac{z}{\lambda})=q(-\dfrac{z}{\lambda})\to q_u(\hat z)\leq q(-\dfrac{z}{\lambda}) →qu(−λz)=q(−λz)→qu(z^)≤q(−λz)
因为 π ≤ q u ( z ^ ) → π ≤ q ( − z λ ) → λ π + q ( z ) ≥ 0 \pi\leq q_u(\hat z)\to\pi\leq q(-\dfrac{z}{\lambda})\to\lambda\pi+q(z)\geq0 π≤qu(z^)→π≤q(−λz)→λπ+q(z)≥0
由于 Q ( y ) = Q ( z + λ z ^ ) = q ( z ) + λ π ≥ 0 Q(y)=Q(z+\lambda\hat z)=q(z)+\lambda\pi\geq0 Q(y)=Q(z+λz^)=q(z)+λπ≥0,所以得到 Q ( y ) ≥ 0 Q(y)\geq0 Q(y)≥0。
λ = 0 , y = z ≥ 0 , Q ( y ) = q ( z ) ≥ 0 \lambda=0,y=z\geq0,Q(y)=q(z)\geq0 λ=0,y=z≥0,Q(y)=q(z)≥0
if NA(等价地,if payoff functional q ( z ) q(z) q(z) is strictly positive),then by Proposition 3, 选择有限值 π \pi π 满足:
q l ( z ^ ) < π < q u ( z ^ ) q_l(\hat z)<\pi< q_u(\hat z) ql(z^)<π<qu(z^)
Propositon 5 : Q : N → R Q:N\to R Q:N→R is strictly positive if q : M → R q:M\to R q:M→R is strictly positive
proof:类似于上述。
唯一性(Uniqueness):
Theorem:Assume NA, Market is complete iff only ∃ \exist ∃ one Q ( z ) > 0 Q(z)>0 Q(z)>0(strictly positive).
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