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以股票的形式推導Black Scholes PDE

有很多方法可以推導出Black Scholes PDE。鞅的方式是要求期權價格根據特定的措施無漂移。下面我使用銀行賬戶作為計數器導出正確的PDE,但在將股票用作計價器時未能獲得正確的PDE。我希望有人能夠指出我做錯了什麽。

使用銀行賬戶作為計價來推導Black Scholes PDE

推導Black-Scholes方程的方法之一是將銀行賬戶$ B_t $作為計價器,然後要求$ d \ frac {C_t} {B_t} $無漂移。下面我保留下標表示時間隱含。

具體而言,在這個計價單$ W_B $(其中$ B $代表銀行賬戶)

$$ dS=S r dt + S \sigma dW_B \\ dB=B r dt $$ so we simply get $$ \begin{eqnarray} d\frac{C}{B} &=& \frac{\partial_t C dt + \partial_S CdS + \frac{1}{2} \partial_{S,S} CdS^2 }{B}-\frac{CdB}{B^2} \\ &=& \frac{\partial_t C + r S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C -rC}{B} dt + \frac{\sigma S \partial_S C}{B} dW_B + \mathcal O({dt}^{3/2}) \end{eqnarray} $$ and demanding that $\frac{C}{B}$ be a Martingale requires the vanishing of the drift term and we get the Black Scholes PDE: $$ \partial_t C + r S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C -rC=0 $$

嘗試使用股票作為計數來推導Black Scholes PDE

Now I try to do the same while taking the Stock as a numeraire. I will demand, as usual, that $d \frac{C}{S}$ is a Martingale under this measure. Under this measure we have $$ dS = S(r+\sigma^2) dt + S \sigma dW_S $$ so we get $$ \begin{eqnarray} d\frac{C}{S} &=& \frac{\partial_t C dt + \partial_S CdS + \frac{1}{2} \partial_{S,S} CdS^2 }{S}-\frac{CdS}{S^2} + \frac{CdS^2}{S^3} \\ &=& \frac{\partial_t C + (r+\sigma^2) S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C}{S} dt + \frac{\sigma S\partial_S C dW_S}{S} - \frac{C}{S}\big((r+\sigma^2) dt + \sigma dW_S \big)+\frac{C}{S}\sigma^2 dt +\mathcal O(dt^{3/2}) \\ &=& \frac{\partial_t C + (r+\sigma^2) S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C -rC}{S} dt + \frac{\sigma S \partial_S C-C}{S} dW_S +\mathcal O({dt}^{3/2}) \end{eqnarray} $$

現在要求漂移項為零給我一個額外的術語 $$ \ partial_t C +(r + \ color {red} {\ sigma ^ 2})S \ partial_S + + frac {1} {2} \ sigma ^ 2 S ^ 2 \ partial_ {S,S} C -rC = 0 $$

最佳答案

您錯過了用於表達$ d \ left(\ frac {C_t} {S_t} \ right)$的Ito公式中的交叉導數項。更具體地說(見下面的 [備註] ),

$$ d \ left(\ frac {C_t} {S_t} \ right)= \ frac {1} {S_t} dC_t - \ frac {C_t} {S_t ^ 2} dS_t + \ frac {C_t} {S_t ^ 3} d \ langle S_t,S_t \ rangle {\ color {green} { - \ frac {1} {S_t ^ 2} d \ langle C_t,S_t \ rangle}} $$

這最後一個術語評估為 $$ - \ partial_S C_t \ sigma ^ 2 dt $$

意思是人們可以寫:

$$ d \ left(\ frac {C_t} {S_t} \ right)= \ frac {1} {S_t}(\ partial_t C_t dt + \ partial_S C_t dS_t + \ frac {1} {2} \ partial_ {SS} C_t \ sigma ^ 2 S_t ^ 2 dt) - \ frac {1} {S_t} \ left((r + \ sigma ^ 2)C_t dt + \ sigma C_t dW_t \ right)+ \ frac {1} {S_t} \ sigma ^ 2 C_t dt - \ frac {1} {S_t} \ partial_S_t \ sigma ^ 2 S_t dt $$ 或等同於重新安排一些條款 $$ d \ left(\ frac {C_t} {S_t} \ right)= \ frac {1} {S_t}(\ partial_t C_t + r S_t \ partial_S C_t + \ frac {1} {2} \ partial_ {SS} C_t \ sigma ^ 2 S_t ^ 2 - rC_t)dt +(。)dW_t $$

因此Black-Scholes pde來自鞅表示定理。

[Remark] This result simply comes from applying the bidimensional version of Ito's lemma $$df = (\partial_t f) dt + (\partial_X f) dX_t + \frac {1}{2} (\partial_{XX} f) d\langle X_t \rangle + (\partial_Y f) dY_t + \frac {1}{2} (\partial_{YY} f) d\langle Y_t \rangle + (\partial_{XY} f) d\langle X_t, Y_t \rangle$$

函數$ f(t,X_t,Y_t)= \ frac {X_t} {Y_t} $

轉載註明原文: 以股票的形式推導Black Scholes PDE