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Black-Scholes公式的連續性

How to proof B&S pricing formula is continuous in time $t$ (or it is not?).

The general pricing formula is $$ C_t = e^{-r(T-t)} \mathbb{E}^*[(S_T-K)^+ | \mathcal{F}_t] \hspace{1cm} 0\leq t\leq T $$ Then for time at maturity $t=T$ $$ C_T = \mathbb{E}^*[(S_T-K)^+ | \mathcal{F}_T] = (S_T-K)^+ $$ which is logic. For other anterior time $t

最佳答案

首先,讓我們記住,通過$ \ mathcal {N} $作為累積分布函數的屬性,我們得到$$ \ lim_ {x \ to - \ infty} \ mathcal {N}(x)= 0 $$和$ $ \ lim_ {x \ to \ infty} \ mathcal {N}(x)= 1 $$。

現在,讓我們看看Black-Scholes價格隨著時間的推移如何表現。首先,我們有

$$ d \ pm = \ left(\ frac {\ text {ln} \ frac {S_t} {K}} {\ sigma \ sqrt {Tt}} +(r \ pm \ frac {\ sigma ^ 2} {2}) \ SQRT {Tt的} \右) $$

$$ C_t = \ left(S_t-e ^ { - r(Tt)} K \ right)\ cdot \ mathcal {N}(d +)+ e ^ { - r(Tt)} K \ cdot \ left(\ mathcal {N}(d +) - \ mathcal {N}(d - )\右)$$

可以進一步改寫為

$$ C_t = \ left(S_t-e ^ { - r(Tt)} K \ right)\ cdot \ mathcal {N}(d +)+ \ frac {e ^ { - r(Tt)} K} {\ sqrt {2 \ pi}} \ int_ {d - } ^ {(d - )+ \ sigma ^ 2 \ sqrt {Tt}} e ^ { - \ frac {z ^ 2} {2}} \,dz $$

首先,註意到$ z \ mapsto e ^ { - \ frac {z ^ 2} {2}} $限制在$ \ mathbb {R} $之上,我們可以推斷出第二個期限為$ 0 $ $ $ $因為積分間隔變得任意小,所以轉到$ T $。

因此,$ C_t $的收斂由第一個期限的收斂決定,該期限具有不同的限制,具體取決於$ S_T/K $:

  • if $S_T > K$, then $\ln\frac{S_T}{K} > 0$ 和 $\lim d+ = \infty$, so $\lim C_t = \left(S_T - K\right)\cdot\mathcal{N}(\infty) = S_T - K$
  • if $S_T < K$, then $\ln\frac{S_T}{K} < 0$ 和 $\lim d+ = -\infty$, so $\lim C_t = \left(S_T - K\right)\cdot\mathcal{N}(-\infty) = 0$
  • if $S_T = K$, then $S_t$ goes to $K$, 和 remembering that $\mathcal{N}$ is bounded by $0$ 和 $1$, we have $\lim C_t = 0$

轉載註明原文: Black-Scholes公式的連續性