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Black-Scholes美國看跌期權

這是我的問題:這是一個關於Black-Scholes模型的問題,但它可能適用於更復雜的模型。在整個討論中,執行價格K $,利率$ r $和波動率$ \ sigma $將被假定為不變。我們感興趣的是選項的時間衰減。

首先考慮一個永久性的美式問題。這有一個最佳的運動水平,我稱之為$ b $,這是由$ b = \ frac {K} {1+ \ sigma ^ 2/2r} $給出的

Now consider any $x>b$, this $x$ is a level, for which it would not be optimal to exercise in a perpetual option. In this case, does this necessarily mean, there exist a $T$ such that $t\in (0,T)$, the equation below holds

$\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ > (K-x)^+$

這裏$ \ tau_b $表示$ b $等級的第一個點擊時間。

我想我已經證明這是$ t $的情況。這是我的證據:

Denote $V(x)=\mathbb{E}_x e^{-r\tau_b}(K-X_{\tau_b})^+$. It is widely known that $V(x)>(K-x)^+$. Assume the statement above is false, then

$\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ \leq (K-x)^+$ for all $t>0$.

然後我們可以在左側采用$ t \ uparrow \ infty $的限制並應用支配收斂定理,我們看到

$ V(x)\ leq(K-x)^ + $

Contradicting the inequality $V(x)>(K-x)^+$

However, I have struggled to show $\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ > (K-x)^+$ for small value of $t$. I believe this conjecture is true for all $t>0$ but struggled to prove it. Anyone has any ideas?

最佳答案

我有一些想法。這種說法是錯誤的。原因如下。

Consider a finite horizon problem with the time to maturity $t$, then there exists an optimal stopping boundary $B(t)$, such that if the stock price is at this level, we should exercise the option. Notice, this $B(t)>b$ and it is an increasing function of $t$.

If we start our process, at $B(t)$ with amount of time $t$ left to run, then since we started at the optimal stopping boundary so $(K-x)^+ > \mathbb{E}_{B(t)} (K-X_\tau)^+$ for any stopping time $\tau

In fact, the result I asked for is totally ridiculous, because as $t$ goes to $0$, $B(t) \uparrow K$. For small times, it is a lot better to stay where you are than let the process run on.

轉載註明原文: Black-Scholes美國看跌期權