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0

Here is a good paper which can help you. https://www.rocq.inria.fr/mathfi/Premia/free-version/doc/premia-doc/pdf_html/mc_jourdainlelong_doc.pdf


2

I think the chain of logic should be as follows: We have put value >= intrinsic. Therefore either put value > intrinsic or put value= intrinsic. If put value > intrinsic, then it is not optimal to exercise. If put value = intrinsic , it may be optimal to exercise.. Hence there is no contradiction.


3

So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value. Yes indeed However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to indicate that it would be an unwise decision to ever ...


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If you have a formula for options on stock, you can turn it into a formula for options on futures by using the relation $S=F e^{-(r-d)T}$. In other words you observe the price of the future F, you turn it into S by this relation and then you pass this pseudo stock price to the options on stock function or program that you have. When you do this to the Black-...


1

Assuming deterministic interest rates, the price of an American call option struck at $K$ and expiring at $T$ is given by $$ V_0 = \text{sup}_{\tau \in \mathcal{T}[0,T]} \mathbb{E}_0^\mathbb{Q}\left[ e^{-r\tau} \max(S_{\tau}-K, 0) \right] $$ where $\mathcal{T}[0,T]$ denotes a family of stopping times with values in $[0,T]$ and where, under the risk-neutral ...


0

I would suggest to use : $$f(t,S^\theta_t)=\max(K-S^\theta_t,0)\exp(-\theta W_t\color{red}{\mathbf{-}}\frac{1}{2}\theta^2t)$$ where $dS^\theta_t=(r+\sigma\theta)S^\theta_t dt + \sigma S^\theta_t dW_t$


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With respect I think that this issue was associated Martingale properties AND dominated convergence theorem.(May be Wrong) Let $L\in(0,K)$ a fixed price, we can consider the following choices for the exercise of a put option with strike price $K$: If $S_t\le K$, then we exercise contract at time $t$, and were delighted. O.W. we should wait until the ...


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To elaborate on the explanation provided by @Alex, the reasoning is because when we look at the PDE we notice that the $S$ terms appear in pairs with the $\dfrac{\partial}{\partial S}$, i.e. $S\dfrac{\partial}{\partial S}$ and $S^2\dfrac{\partial^2}{\partial S^2}$. What this says it that if we were to try a polynomial function of $S$ then after applying ...


0

I just made things clearer hoping it would help. Let define $\mathbb{Q}_\theta$ as $$\frac{d\mathbb{Q}_\theta}{d\mathbb{P}}|_{\mathcal{F}_t}=\exp(\theta W_t -\frac{1}{2}\theta^2 t)=Z^\theta_t$$ By girsanov, if $W$ is a brownian motion under $\mathbb{P}$, then $W^\theta_t=W_t-\theta t$ is a brownian motion under $\mathbb{Q}^\theta$ $$\begin{split} \mathbb{...



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