5

The Snell envelope is the smallest super-martingale that is greater than $X$. Since $\tau \le N$, it is obvious that $A_N^{\tau} = A_{N\wedge \tau} = A_{\tau}$. For part (b), note that, from the Doob decomposition, $M$ is a martingale, $A$ is increasing, $M_0=Z_0$, and $A_0=0$. If $Z^{\tau}= \{Z_n^{\tau}\}_{n=1}^N$ is also a martingale, then \begin{align*}...


5

The exercise boundary $B_t$ for a finite maturity American put option is not a constant function of time as in your plot. As mentioned in the excerpt, $B_T = K$ at maturity. But for $t < T$, we have $B_t < K$ as you would never pre-maturely exercise to receive a zero payoff. Below is a plot of the early exercise boundary that I once produced for a ...


3

Assuming $\theta>0$ (take $\tilde{X}=\mu-X$ if it is not the case) Let us denote $\text{erfi}(x)$ the imaginary error function Let us denote $\tau_L$,resp.$\tau_U$ the hitting time of $L$resp.$U$ where $L<U$ 1) Using Ito's lemma, prove that : $$Y_t = \text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(X_t-\mu\right)\right) \text{ is a martingale}$$ ...


3

What do you mean by annotation date, there is a declaration(announcement) date, ex-date, record date but I've never heard of an annotation date. Dividends are not decided always at the fiscal year end, in some countries they are approved by the shareholders general meeting which can happen at any time during the year, some companies pay quarterly, others ...


1

As is often the case, there are at least two solution strategies here. (Probabilistic) You explicitly solve for the expected discount factor at the first passage time $\nu$ of $S$ to the level $B$ under the risk-neutral probability measure $\mathbb{P}^*$, i.e. \begin{equation} V_0 = \mathbb{E}_{\mathbb{P}^*} \left[ e^{-r \nu} \mathrm{1} \left\{ \nu \leq T \...


1

I think the proof has already been provided at the end of the proof in Shreve's Theorem 4.4.5. Specifically, note that, since \begin{align*} \frac{1}{(1+r)^{n \wedge \tau^*}}V_{n \wedge \tau^*}. \end{align*} is a martingale, \begin{align*} \tilde{\mathbb{E}}\left(\frac{1}{(1+r)^{N \wedge \tau^*}}V_{N \wedge \tau^*}\right) &= V_0 = \max_{\tau \in S_0} \...


1

Idea Let $B$ be a standard brownian motion starting from $x_0=0$, $m_T = \inf_{u\leq T}B_u$ and $M_T =\sup_{u\leq T}B_u$. Let's define if it exists for $A\in\sigma(B_u,u\leq T)$, $\mathbb{P}(A | B_T=x_T)\stackrel{\rm def}{=}\lim_{\varepsilon\to 0}\mathbb{P}(A|B_T\in(x_T-\varepsilon,x_T+\varepsilon))$ $$\begin{split} \mathbb{P}(\tau_U\leq T \cap \tau_U\leq ...


1

In that case, the problem becomes a non-trivial stopping time problem. Consider a filtered probability space $(\Omega, \mathcal{F}, \mathbb{P})$ equipped with the natural filtration of a standard Brownian motion $W_t^\mathbb{P}$. Assuming a geometric Brownian motion for the underlying asset, one gets $$ S_t = S_0 \exp\left((\mu-\frac{1}{2}\sigma^2)t + \...


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