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11

Although Math SE might be a bit more suited for this one, I wanted to give it a try. The answer relies on the law of total expectation, the law of total variance, and the relationship between Euler's number $e$ and series involving the factorial function $n!$. In the first half of the answer, I will present the derivation of $E(n)$ and $Var(n)$, i.e. mean ...

6

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 ...

5

I think you are absolutely correct if the hazard rate is deterministic, although I think you are forgetting a discounting factor in your example. But sometimes the hazard rate cannot be assumed to be deterministic (e.g. when pricing CVA and DVA). Here the hazard rate is instead assumed to follow a stochastic process itself, such that $\mathbf{1}_{\{\tau>t\... 5 As shown in Credit Risk Modeling Notes (Bielecki, Jeanblanc, Rutkowski), Corollary 1.3.1, for$t < s, we have: $$P(\tau \leq s | {\cal F}_t) = N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right ) + {\rm e}^{-2\nu \sigma^{-2}Y_t} N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right ),$$ where Y_t = y_0+ \nu t +\sigma W_t, \: \sigma >0,... 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*}... 4 Your {\cal F} is actually {\cal G}, that is the already enlarged filtration/probability space. So, the claim here seems to be that we do not have to consider the smaller, market filtration, {\cal F}. But, before we invoke Hypothesis (H), only this is true: E\left[1_{\tau>T}|{\cal G}_t\right] = 1_{\tau>t} E\left[e^{\Gamma_t -\Gamma_T}|{\cal F}... 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 ... 2 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$\mathbb{P}^*, i.e. \begin{equation} V_0 = \mathbb{E}_{\mathbb{P}^*} \left[ e^{-r \nu} \mathrm{1} \left\{ \nu \leq T \... 1 There is a method called stochastic mesh that has been proposed in the literature but it is not much used in practice https://www0.gsb.columbia.edu/faculty/pglasserman/Other/bgh.pdf There are numerical methods to make it faster (fast gauss transform for instance), but in the end not a lot of advantage compared to using good old LS in my experience. Cheers 1 I think the proof has already been provided at the end of the proof in Shreve's Theorem 4.4.5. Specifically, note that, since \begin{align*} \frac{1}{(1+r)^{n \wedge \tau^*}}V_{n \wedge \tau^*}. \end{align*} is a martingale, \begin{align*} \tilde{\mathbb{E}}\left(\frac{1}{(1+r)^{N \wedge \tau^*}}V_{N \wedge \tau^*}\right) &= V_0 = \max_{\tau \in S_0} \... 1 Idea LetB$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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