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Roughly speaking, we can express the difference between a Markov process and a martingale as follows: A Markov process is one for which conditioning its future value on its history is the same as conditioning its future value on its present value, so that $E(h(X_t)\,|\,X_u,\,u\leq s)=E(h(X_t)\,|\,X_s)$, for any appropriate function $h$; A martingale is a ...

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Here, we assume that \begin{align*} g(t, x) = \mathbb{E}\left(h(X_T) \mid X_t = x \right). \end{align*} Note that, by Shiryaev, $g(t, x)$ is a Borel measurable function such that, for any Borel measurable set $A$, \begin{align*} \int_{\{X_t \in A\}} h(X_T) d\mathbb{P} &= \int_A g(t, x) \mathbb{P}_{X_t}(dx), \end{align*} where $\mathbb{P}_{X_t}(dx)$ is ...

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You are right, the rules to time-scale a T-years transition matrix $M_T$ are: $M_{k·T} = M_T^k$ $M_{T/k} = \sqrt[k]{M_T}$ The root of a matrix M can be obtained using the spectral decomposition: $M = P·D·P^{-1} \Longrightarrow M^k = P·D^k·P^{-1}$ where $P$ and $D$ are the eigenvectors and eigenvalue matrices of $M_T$. Note: The Perron-Frobenius tells ...

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Your equations are for cum-dividend prices, i.e. the price plus dividend today. The paper refers to ex-dividend prices. The correct two equations for investor group $a=1$ are \begin{align} p^1(0) =&\ \frac{3}{4} \left(\frac{1}{2}p^1(0) + \frac{1}{2}(1+p^1(1))\right) \\ p^1(1) =&\ \frac{3}{4} \left(\frac{2}{3}p^1(0) + \frac{1}{3}(1+p^1(1))\right) \end{...

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I don't think you can have an explicit form. Let $Y_t= e^{at}X_t$ then : $$Y_t -Y_0 =\sum_{i=1}^{N_t}e^{aT_i}$$ where $(T_i)_{i=1...N_t}$ are the jump times of your poisson process. then $$P(Y_t\leq x)=\sum_{n\geq 0}\frac{(mt)^n}{n!}e^{-mt}P(\sum_{i=1}^{N_t}e^{aT_i}\leq x|N_t=n)$$ $$P(\sum_{i=1}^{N_t}e^{aT_i}\leq x|N_t=n) =\int_{[0,+\infty]^n}\mathbf{1}... 4 The clearest and most intuitive article I have seen so far is Kritzman et al., Regime Shifts: Implications for Dynamic Strategies in FAJ (May / June 2012) It not only shows how you can use HMM for financial modelling but it also goes through the actual estimation algorithm (Baum-Welch) step-by-step and even gives full Matlab-code. From the abstract: ... 3 This is a corollary of Feynman-Kac theorem. For self-containedness, I re-produce the proof as follows. Assume that there exists a C^{1,2}-function F=F(t,x) defined on [0,T]\times\mathbb{R} that satisfies the PDE on the interior$$ F_{t}+\beta F_{x}+\frac{1}{2}\sigma^{2}F_{xx}=0, $$and the boundary condition: F(T,x)=g(x). Consider the process \left(... 3 Your decision. You can define the states to be "positive return" or "negative return". So if the return is negative, then its state would be that "negative return". Or, you could even model them with continuous emission so as to define a state with a specific mu and variance. So that given the return at time T, you can say "it could technically be possible ... 3 The initial condition for the backward Kolmogorov PDE is that$$ u(0,x) = g(x)  for all $x$ in the relevant domain and not just at a particular point. So if your functions $f$ and $g$ agree only at a single point the initial conditions are in fact different.

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You have been given good answers above. Basically, a stochastic process ${X_t}$ is a Markov process if $P(\{X_{t} \leq x\} | \mathcal{F}_{s}) = P(\{X_{t} \leq x\} | X_{s})$, for $s \leq t$. Here $\mathcal{F}_{s}$ is a $\sigma$-algebra, a special collection of subsets of the underlying sample space $\Omega$, containing all information about the process $\{X_t\... 2 This is the answer to the first version of the question which asked whether a stationary process has an increasing variance over time. No the definition of (weakly) stationary (http://en.wikipedia.org/wiki/Stationary_process) is that the variance is the same for each point in time. In the literature it is often dealt with the covariance function. For a ... 2 Just to give you two examples. Note that$dX_t =a \; dt + dW_t$is Markov but is not a martingale.$dX_t=(\int_0^t X_s ds) \; dW_t$is a martingale but is not Markov. 2 it's difficult to say that they are not popular. Some people definitely use them for live pricing. I'd say the real question is "why are they not popular in the academic literature"? One answer would simply be that most the questions that arise in their use are ones of fiddliness which do not make good papers. 2 I think I might have found the solution to my own question. The Markov property as stated above has no direct relation with the recombination of the approximating lattice. However, if we consider the "traditional" meaning of Markovness, that is being memoryless, things become clearer. Consider a binomial tree, where the random variable$Xcan ... 2 Two things to note: First you are assuming that stock returns follow some type of AR(1) which I do not think is a reasonable model; Casting consideration (1) aside, you can estimate what you want by doing: $$Y_t = \alpha + \alpha_{mon} + \alpha_{tue} + \alpha_{wed} + \alpha_{thu} + b Y_{t-1} + error$$ This will give you a first ... 1 Regime detection with hidden Markov model: http://scikit-learn.sourceforge.net/stable/modules/hmm.html 1 The weak form of the efficient market hypothesis (EMH) just says that the market is efficient to all prior information contained within price. By definition, the weak form of EMH obeys the Markov property such that the current state contains more information about the future state than all prior states combined. Thus, the weak form EHM is alone sufficient to ... 1 Based on the form of your equation, we can consider the SDE \begin{align*} dX_t = \sigma dW_t, \end{align*} whereW$is a standard Brownian motion. Since, for$0 \leq t \leq T\$, \begin{align*} X_T = X_t + \sigma (W_T-W_t), \end{align*} based on Feynman–Kac formula, the solution is given by \begin{align*} F(t, x) &= E\left(X_T^2 \mid X_t = x\right)\\ &...

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In context of Bermudan Options, I believe that since the model determines everything exogenously, calibrating to swaptions may give you cases where the implied forward rate is negatively correlated to swap rates. Note this will never happen in an endogenous model where the short rate equation constrains this possibility. This will obviously distort ...

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