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Is my reasoning correct? assume $\mathcal{F}$ be an infinite σ-algebra on $\Omega$ .Prove that $\mathcal{F}$ is not countable. Let $\mathcal{F}=\{A_{i}\}_{i=1}^{\infty}$. For each $w\in \Omega$ define $B_{\omega}:=\bigcap_{\omega\in A_{i}}A_{i}$. Note that $B_{\omega}\in \mathcal{F}$ since this is a countable intersection. lemma: If $B_{\omega_1}\cap ... 1 Here is a sketch for an argument for 'yes': Let$\Omega$have$n$elements. For each extra element we add, the smallest$\sigma$-algebra containing$\Omega$,$\sigma(\Omega)$, will add some finite number of elements. Now consider the rational numbers. Between each pair of consecutive natural numbers there are infinitely many rational numbers, so each of the ... 2 I would argue that there is some path-dependency involved. The BS model is considered the big breakthrough and it presented the world with some kind of tractable toy model. After that people saw that you had to adjust the model to account for all kinds of stylized facts (e.g. non-constant volatility for different strikes, over time and so on). Yet finite ... 4 Mean reversion speed$\kappa$is better interpreted with the concept of half-life, which can be calculated from$\text{HL} = \ln(2) / \kappa$. For example, if the mean reversion coefficient is$\kappa = 1.5$, then the half-life of the process is$\ln(2) / 1.5 = 0.46209812years, or about 6 months. Let's assume that the current interest rate is 1% and the ... 0 On a pure technical aspect, a model does not need to have a finite variance. In the context of option pricing, what you need it a way to replicate the behaviour of the stock price. Once you have it you need to find a corresponding risk-neutral measure. There you will have the first difficulty, with infinite variance, the corresponding hedging strategy is ... 3 In the Mean-Reverting Models like C.I.R \begin{align} &dr_t=\kappa(\theta-r_t)dt+\sigma\sqrt {r_t} dW_t \end{align} speed of mean reversion (\kappa$) is not negative.If the condition$2\kappa\theta> \sigma^2$holds, then the drift is sufficiently large for the process to be guaranteed positive and not reach zero. This condition is known as the Feller ... 2 If$\tau$is finite then from the strong Markov property both the paths$X_t = \{W_{t+\tau} −W_\tau ∶ t\geq 0\}$and$−X_t = \{−(W_{t+\tau} − W_\tau) ∶ t \geq 0\}$are standard Wiener processes and independent of$Y_t = \{W_t ∶ 0 \leq t \leq \tau\}$, and hence both$(X_t, Y_t)$and$(X_t ,−Y_t)$have the same distribution. Given the two processes defined on ... 4 Write$X_t = A_t B_t$with$A_t = e^{(\lambda - \eta)t}$and$B_t = \left(\frac{\eta}{\lambda} \right)^{N_t}$. Then$dX_t = A_t dB_t + B_t dA_t$by the product rule of calculus. There are no second order terms since both$A_t$and$B_t$are finite variation (i.e.$\langle A_t, B_t\rangle$= 0). Next,$dA_t = (\lambda - \eta)A_t dt$, and$dB_t = B_t \cdot ...

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First note that paths are a.s continuous. Then by strong Markov property and reflection principle, $(W_\tau - W_t)$ is a Brownian motion independant of the before tau part. Then you can verify that increments are independent and gaussian by decomposing them in before and after tau part. Or you can décompose the quadratic variation and use Lévy 's ...

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By Ito's lemma, \begin{align} dX_t=\frac{\partial X_t}{\partial t}dt+\frac{\partial X_t}{\partial N(t)}dN_t+\frac{1}{2!}\frac{\partial^2 X_t}{\partial N^2_t}(dN_t)^2+\frac{\partial^2 X_t}{\partial N_t\partial t}{}dN_tdt+\frac{1}{3!}\frac{\partial^3 X_t}{\partial N^3_t}(dN_t)^3+... \end{align} Since $dN_t\,dt = 0, (dN_t)^2 = (dN_t)^3 = . . . = dN_t$, we have ...

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There are three main issues. As per my comment, one is the lack of specification for the distribution of the jumps (I'll assume that there is a $J_0 = 0$ at time 0 (otherwise, the process doesn't account for no jumps). Unless $P (J \leq -1) = 0$, your price process is problematic, and the Girsanov theorem is not applicable. To see why: S_t = S_0 e^{\sigma ... 8 Milstein Scheme This scheme is described in Glasserman (2003) and in Kloeden and Platen (1992) for general processes.Hence, for simplicity, we can assume that the Stochastic Process is driven by the SDE \begin{align} &dX_t=\Xi(t,X_t)dt+\Sigma(t,X_t)dW_t\\ \end{align} Milstein discretization is, \begin{align} dX_{t+\Delta ... 2 I would say the following: the tripple(\Omega,\mathcal{F},P)is an abstract probability space with all the properties that I assume that you know. then we can define random variables as mappings from this probability space to the real numbers $$X: \omega \mapsto X(\omega) \in \mathbb{R}.$$ But we want to study processes (X_t)_{t \ge 0}: \omega ... 0 You have to make further assumptions on the distribution of J_is. For example, if J_is are iid normal, your option pricing problem becomes that of Merton (1976) and the solution to it is an infinite sum. If J_is are assumed to be double exponential, you end up with Kou (2004) model and it has an analytical solution. Furthermore, there are three ... 4 You Know that dB_t=r_tB(t)dt . Ito's formula give us \begin{align} dZ(t)=\frac{1}{B(t)}d\,\Pi(t)-\frac{\Pi(t)}{B\,^2(t)}dB(t)+0 \end{align} As your teacher mentioned, d\Pi(t)=r(t)\Pi(t)dt+\sigma(\Pi(t),t)dW(t),Thus we have \begin{align} & dZ(t)=\frac{1}{B(t)}[r(t)\Pi(t)dt+\sigma(\Pi(t),t)dW(t)]-\frac{\Pi(t)}{B\,^2(t)}r(t)B(t)dt\\ & ... 6 You know that E\left[\int_{0}^{s}W_udu\right]=E\left[\int_{0}^{t}W_vdv\right]=0. By definition \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]-0 \end{align} then \begin{align} & ... 0 Let S_t and B_t be respectively the stock price and the money market account value at time t. Then S_t/B_t is called the discounted stock price. Note that \begin{align*} E\left(\frac{S_N}{S_0}\right) &= E\left(\frac{S_N}{B_N} \frac{B_N}{B_0}\right)\frac{B_0}{S_0}\\ &= E\left(\frac{S_N}{B_N}\right) E\left(\frac{B_N}{B_0}\right)\frac{B_0}{S_0} ... 4 In this case it is just the notion that your payoff function should not explode at some point - made mathematically rigorous. Have a look at the following picture from wikipedia: Intuitively the Lipschitz condition (or Lipschitz continuity) ensures that your payoff function always remains entirely outside the white cone, so it cannot e.g. become ... 2 For the general solution in the case where f is not a constant, note that, from the SDE \begin{align*} dx_t = \theta(f(t)-x_t)dt + \sigma dW_t, \end{align*} we obtain that \begin{align*} d\big(e^{\theta t} x_t \big) = \theta e^{\theta t} f(t)dt + \sigma e^{\theta t} dW_t. \end{align*} Then \begin{align*} e^{\theta t} x_t = x_0 + \int_0^t \theta e^{\theta ... 2 You can just take expectations on both sides of your SDE/corresponding integral equation and obtain an ODE on the expectation function m_t = \Bbb E[x_t]: \dot m = \theta(f - m) $$which you can easily solve using ansatz m_t = c_t \mathrm e^{-\theta t} which brings you to$$ m_t = x_0\mathrm e^{-\theta t} + \theta\cdot\int_0^tf(s)\mathrm ... 0 You can start with a deterministic basis spread. There are several attempts to model the basis spread both modeling the spreads separately with positive stochastic processes and by modelling the different indexes. You are right: if you model the indices they could cross and it is hard to enforce abscence of twists. Probably every paper on this subject ... 1 Yes, this is trivially true once you know that every continuous local martingale is a time-changed brownian motion. Therefore, if you change your time variablet$in$dX=a\,dt+b\,dW(t)$to the right$t^\prime$you can get a standard tree representation. Now, the correct time change may be difficult or impossible to figure out, so this theorem is of ... 4 For a martingale$dX=a(X,t)\,dt+b(X,t) dW(t)$where$a$and$b\$ are not constant, your tree will not recombine in general . This is the main issue. See for instance: Florescu, I. and F. G. Viens (2008, March). Stochastic volatility: Option pricing using a multinomial recombining tree. Applied Mathematical Finance 15 (2), 151-181. It deals with the case ...

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