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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 ...


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} & ...


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 [edit]. 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 ...


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\\ & ...


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 ...


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 ...


4

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 ...


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 $\ln(2) / 1.5 = 0.46209812$ years, or about 6 months. Let's assume that the current interest rate is 1% and the ...


3

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 ...


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 ...


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 ...


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 ...


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 ...


1

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 ...


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 variable $t$ 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 ...



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