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2

The Feynman-Kac theorem can be used in both directions. That is, If we know that $r_t$ follows the Ito process as described by the following stochastic differential equation \begin{align} d{{r}_{t}}=\mu ({{r}_{t}},t)dt+\sigma ...


3

C.I.R Process belongs the class of affine diffusion processes.For processes within this class, a closed form solution of the characteristic function exists(Duffie,et al). For more details, Suppose we have given a scalar SDEs, i.e., $$dX_t=\mu(X_t,t)dt+\sigma(X_t,t)dW_t$$ this process ($\{X_t\}_{0\leq t\leq T}$) is said to be of the affine form if ...


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I would also say that the pricing of some exotic products require to compute expectations of functions of the random variable at consideration, and these functions may grow more than linearly : you need finite moments in order for the prices of these exotic derivatives to be bounded.


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EDIT: My reasoning below seems to be wrong. The process as you write it tends to infinity if $a$ is big enough and positive and if $\lambda_0$ is positive. I would not call this process non-meanreverting OU. It is just an Ito process of a simple form. If we remove the stochastic part then we get $$ d\lambda_t = a \lambda_t dt $$ with the solution (if ...


0

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 ...


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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 ...


1

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 ...


0

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 ...


0

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 ...


2

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 ...


2

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 ...


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 ...


4

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 ...



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